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Chapters 1, 2 Making Decisions Using Models

Why study the Science of Making Decisions? § Intuition or Reason-based decisions? – Complex situations – Changing environments – Wide range of options

Making Decisions Using Models. Introduction § Numerous complex problems – Make good decisions in life & business. – Increase opportunities § Complexity => Difficulty in making good decisions § Analyze and evaluate Alternatives – Choose best action. How? § Use computers to build models and analyze alternatives – Computer model: set of mathematical relationships and assumptions – Represent real world decision problems. – Spreadsheets: most convenient business tool to analyze and implement models.

Modeling in Decision Science § Scientific approach to assist executives § For better decisions and results § Applied via spreadsheets

Data, Information, and Decisions DATA Prediction Models Help you extract information from data INFORMATION Decision Models Help you exploit information to make better decisions ACTION

Learning Objectives § Introduction to business modeling § Familiarity with analytical software tools § Critical thinking skills § Sensitivity analysis of changing business scenarios

Maximizing Learning Value Course philosophy: Learning by Doing

Making Decisions Using Models. Introduction § In this course: variety of techniques in field of Management Science – Applied via spreadsheets – Assist in Decision Making § What is Management Science? – A field of study that uses computers, statistics, and mathematics to solve business problems. – Also known as: ØOperations research ØDecision science

Who Uses Management Science? § NOT targeted for who know advanced mathematics – Required is LOGIC & a business problem, not algebra … § § For BUSINESSES Spreadsheets are easy to use Powerful and easy tools for managers We use MS every day!

A Framework for Decision Making IDENTIFY PROBLEM DEVELOP MODEL USE TOOLS MAKE DECISIONS ANALYZE RESULTS

Management Science Applied in many Real Life Situations § Motorola – Procurement of goods and services account for 50% of its costs – Developed an Internet-based system for negotiations with suppliers – The system optimized multiproduct, multi-vendor contract awards – Benefits: $600 million in savings § Zara Distribution co. – Over 100, 000 vehicles – Developed vehicle routing optimization system – Benefits: Eliminated thousands of routes Annual savings millions of dollars Saves Millions of Dollars !!

Management Science Applied in many Real Life Situations § Hong Kong International Port – Busiest container terminal in the world – Thousands of trucks move containers in & out of storage yard – Used DSS to optimize operational decisions involving trucks, cranes & storage locations – Benefits: Ø 35% reduction in container handling costs Ø 50% increase in throughput Ø 30% improvement in vessel turnaround time § Starbucks Operations Management – product-stocking locations – Demand is highly seasonal and erratic – Developed inventory system to optimize stocking levels Benefits: Ø Billions of dollars in reduced inventory Ø Improved customerservice levels

More Applications … Linear Programming was used by Burger King to find how to best blend cuts of meat to minimize costs. Integer Linear Programming model was used by American Air Lines to determine an optimal flight schedule. The Shortest Route Algorithm was implemented by the Sony Corporation to develop an onboard car navigation system. Queuing models are incorporated into the overall design plans for Disneyland Disney World, which lead to development of ‘waiting line entertainment’ to improve customer satisfaction.

The Modeling Approach to Decision Making § Everyone uses models to make decisions. § Types of models to Analyze & Decide: – Mental – arranging furniture: different arrangements – Visual: for more complex decision problems – Blueprints for a house/building: various structure – Road maps: various routes – Physical/Scale Models – Auto engineers to study aerodynamics of various car/plane/building designs – Mathematical – what we’ll be studying – Mathematical relationships to represent a problem

Characteristics of Models § Models are usually simplified versions of the things they represent – Ex: no need to build entire car with small details… A Model will do! § A valid model accurately represents the relevant characteristics of the object or problem being studied

Benefits of Modeling § Economy - It is often less costly to analyze decision problems using models and correct mistakes if any. – Ex: correcting a mistakes in aircraft wing scale model versus testing a real aircraft will result in a crash. § Timeliness - Models often deliver needed information more quickly than their real-world counterparts. – Ex: Scale models of cars and airplanes quicker to develop § Feasibility - Models can be used to do things that would be impossible. – Crash tests on human models § Models give us insight & understanding about a decision problem which improves decision making (As we will see !)

Example of a Simple Mathematical Model Profit = Revenue - Expenses or Profit = f(Revenue, Expenses) or Y = f(X 1, X 2) What is X, Y, f(. )? Interpretation? What does this expression mean?

A Generic Mathematical Model Where: Y = f(X 1, X 2, …, Xn) Y = dependent variable (aka bottom-line performance measure) Xi = independent variables (inputs having an impact on Y) f(. ) = function (might be very complex) defining the relationship between the Xi & Y Many business problems can be represented by such mathematical models.

Mathematical Models & Spreadsheets § Most spreadsheet models are very similar to our generic mathematical model: Y = f(X 1, X 2, …, Xn) § Most spreadsheets have input cells (representing Xi) to which mathematical functions ( f(. )) are applied to compute a bottom-line performance measure (or Y) in an output cell. § Ex: Car loan monthly payment = f (price, down payment, term load, interest rate, number of periods)

How Modeling fits in. The Problem Solving Process Our Focus in this course Identify Problem Formulate & Implement Model Analyze Model Test Results Implement Solution unsatisfactory results All steps are important. Starting with the wrong problem implies getting the wrong answer!

Chapter 2 Overview of Optimization

Introduction § We all face decision about how to use limited resources such as: – Oil – Land – Time – Money – Workers October 3 DCSN 200 section 4 22

Mathematical Programming. . . § MP is a field of management science that finds the optimal, or most efficient, way of using limited resources to achieve the objectives of an individual of a business. § a. k. a. Optimization

Applications of Optimization § Determining Product Mix – Variety of products requiring different amounts of raw material and labor and generating different profits § Manufacturing § Routing and Logistics – Cost of shipping and delivering changes with amount needed at the retailer § Financial Planning

Characteristics of Optimization Problems § Decisions § Constraints § Objectives

General Form of an Optimization Problem MAX (or MIN): f 0(X 1, X 2, …, Xn) Subject to: f 1(X 1, X 2, …, Xn)<=b 1 : fk(X 1, X 2, …, Xn)>=bk : fm(X 1, X 2, …, Xn)=bm Note: If all the functions in an optimization are linear, the problem is a Linear Programming (LP) problem

Linear Programming (LP) Problems MAX (or MIN): c 1 X 1 + c 2 X 2 + … + cn. Xn Subject to: a 11 X 1 + a 12 X 2 + … + a 1 n. Xn <= b 1 : ak 1 X 1 + ak 2 X 2 + … + akn. Xn >=bk : am 1 X 1 + am 2 X 2 + … + amn. Xn = bm c_i: marginal profit/cost a_ij: coefficient of ith constraint for jth variable

An Example LP Problem Blue Ridge Hot Tubs produces two types of hot tubs: Aqua-Spas & Hydro-Luxes. Pumps Labor Tubing Unit Profit Aqua-Spa 1 9 hours 12 feet $350 Hydro-Lux 1 6 hours 16 feet $300 There are 200 pumps, 1566 hours of labor, and 2880 feet of tubing available.

5 Steps In Formulating LP Models: 1. Understand the problem. 2. Identify the decision variables. X 1=number of Aqua-Spas to produce X 2=number of Hydro-Luxes to produce 3. State the objective function as a linear combination of the decision variables. MAX: 350 X 1 + 300 X 2

5 Steps In Formulating LP Models (continued) 4. State the constraints as linear combinations of the decision variables. 1 X 1 + 1 X 2 <= 200} pumps 9 X 1 + 6 X 2 <= 1566 } labor 12 X 1 + 16 X 2 <= 2880 } tubing 5. Identify any upper or lower bounds on the decision variables. X 1 >= 0 X 2 >= 0

LP Model for Blue Ridge Hot Tubs MAX: 350 X 1 + 300 X 2 S. T. : 1 X 1 + 1 X 2 <= 200 9 X 1 + 6 X 2 <= 1566 12 X 1 + 16 X 2 <= 2880 X 1 >= 0 X 2 >= 0

Solving LP Problems: An Intuitive Approach § Idea: Each Aqua-Spa (X 1) generates the highest unit profit ($350), so let’s make as many of them as possible! § How many would that be? – Let X 2 = 0 Ø 1 st constraint: 1 X 1 <= 200 Ø 2 nd constraint: 9 X 1 <=1566 or X 1 <=174 Ø 3 rd constraint: 12 X 1 <= 2880 or X 1 <= 240 § If X 2=0, the maximum value of X 1 is 174 and the total profit is $350*174 + $300*0 = $60, 900 § This solution is feasible, but is it optimal? § No!

Solving LP Problems: A Graphical Approach § The constraints of an LP problem defines its feasible region. § The best point in the feasible region is the optimal solution to the problem. § For LP problems with 2 variables, it is easy to plot the feasible region and find the optimal solution.

Plotting the First Constraint X 2 250 (0, 200) 200 boundary line of pump constraint X 1 + X 2 = 200 150 100 50 (200, 0) 0 0 50 100 150 200 250 X 1

Plotting the Second Constraint X 2 (0, 261) 250 boundary line of labor constraint 9 X 1 + 6 X 2 = 1566 200 150 100 50 (174, 0) 0 0 50 100 150 200 250 X 1

Plotting the Third Constraint X 2 250 (0, 180) 200 150 boundary line of tubing constraint 12 X 1 + 16 X 2 = 2880 100 Feasible Region 50 (240, 0) 0 0 50 100 150 200 250 X 1

Plotting A Level Curve of the Objective Function X 2 250 200 (0, 116. 67) 150 objective function 350 X 1 + 300 X 2 = 35000 100 (100, 0) 50 0 0 50 100 150 200 250 X 1

A Second Level Curve of the Objective Function X 2 250 (0, 175) 200 objective function 350 X 1 + 300 X 2 = 35000 objective function 350 X 1 + 300 X 2 = 52500 150 100 (150, 0) 50 0 0 50 100 150 200 250 X 1

Using A Level Curve to Locate the Optimal Solution X 2 250 objective function 350 X 1 + 300 X 2 = 35000 200 150 optimal solution 100 objective function 350 X 1 + 300 X 2 = 52500 50 0 0 50 100 150 200 250 X 1

Calculating the Optimal Solution § The optimal solution occurs where the “pumps” and “labor” constraints intersect. § This occurs where: X 1 + X 2 = 200 (1) and 9 X 1 + 6 X 2 = 1566 (2) § From (1) we have, X 2 = 200 -X 1 (3) § Substituting (3) for X 2 in (2) we have, 9 X 1 + 6 (200 -X 1) = 1566 which reduces to X 1 = 122 § So the optimal solution is, X 1=122, X 2=200 -X 1=78 Total Profit = $350*122 + $300*78 = $66, 100

Enumerating The Corner (Extreme) Points X 2 250 obj. value = $54, 000 (0, 180) 200 Note: This technique will not work if the solution is unbounded. obj. value = $64, 000 150 (80, 120) obj. value = $66, 100 (122, 78) 100 50 obj. value = $60, 900 (174, 0) obj. value = $0 (0, 0) 0 0 50 100 150 200 250 X 1

Summary of Graphical Solution to LP Problems 1. Plot the boundary line of each constraint 2. Identify the feasible region 3. Locate the optimal solution by either: a. Plotting level curves b. Enumerating the extreme points

Special Conditions in LP Models § A number of anomalies can occur in LP problems: – Alternate Optimal Solutions – Redundant Constraints – Unbounded Solutions – Infeasibility

Alternate Optimal Solutions MAX: 450 X 1 + 300 X 2 S. T. : 1 X 1 + 1 X 2 <= 200 9 X 1 + 6 X 2 <= 1566 12 X 1 + 16 X 2 <= 2880 X 1 >= 0 X 2 >= 0

Example of Alternate Optimal Solutions X 2 250 objective function level curve 450 X 1 + 300 X 2 = 78300 200 150 100 alternate optimal solutions 50 0 0 50 100 150 200 250 X 1

Redundant Constraint MAX: 350 X 1 + 300 X 2 S. T. : 1 X 1 + 1 X 2 <= 225 9 X 1 + 6 X 2 <= 1566 12 X 1 + 16 X 2 <= 2880 X 1 >= 0 X 2 >= 0

Example of a Redundant Constraint X 2 250 boundary line of tubing constraint 200 boundary line of pump constraint 150 boundary line of labor constraint 100 Feasible Region 50 0 0 50 100 150 200 250 X 1

Unbounded Solution MAX: X 1 + X 2 S. T. : X 1 + X 2 >= 400 -X 1 + 2 X 2 <= 400 X 1 >= 0 X 2 >= 0

Example of an Unbounded Solution X 2 1000 objective function X 1 + X 2 = 600 800 -X 1 + 2 X 2 = 400 objective function X 1 + X 2 = 800 600 400 200 X 1 + X 2 = 400 0 0 200 400 600 800 1000 X 1

Infeasibility MAX: X 1 + X 2 S. T. : X 1 + X 2 <= 150 X 1 + X 2 >= 200 X 1 >= 0 X 2 >= 0

Example of Infeasibility X 2 250 200 X 1 + X 2 = 200 feasible region for second constraint 150 100 feasible region for first constraint 50 X 1 + X 2 = 150 0 0 50 100 150 200 250 X 1

End of Chapter 2