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1 Spreadsheet Modeling & Decision Analysis: A Practical Introduction to Management Science 3d edition by Cliff Ragsdale.

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Presentation on theme: "1 Spreadsheet Modeling & Decision Analysis: A Practical Introduction to Management Science 3d edition by Cliff Ragsdale."— Presentation transcript:

1 1 Spreadsheet Modeling & Decision Analysis: A Practical Introduction to Management Science 3d edition by Cliff Ragsdale

2 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-2 Introduction to Optimization and Linear Programming Chapter 2

3 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-3 Introduction u We all face decision about how to use limited resources such as: –Oil in the earth –Land for dumps –Time –Money –Workers

4 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-4 Mathematical Programming... u 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. u a.k.a. Optimization

5 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-5 Applications of Mathematical Optimization u Determining Product Mix u Manufacturing u Routing and Logistics u Financial Planning

6 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-6 Characteristics of Optimization Problems u Decisions u Constraints u Objectives

7 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-7 General Form of an Optimization Problem MAX (of MIN): f 0 (X 1, X 2, …, X n ) Subject to: f 1 (X 1, X 2, …, X n )<=b 1 : f k (X 1, X 2, …, X n )>=b k : f m (X 1, X 2, …, X n )=b m Note: If all the functions in an optimization are linear, the problem is a Linear Programming (LP) problem

8 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-8 General Form of a Linear Programming (LP) Problem MAX (or MIN): c 1 X 1 + c 2 X 2 + … + c n X n Subject to:a 11 X 1 + a 12 X 2 + … + a 1 n X n <= b 1 : a k 1 X 1 + a k 2 X 2 + … + a kn X n >=b k : a m 1 X 1 + a m 2 X 2 + … + a mn X n = b m

9 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-9 An Example LP Problem Blue Ridge Hot Tubs produces two types of hot tubs: Aqua-Spas & Hydro-Luxes. There are 200 pumps, 1566 hours of labor, and 2880 feet of tubing available. Aqua-SpaHydro-Lux Pumps11 Labor 9 hours6 hours Tubing12 feet16 feet Unit Profit$350$300

10 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-10 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: 350X 1 + 300X 2

11 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-11 5 Steps In Formulating LP Models (continued) 4. State the constraints as linear combinations of the decision variables. 1X 1 + 1X 2 <= 200} pumps 9X 1 + 6X 2 <= 1566} labor 12X 1 + 16X 2 <= 2880} tubing 5. Identify any upper or lower bounds on the decision variables. X 1 >= 0 X 2 >= 0

12 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-12 Summary of the LP Model for Blue Ridge Hot Tubs MAX: 350X 1 + 300X 2 S.T.:1X 1 + 1X 2 <= 200 9X 1 + 6X 2 <= 1566 12X 1 + 16X 2 <= 2880 X 1 >= 0 X 2 >= 0

13 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-13 Solving LP Problems: An Intuitive Approach u Idea: Each Aqua-Spa (X 1 ) generates the highest unit profit ($350), so let’s make as many of them as possible! u How many would that be? –Let X 2 = 0 v 1st constraint:1X 1 <= 200 v 2nd constraint:9X 1 <=1566 or X 1 <=174 v 3rd constraint:12X 1 <= 2880 or X 1 <= 240 u If X 2 =0, the maximum value of X 1 is 174 and the total profit is $350*174 + $300*0 = $60,900 u This solution is feasible, but is it optimal? u No!

14 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-14 Solving LP Problems: A Graphical Approach u The constraints of an LP problem defines its feasible region. u The best point in the feasible region is the optimal solution to the problem. u For LP problems with 2 variables, it is easy to plot the feasible region and find the optimal solution.

15 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-15 X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 (0, 200) (200, 0) boundary line of pump constraint X 1 + X 2 = 200 Plotting the First Constraint

16 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-16 X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 (0, 261) (174, 0) boundary line of labor constraint 9X 1 + 6X 2 = 1566 Plotting the Second Constraint

17 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-17 X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 (0, 180) (240, 0) boundary line of tubing constraint 12X 1 + 16X 2 = 2880 Feasible Region Plotting the Third Constraint

18 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-18 X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 (0, 116.67) (100, 0) objective function 350X 1 + 300X 2 = 35000 Plotting A Level Curve of the Objective Function

19 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-19 X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 (0, 175) (150, 0) objective function 350X 1 + 300X 2 = 35000 A Second Level Curve of the Objective Function objective function 350X 1 + 300X 2 = 52500

20 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-20 X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 objective function 350X 1 + 300X 2 = 35000 Using A Level Curve to Locate the Optimal Solution objective function 350X 1 + 300X 2 = 52500 optimal solution

21 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-21 Calculating the Optimal Solution u The optimal solution occurs where the “pumps” and “labor” constraints intersect. u This occurs where: X 1 + X 2 = 200 (1) and 9X 1 + 6X 2 = 1566(2) u From (1) we have, X 2 = 200 -X 1 (3) u Substituting (3) for X 2 in (2) we have, 9X 1 + 6 (200 -X 1 ) = 1566 which reduces to X 1 = 122 u So the optimal solution is, X 1 =122, X 2 =200-X 1 =78 Total Profit = $350*122 + $300*78 = $66,100

22 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-22 X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 (0, 180) (174, 0) Enumerating The Corner Points (122, 78) (80, 120) (0, 0) obj. value = $54,000 obj. value = $64,000 obj. value = $66,100 obj. value = $60,900 obj. value = $0 Note: This technique will not work if the solution is unbounded.

23 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-23 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

24 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-24 Special Conditions in LP Models u A number of anomalies can occur in LP problems: –Alternate Optimal Solutions –Redundant Constraints –Unbounded Solutions –Infeasibility

25 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-25 X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 450X 1 + 300X 2 = 78300 Example of Alternate Optimal Solutions objective function level curve alternate optimal solutions

26 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-26 X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 boundary line of tubing constraint Feasible Region Example of a Redundant Constraint boundary line of pump constraint boundary line of labor constraint

27 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-27 X2X2 X1X1 1000 800 600 400 200 0 0 400 600 8001000 Example of an Unbounded Solution X 1 + X 2 = 400 X 1 + X 2 = 600 objective function X 1 + X 2 = 800 objective function -X 1 + 2X 2 = 400

28 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-28 X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 X 1 + X 2 = 200 Example of Infeasibility X 1 + X 2 = 150 feasible region for second constraint feasible region for first constraint

29 Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning. 2-29 End of Chapter 2


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