1. 1-1 Introduction to Optimization and Linear Programming Chapter 1

2. 1-2 Introduction We all face decision about how to use limited resources such as: –Oil in the earth –Land for waste dumps –Time –Money –Workers

3. 1-3 Mathematical Programming... MP is a field of operations research that finds the optimal, or most efficient, way of using limited resources to achieve the objectives of an individual or a business. a.k.a. Optimization

4. 1-4 Applications of Optimization Determining Product Mix Manufacturing Routing and Logistics Financial Planning

5. 1-5 Characteristics of Optimization Problems Decisions Constraints Objectives

6. 1-6 General Form of an Optimization Problem MAX (or 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

7. 1-7 Linear Programming (LP) Problems 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

8. 1-8 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

9. 1-9 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

10. 1-10 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

11. 1-11 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

12. 1-12 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 1st constraint:1X 1 <= 200 2nd constraint:9X 1 <=1566 or X 1 <=174 3rd constraint:12X 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!

13. 1-13 Solving LP Problems: A Graphical Approach The constraints of an LP problem define the 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.

14. 1-14 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

15. 1-15 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

16. 1-16 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

17. 1-17 X2X2 Plotting A Level Curve of the Objective Function X1X1 250 200 150 100 50 0 0 100 150 200250 (0, 116.67) (100, 0) objective function 350X 1 + 300X 2 = 35000

18. 1-18 A Second Level Curve of the Objective Function X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 (0, 175) (150, 0) objective function 350X 1 + 300X 2 = 35000 objective function 350X 1 + 300X 2 = 52500

19. 1-19 Using A Level Curve to Locate the Optimal Solution X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 objective function 350X 1 + 300X 2 = 35000 objective function 350X 1 + 300X 2 = 52500 optimal solution

20. 1-20 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 9X 1 + 6X 2 = 1566(2) From (1) we have, X 2 = 200 -X 1 (3) Substituting (3) for X 2 in (2) we have, 9X 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

21. 1-21 Enumerating The Corner Points X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 (0, 180) (174, 0) (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.

22. 1-22 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

23. 1-23 Special Conditions in LP Models A number of anomalies can occur in LP problems: –Alternate Optimal Solutions –Redundant Constraints –Unbounded Solutions –Infeasibility

24. 1-24 Example of Alternate Optimal Solutions X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 450X 1 + 300X 2 = 78300 objective function level curve alternate optimal solutions

25. 1-25 Example of a Redundant Constraint X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 boundary line of tubing constraint Feasible Region boundary line of pump constraint boundary line of labor constraint

26. 1-26 Example of an Unbounded Solution X2X2 X1X1 1000 800 600 400 200 0 0 400 600 8001000 X 1 + X 2 = 400 X 1 + X 2 = 600 objective function X 1 + X 2 = 800 objective function -X 1 + 2X 2 = 400

27. 1-27 Example of Infeasibility X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 X 1 + X 2 = 200 X 1 + X 2 = 150 feasible region for second constraint feasible region for first constraint

28. 1-28 Important ”Behind the Scenes” Assumptions in LP Models

29. 1-29 Proportionality and Additivity Assumptions An LP objective function is linear; this results in the following 2 implications:  proportionality: contribution to the objective function from each decision variable is proportional to the value of the decision variable. e.g., contribution to profit from making 4 aqua-spas (4@$350) is 4 times the contribution from making 1 aqua-spa ($350)

30. 1-30 Proportionality and Additivity Assumptions (cont.)  Additivity: contribution to objective function from any decision variable is independent of the values of the other decision variables. E.g., no matter what the value of x 2, the manufacture of x 1 aqua-spas will always contribute 350 x 1 dollars to the objective function.

31. 1-31 Proportionality and Additivity Assumptions (cont.) Analogously, since each constraint is a linear inequality or linear equation, the following implications result:  proportionality: contribution of each decision variable to the left-hand side of each constraint is proportional to the value of the variable. E.g., it takes 3 times as many labor hours (9@3=27 hours) to make 3 aqua-spas as it takes to make 1 aqua-spa (9@1=9 hours) [No economy of scale]

32. 1-32 Proportionality and Additivity Assumptions (cont.)  Additivity: the contribution of a decision variable to the left-hand side of a constraint is independent of the values of the other decision variables. E.g., no matter what the value of x 1 (no. of aqua- spas produced), the production of x 2 hydro-luxes uses x 2 pumps, 6x 2 hours of labor, 16x 2 feet of tubing.

33. 1-33 More Assumptions Divisibility Assumption: each decision variable is allowed to assume fractional values Certainty Assumption: each parameter (objective function coefficient cj, right-hand side constant bi of each constraint, and technology coefficient aij) is known with certainty.

34. 1-34 End of Chapter 1