1. Chapter 2 Introduction to Linear Programming n Linear Programming Problem n Problem Formulation n A Maximization Problem n Graphical Solution Procedure n Extreme Points and the Optimal Solution n Computer Solutions n A Minimization Problem n Special Cases

2. Linear Programming (LP) Problem n The maximization or minimization of some quantity is the objective in all linear programming problems. n All LP problems have constraints that limit the degree to which the objective can be pursued. n A feasible solution satisfies all the problem's constraints. n An optimal solution is a feasible solution that results in the largest possible objective function value when maximizing (or smallest when minimizing). n A graphical solution method can be used to solve a linear program with two variables.

3. Linear Programming (LP) Problem n If both the objective function and the constraints are linear, the problem is referred to as a linear programming problem. n Linear functions are functions in which each variable appears in a separate term raised to the first power and is multiplied by a constant (which could be 0). n Linear constraints are linear functions that are restricted to be "less than or equal to", "equal to", or "greater than or equal to" a constant.

4. Problem Formulation n Problem formulation or modeling is the process of translating a verbal statement of a problem into a mathematical statement.

5. Guidelines for Model Formulation n Understand the problem thoroughly. n Describe the objective. n Describe each constraint. n Define the decision variables. n Write the objective in terms of the decision variables. n Write the constraints in terms of the decision variables.

6. Example 1: A Maximization Problem n LP Formulation Max 5 x 1 + 7 x 2 Max 5 x 1 + 7 x 2 s.t. x 1 < 6 s.t. x 1 < 6 2 x 1 + 3 x 2 < 19 2 x 1 + 3 x 2 < 19 x 1 + x 2 < 8 x 1 + x 2 < 8 x 1, x 2 > 0 x 1, x 2 > 0

7. Example 1: Graphical Solution n Constraint #1 Graphed x 2 x 2 x1x1x1x1 x 1 < 6 (6, 0) 87654321 1 2 3 4 5 6 7 8 9 10

8. Example 1: Graphical Solution n Constraint #2 Graphed 2 x 1 + 3 x 2 < 19 x 2 x 2 x1x1x1x1 (0, 6 1/3 ) (9 1/2, 0) 87654321 1 2 3 4 5 6 7 8 9 10

9. Example 1: Graphical Solution n Constraint #3 Graphed x 2 x 2 x1x1x1x1 x 1 + x 2 < 8 (0, 8) (8, 0) 87654321 1 2 3 4 5 6 7 8 9 10

10. Example 1: Graphical Solution n Combined-Constraint Graph 2 x 1 + 3 x 2 < 19 x 2 x 2 x1x1x1x1 x 1 + x 2 < 8 x 1 < 6 87654321 1 2 3 4 5 6 7 8 9 10

11. 87654321 Example 1: Graphical Solution n Feasible Solution Region x1x1x1x1 FeasibleRegion x 2 x 2

12. 87654321 1 2 3 4 5 6 7 8 9 10 Example 1: Graphical Solution n Objective Function Line x1x1x1x1 x 2 x 2 (7, 0) (0, 5) Objective Function 5 x 1 + 7x 2 = 35 Objective Function 5 x 1 + 7x 2 = 35

13. 87654321 1 2 3 4 5 6 7 8 9 10 Example 1: Graphical Solution n Optimal Solution x1x1x1x1 x 2 x 2 Objective Function 5 x 1 + 7x 2 = 46 Objective Function 5 x 1 + 7x 2 = 46 Optimal Solution ( x 1 = 5, x 2 = 3) Optimal Solution ( x 1 = 5, x 2 = 3)

14. Summary of the Graphical Solution Procedure for Maximization Problems n Prepare a graph of the feasible solutions for each of the constraints. n Determine the feasible region that satisfies all the constraints simultaneously.. n Draw an objective function line. n Move parallel objective function lines toward larger objective function values without entirely leaving the feasible region. n Any feasible solution on the objective function line with the largest value is an optimal solution.

15. Slack and Surplus Variables n A linear program in which all the variables are non- negative and all the constraints are equalities is said to be in standard form. n Standard form is attained by adding slack variables to "less than or equal to" constraints, and by subtracting surplus variables from "greater than or equal to" constraints. n Slack and surplus variables represent the difference between the left and right sides of the constraints. n Slack and surplus variables have objective function coefficients equal to 0.

16. Example 1: Standard Form Max 5 x 1 + 7 x 2 + 0 s 1 + 0 s 2 + 0 s 3 Max 5 x 1 + 7 x 2 + 0 s 1 + 0 s 2 + 0 s 3 s.t. x 1 + s 1 = 6 s.t. x 1 + s 1 = 6 2 x 1 + 3 x 2 + s 2 = 19 2 x 1 + 3 x 2 + s 2 = 19 x 1 + x 2 + s 3 = 8 x 1 + x 2 + s 3 = 8 x 1, x 2, s 1, s 2, s 3 > 0 x 1, x 2, s 1, s 2, s 3 > 0

17. Extreme Points and the Optimal Solution n The corners or vertices of the feasible region are referred to as the extreme points. n An optimal solution to an LP problem can be found at an extreme point of the feasible region. n When looking for the optimal solution, you do not have to evaluate all feasible solution points. n You have to consider only the extreme points of the feasible region.

18. 87654321 1 2 3 4 5 6 7 8 9 10 Example 1: Extreme Points x1x1x1x1 FeasibleRegion 1122 33 44 55 x 2 x 2

19. Computer Solutions n Computer programs designed to solve LP problems are now widely available. n Most large LP problems can be solved with just a few minutes of computer time. n Small LP problems usually require only a few seconds. n Linear programming solvers are now part of many spreadsheet packages, such as Microsoft Excel.

20. Interpretation of Computer Output n In this chapter we will discuss the following output: objective function value objective function value values of the decision variables values of the decision variables reduced costs reduced costs slack/surplus slack/surplus n In the next chapter we will discuss how an optimal solution is affected by a change in: a coefficient of the objective function a coefficient of the objective function the right-hand side value of a constraint the right-hand side value of a constraint

21. Example 1: Spreadsheet Solution n Partial Spreadsheet Showing Problem Data

22. Example 1: Spreadsheet Solution n Partial Spreadsheet Showing Solution

23. Example 1: Spreadsheet Solution n Interpretation of Computer Output We see from the previous slide that: Objective Function Value = 46 Objective Function Value = 46 Decision Variable #1 ( x 1 ) = 5 Decision Variable #1 ( x 1 ) = 5 Decision Variable #2 ( x 2 ) = 3 Decision Variable #2 ( x 2 ) = 3 Slack in Constraint #1 = 1 (= 6 - 5) Slack in Constraint #1 = 1 (= 6 - 5) Slack in Constraint #2 = 0 (= 19 - 19) Slack in Constraint #2 = 0 (= 19 - 19) Slack in Constraint #3 = 0 (= 8 - 8) Slack in Constraint #3 = 0 (= 8 - 8)

24. Reduced Cost n The reduced cost for a decision variable whose value is 0 in the optimal solution is the amount the variable's objective function coefficient would have to improve (increase for maximization problems, decrease for minimization problems) before this variable could assume a positive value. n The reduced cost for a decision variable with a positive value is 0.

25. Example 1: Spreadsheet Solution n Reduced Costs

26. Example 2: A Minimization Problem n LP Formulation Min 5 x 1 + 2 x 2 Min 5 x 1 + 2 x 2 s.t. 2 x 1 + 5 x 2 > 10 s.t. 2 x 1 + 5 x 2 > 10 4 x 1 - x 2 > 12 4 x 1 - x 2 > 12 x 1 + x 2 > 4 x 1 + x 2 > 4 x 1, x 2 > 0 x 1, x 2 > 0

27. Example 2: Graphical Solution n Graph the Constraints Constraint 1: When x 1 = 0, then x 2 = 2; when x 2 = 0, then x 1 = 5. Connect (5,0) and (0,2). The ">" side is above this line. Constraint 1: When x 1 = 0, then x 2 = 2; when x 2 = 0, then x 1 = 5. Connect (5,0) and (0,2). The ">" side is above this line. Constraint 2: When x 2 = 0, then x 1 = 3. But setting x 1 to 0 will yield x 2 = -12, which is not on the graph. Thus, to get a second point on this line, set x 1 to any number larger than 3 and solve for x 2 : when Constraint 2: When x 2 = 0, then x 1 = 3. But setting x 1 to 0 will yield x 2 = -12, which is not on the graph. Thus, to get a second point on this line, set x 1 to any number larger than 3 and solve for x 2 : when x 1 = 5, then x 2 = 8. Connect (3,0) and (5,8). The ">" side is to the right. Constraint 3: When x 1 = 0, then x 2 = 4; when x 2 = 0, then x 1 = 4. Connect (4,0) and (0,4). The ">" side is above this line. Constraint 3: When x 1 = 0, then x 2 = 4; when x 2 = 0, then x 1 = 4. Connect (4,0) and (0,4). The ">" side is above this line.

28. Example 2: Graphical Solution n Constraints Graphed 54321 5544332211 5544332211 54321 5544332211 5544332211 1 2 3 4 5 6 1 2 3 4 5 6 x2x2x2x2 x2x2x2x2 4 x 1 - x 2 > 12 x 1 + x 2 > 4 x 1 + x 2 > 4 4 x 1 - x 2 > 12 x 1 + x 2 > 4 x 1 + x 2 > 4 2 x 1 + 5 x 2 > 10 x1x1x1x1 x1x1x1x1 Feasible Region

29. Example 2: Graphical Solution n Graph the Objective Function Set the objective function equal to an arbitrary constant (say 20) and graph it. For 5 x 1 + 2 x 2 = 20, when x 1 = 0, then x 2 = 10; when x 2 = 0, then x 1 = 4. Connect (4,0) and (0,10). n Move the Objective Function Line Toward Optimality Move it in the direction which lowers its value (down), since we are minimizing, until it touches the last point of the feasible region, determined by the last two constraints.

30. Example 2: Graphical Solution n Objective Function Graphed 54321 54321 54321 54321 1 2 3 4 5 6 x2x2x2x2 x2x2x2x2 Min z = 5 x 1 + 2 x 2 4 x 1 - x 2 > 12 x 1 + x 2 > 4 Min z = 5 x 1 + 2 x 2 4 x 1 - x 2 > 12 x 1 + x 2 > 4 2 x 1 + 5 x 2 > 10 x1x1x1x1 x1x1x1x1

31. n Solve for the Extreme Point at the Intersection of the Two Binding Constraints 4 x 1 - x 2 = 12 4 x 1 - x 2 = 12 x 1 + x 2 = 4 x 1 + x 2 = 4 Adding these two equations gives: Adding these two equations gives: 5 x 1 = 16 or x 1 = 16/5. 5 x 1 = 16 or x 1 = 16/5. Substituting this into x 1 + x 2 = 4 gives: x 2 = 4/5 Substituting this into x 1 + x 2 = 4 gives: x 2 = 4/5 Example 2: Graphical Solution

32. n Solve for the Optimal Value of the Objective Function Solve for z = 5 x 1 + 2 x 2 = 5(16/5) + 2(4/5) = 88/5. Thus the optimal solution is Thus the optimal solution is x 1 = 16/5; x 2 = 4/5; z = 88/5 x 1 = 16/5; x 2 = 4/5; z = 88/5

33. Example 2: Graphical Solution n Optimal Solution 54321 54321 54321 54321 1 2 3 4 5 6 x2x2x2x2 x2x2x2x2 Min z = 5 x 1 + 2 x 2 4 x 1 - x 2 > 12 x 1 + x 2 > 4 Min z = 5 x 1 + 2 x 2 4 x 1 - x 2 > 12 x 1 + x 2 > 4 2 x 1 + 5 x 2 > 10 Optimal: x 1 = 16/5 x 2 = 4/5 x 2 = 4/5 2 x 1 + 5 x 2 > 10 Optimal: x 1 = 16/5 x 2 = 4/5 x 2 = 4/5 x1x1x1x1 x1x1x1x1

34. Example 2: Spreadsheet Solution n Partial Spreadsheet Showing Problem Data

35. Example 2: Spreadsheet Solution n Partial Spreadsheet Showing Formulas

36. Example 2: Spreadsheet Solution n Partial Spreadsheet Showing Solution

37. Feasible Region n The feasible region for a two-variable LP problem can be nonexistent, a single point, a line, a polygon, or an unbounded area. n Any linear program falls in one of three categories: is infeasible is infeasible has a unique optimal solution or alternate optimal solutions has a unique optimal solution or alternate optimal solutions has an objective function that can be increased without bound has an objective function that can be increased without bound n A feasible region may be unbounded and yet there may be optimal solutions. This is common in minimization problems and is possible in maximization problems.

38. Special Cases n Alternative Optimal Solutions In the graphical method, if the objective function line is parallel to a boundary constraint in the direction of optimization, there are alternate optimal solutions, with all points on this line segment being optimal. n Infeasibility A linear program which is overconstrained so that no point satisfies all the constraints is said to be infeasible. A linear program which is overconstrained so that no point satisfies all the constraints is said to be infeasible. n Unboundedness (See example on upcoming slide.)

39. Example: Infeasible Problem n Solve graphically for the optimal solution: Max 2 x 1 + 6 x 2 Max 2 x 1 + 6 x 2 s.t. 4 x 1 + 3 x 2 < 12 s.t. 4 x 1 + 3 x 2 < 12 2 x 1 + x 2 > 8 2 x 1 + x 2 > 8 x 1, x 2 > 0 x 1, x 2 > 0

40. Example: Infeasible Problem n There are no points that satisfy both constraints, hence this problem has no feasible region, and no optimal solution. x2x2x2x2 x1x1x1x1 4 x 1 + 3 x 2 < 12 2 x 1 + x 2 > 8 3 4 4 8

41. Example: Unbounded Problem n Solve graphically for the optimal solution: Max 3 x 1 + 4 x 2 Max 3 x 1 + 4 x 2 s.t. x 1 + x 2 > 5 s.t. x 1 + x 2 > 5 3 x 1 + x 2 > 8 3 x 1 + x 2 > 8 x 1, x 2 > 0 x 1, x 2 > 0

42. Example: Unbounded Problem n The feasible region is unbounded and the objective function line can be moved parallel to itself without bound so that z can be increased infinitely. x2x2x2x2 x1x1x1x1 3x 1 + x 2 > 8 x 1 + x 2 > 5 Max 3x 1 + 4x 2 5 5 8 2.67

43. Solving LP Problem: Blue Ridge Hot Tubs A Graphical Approach 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 X 2 >= 0

44. X2X2X2X2 X1X1X1X1 25 0 20 0 15 0 10 0 50 50 0 0 10 0 15 0 20 0 25 0 (0, 200) (200, 0) boundary line of pump constraint X 1 + X 2 = 200 Plotting the First Constraint

45. X2X2X2X2 X1X1X1X1 25 0 20 0 15 0 10 0 50 50 0 0 10 0 15 0 20 0 25 0 (0, 261) (174, 0) boundary line of labor constraint 9X 1 + 6X 2 = 1566 Plotting the Second Constraint

46. X2X2X2X2 X1X1X1X1 25 0 20 0 15 0 10 0 50 50 0 0 10 0 15 0 20 0 25 0 (0, 180) (240, 0) boundary line of tubing constraint 12X 1 + 16X 2 = 2880 Feasible Region Plotting the Third Constraint

47. X2X2X2X2 Plotting A Level Curve of the Objective Function X1X1X1X1 25 0 20 0 15 0 10 0 50 50 0 0 10 0 15 0 20 0 25 0 (0, 116.67) (100, 0) objective function 350X 1 + 300X 2 = 35000

48. A Second Level Curve of the Objective Function X2X2X2X2 X1X1X1X1 25 0 20 0 15 0 10 0 50 50 0 0 10 0 15 0 20 0 25 0 (0, 175) (150, 0) objective function 350X 1 + 300X 2 = 35000 objective function 350X 1 + 300X 2 = 52500

49. Using A Level Curve to Locate the Optimal Solution X2X2X2X2 X1X1X1X1 25 0 20 0 15 0 10 0 50 50 0 0 10 0 15 0 20 0 25 0 objective function 350X 1 + 300X 2 = 35000 objective function 350X 1 + 300X 2 = 52500 optimal solution

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

51. Example of a Redundant Constraint X2X2X2X2 X1X1X1X1 25 0 20 0 15 0 10 0 50 50 0 0 10 0 15 0 20 0 25 0 boundary line of tubing constraint Feasible Region boundary line of pump constraint boundary line of labor constraint

52. Enumerating The Corner Points X2X2X2X2 X1X1X1X1 25 0 20 0 15 0 10 0 50 50 0 0 10 0 15 0 20 0 25 0 (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.

53. Understanding How Things Change See file Example.xls Example.xls