1. 1 1 Slide Linear Programming (LP) Problem n A mathematical programming problem is one that seeks to maximize an objective function subject to constraints. 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.

2. 2 2 Slide LP Solutions n The maximization or minimization of some quantity is the objective in all linear programming problems. 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. 3 3 Slide Problem Formulation n Problem formulation or modeling is the process of translating a verbal statement of a problem into a mathematical statement.

4. 4 4 Slide Guidelines for Model Formulation n Understand the problem thoroughly. n Write a verbal description of the objective. n Write a verbal description of 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.

5. 5 5 Slide Example 1: A Maximization Problem n LP Formulation Max z = 5 x 1 + 7 x 2 Max z = 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

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

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

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

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

10. 10 Slide Example 1: Graphical Solution n Feasible Solution Region 87654321 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 x1x1x1x1 FeasibleRegion x 2 x 2

11. 11 Slide Example 1: Graphical Solution n Objective Function Line 87654321 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 x1x1x1x1 x 2 x 2 (7, 0) (0, 5) 5x 1 + 7x 2 = 35

12. 12 Slide Example 1: Graphical Solution n Optimal Solution 87654321 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 x1x1x1x1 x 2 x 2 5x 1 + 7x 2 = 46 Optimal Solution

13. 13 Slide 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.

14. 14 Slide 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.

15. 15 Slide Example 1 n Standard Form Max z = 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

16. 16 Slide 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.

17. 17 Slide Example 1: Graphical Solution n The Five Extreme Points 87654321 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 x1x1x1x1 FeasibleRegion 1122 33 44 55

18. 18 Slide 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.

19. 19 Slide Interpretation of Computer Output n In this chapter we will discuss the following output: objective function valueobjective function value values of the decision variablesvalues of the decision variables reduced costsreduced costs slack/surplusslack/surplus n In Chapter 3 we will discuss how an optimal solution is affected by a: change in a coefficient of the objective functionchange in a coefficient of the objective function change in the right-hand side value of a constraintchange in the right-hand side value of a constraint

20. 20 Slide Example 1: Spreadsheet Solution n Partial Spreadsheet Showing Problem Data

21. 21 Slide Example 1: Spreadsheet Solution n Partial Spreadsheet Showing Solution

22. 22 Slide Example 1: Spreadsheet Solution n Interpretation of Computer Output We see from the previous slide that: Objective Function Value = 46Objective Function Value = 46 Decision Variable #1 ( x 1 ) = 5Decision Variable #1 ( x 1 ) = 5 Decision Variable #2 ( x 2 ) = 3Decision 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)

23. 23 Slide 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.

24. 24 Slide Example 1: Spreadsheet Solution n Reduced Costs

25. 25 Slide Example 2: A Minimization Problem n LP Formulation Min z = 5 x 1 + 2 x 2 Min z = 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

26. 26 Slide 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 x 1 = 5, then x 2 = 8. Connect (3,0) and (5,8). The ">" side is to the right. 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.

27. 27 Slide 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 4x 1 - x 2 > 12 x 1 + x 2 > 4 x 1 + x 2 > 4 4x 1 - x 2 > 12 x 1 + x 2 > 4 x 1 + x 2 > 4 2x 1 + 5x 2 > 10 x1x1x1x1 x1x1x1x1 Feasible Region

28. 28 Slide 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.

29. 29 Slide Example 2: Graphical Solution n Objective Function Graphed 54321 54321 54321 54321 1 2 3 4 5 6 x2x2x2x2 x2x2x2x2 Min z = 5x 1 + 2x 2 4x 1 - x 2 > 12 x 1 + x 2 > 4 Min z = 5x 1 + 2x 2 4x 1 - x 2 > 12 x 1 + x 2 > 4 2x 1 + 5x 2 > 10 x1x1x1x1 x1x1x1x1

30. 30 Slide Example 2: Graphical Solution 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 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

31. 31 Slide Example 2: Graphical Solution n Optimal Solution 54321 54321 54321 54321 1 2 3 4 5 6 x2x2x2x2 x2x2x2x2 Min z = 5x 1 + 2x 2 4x 1 - x 2 > 12 x 1 + x 2 > 4 Min z = 5x 1 + 2x 2 4x 1 - x 2 > 12 x 1 + x 2 > 4 2x 1 + 5x 2 > 10 Optimal: x 1 = 16/5 x 2 = 4/5 x 2 = 4/5 2x 1 + 5x 2 > 10 Optimal: x 1 = 16/5 x 2 = 4/5 x 2 = 4/5 x1x1x1x1 x1x1x1x1

32. 32 Slide Example 2: Spreadsheet Solution n Partial Spreadsheet Showing Problem Data

33. 33 Slide Example 2: Spreadsheet Solution n Partial Spreadsheet Showing Formulas

34. 34 Slide Example 2: Spreadsheet Solution n Partial Spreadsheet Showing Solution

35. 35 Slide Feasible Region n The feasible region for a two-variable linear programming 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 infeasibleis infeasible has a unique optimal solution or alternate optimal solutionshas a unique optimal solution or alternate optimal solutions has an objective function that can be increased without boundhas 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.

36. 36 Slide 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 Unbounded (See example on upcoming slide.)

37. 37 Slide Example: Infeasible Problem n Solve graphically for the optimal solution: Max z = 2 x 1 + 6 x 2 Max z = 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

38. 38 Slide 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 4x 1 + 3x 2 < 12 2x 1 + x 2 > 8 3 4 4 8

39. 39 Slide Example: Unbounded Problem n Solve graphically for the optimal solution: Max z = 3 x 1 + 4 x 2 Max z = 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

40. 40 Slide 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