1. Management Science – MNG221 Linear Programming: Graphical Solution

2. Linear Programming: Introduction Most Firms Objectives - Maximize profit(Overall Org.) - Minimize cost (Individual Depts.) Constraints/Restrictions - Limited Resources, - Restrictive Guidelines Linear Programming is a model that consists of linear relationships representing a firm’s decision(s), given an objective and resource constraints.

3. Linear Programming: Introduction Steps in applying the linear programming technique 1.Problem must be solvable by linear programming 2. The unstructured problem must be formulated as a mathematical model. 3. Problem must be solved using established mathematical techniques.

4. Linear Programming: Introduction The linear programming technique derives its name from the fact that: 1. the functional relationships in the mathematical model are linear (Capable of being represented by a straight line), 2.and the solution technique consists of predetermined mathematical steps that is, a Program (a system of procedures or activities that has a specific purpose).

5. Linear Programming: Model Formulation A linear programming model consists of: Objective function: reflects the objective of the firm in terms of the decision variables Always consists of: Maximizing profit Minimizing cost

6. Linear Programming: Model Formulation Constraints: a restriction on decision making placed on the firm by the operating environment E.g. Raw materials, labour, market size etc. The Objective Function and Constraints consists of: Decision variables: mathematical symbols that represent levels of activity e.g. x 1, x 2, x 3 etc. Parameters: numerical values that are included in the objective functions and constraints E.g. 40 hrs

7. Linear Programming: Model Formulation Objective of Firm: Maximize Profits Beaver Creek Pottery Company

8. Linear Programming: Maximization Problem Model Formulation Step 1: Define the decision variables  x 1 – number of bowls  x 2 – number of mugs Step 2: Define the objective function Maximize profit Step 3: Define the constraints  Clay - A total 120Lbs  Labour – A total 40hrs

9. Linear Programming: Maximization Problem Resource Requirements ProductLabour (Hr/Unit) Clay (Lb/Unit) Profit $/Unit Bowl1440 Mug2350 Step 1: Define the decision variables x 1 – number of bowls x 2 – number of mugs Step 2: Define the objective function Z = 40 x 1 + 50 x 2 Step 3: Define the constraints x 1 + 2 x 2 ≤ 40 4 x 1 + 3 x 2 ≤ 120 Non-negativity constraints x 1, x 2 ≥ 0 Step 4: Solve the problem There are 40 labour hours and 120 pounds of clay available

10. Linear Programming: Maximization Problem The complete linear programming model for this problem can now be summarized as follows: Maximize Z = 40x 1 + 50x 2 Where x 1 + 2x 2 ≤ 40 4x 1 + 3x 2 ≤ 120 x 1, x 2 ≥ 0

11. Linear Programming: Maximization Problem The solution of this model will result in numeric values for x 1 and x 2 that will maximize total profit, Z, but should not be infeasible A feasible solution does not violate any of the constraints. E.g. x 1 = 5, x 2 = 10 An infeasible problem violates at least one of the constraints. E.g. x 1 = 10, x 2 = 20

12. Linear Programming: Graphical Solution The next stage in the application of linear programming is to find the solution of the model A common solution approach is to solve algebraically:  Manually  Computer Program

13. Linear Programming: Graphical Solution Graphical Solutions are limited to linear programming problems with only two decision variables. The graphical method provides a picture of how a solution is obtained for a linear programming problem

14. Linear Programming: Graphical Solution 1 st Step - Plot constraint lines as equations

15. Linear Programming: Graphical Solution Plotting Line Determine two points that are on the line and then draw a straight line through the points. One point can be found by letting x 1 = 0 and solving for x 2 : A second point can be found by letting x 2 = 0 and solving for x 1 :

16. Linear Programming: Graphical Solution 2 nd - Identify Feasible Solution

17. Linear Programming: Graphical Solution 3 rd Step - Identify the Optimal Solution Point

18. Linear Programming: Graphical Solution 3 rd Step - Identify the Optimal Solution Point To find point B, we place a straightedge parallel to the objective function line $800 = 40 x 1 + 50 x 2 in Figure 2.10 and move it outward from the origin as far as we can without losing contact with the feasible solution area. Point B is referred to as the optimal (i.e., best) solution. Figure 2.10 optimal (i.e., best) solution.

19. Linear Programming: Graphical Solution 4 th Step - Solve for the values of x 1 and x 2

20. Linear Programming: Graphical Solution The Optimal Solution Point is the last point the objective function touches as it leaves the feasible solution area. Extreme Points are corner points on the boundary of the feasible solution area. E.g. A, B or C

21. Linear Programming: Graphical Solution Constraint Equations are solved simultaneously at the optimal extreme point to determine the variable solution values.  First, convert both equations to functions of x 1 :  Now let x 1 in the 1st eq. equal x 1 in the 2 nd eq. 40 - 2 x 2 = 30 - (3 x 2 /4)

22. Linear Programming: Graphical Solution  And solve for x 2:  Substituting x 2 = 8 in one the original equations:

23. Linear Programming: Graphical Solution The optimal solution is point B Where x 1 = 24 and x 2 = 8.

24. Linear Programming: Minimization Model Objective of Firm: Minimization of Cost A Famer’s Field

25. Linear Programming: Graphical Solution A farmer is preparing to plant a crop in the spring and needs to fertilize a field. There are two brands of fertilizer to choose from, Super-gro and Crop-quick. Each brand yields a specific amount of nitrogen and phosphate per bag, as follows: Chemical Contribution BrandNITROGEN (LB./BAG) PHOSPHATE (LB./BAG) Super-gro24 Crop-quick43

26. Linear Programming: Graphical Solution The farmer's field requires at least  16 pounds of nitrogen and  24 pounds of phosphate. Super-gro costs $6 per bag, and Crop-quick costs $3. The farmer wants to know how many bags of each brand to purchase in order to minimize the total cost of fertilizing.

27. Linear Programming: Minimization Problem Model Formulation Step 1: Define the Decision Variables  How many bags of Super-gro and Crop-quick to buy Step 2: Define the Objective Function  Minimize cost Step 3: Define the Constraints  The field requirements for nitrogen and phosphate

28. Linear Programming: Minimization Problem Chemical Contribution BrandNITROGEN (LB./BAG) PHOSPHATE (LB./BAG) Super-gro24 Crop-quick43 Step 1: Define the decision variables x 1 = bags of Super-gro x 2 = bags of Crop-quick Step 2: Define the objective function Minimize Z = $6 x 1 + 3 x 2 Step 3: Define the constraints 2 x 1 + 4 x 2 ≥ 16 lb. 4 x 1 + 3 x 2 ≥ 24 lb Minimum Requirement Non-negativity constraints x 1, x 2 ≥ 0 Step 4: Solve the problem The field requires at least 16 pounds of nitrogen and 24 pounds of phosphate.

29. Linear Programming: Maximization Problem The complete model formulation for this minimization problem is: Minimize Z = $6x 1 + 3x 2 Where 2x 1 + 4x 2 ≥ 16 lb. 4x 1 + 3x 2 ≥ 24 lb x 1, x 2 ≥ 0

30. Linear Programming: Graphical Solution 1 st Step - Plot constraint lines as equations

31. Linear Programming: Graphical Solution 2 nd Step – Identify the feasible solution to reflect the inequalities in the constraints

32. Linear Programming: Graphical Solution 3 rd Step - Locate the optimal point.

33. Linear Programming: Graphical Solution The Optimal Solution of a minimization problem is at the extreme point closest to the origin. Extreme Points are corner points on the boundary of the feasible solution area. E.g. A, B Or C As the Objective Function edges toward the origin, the last point it touches in the feasible solution area is A. In other words, point A is the closest the objective function can get to the origin without encompassing infeasible points.

34. Linear Programming: Graphical Solution  And solve for x 2:  Given that the optimal solution is x 1 = 0, x 2 = 8, the minimum cost, Z, is