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Management Science – MNG221

Management Science – MNG221. Linear Programming: Graphical Solution. Linear Programming: Introduction. Most Firms Objectives - Maximize profit (Overall Org.) - Minimize cost (Individual Depts.) Constraints/Restrictions - Limited Resources, - Restrictive Guidelines

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Management Science – MNG221

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  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 • Problem must be solvable by linear programming • The unstructured problemmust be formulated as a mathematical model. • Problem must be solved using established mathematical techniques.

  4. Linear Programming: Introduction • The linear programming technique derives its name from the fact that: • the functional relationships in the mathematical model are linear (Capable of being represented by a straight line), • 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. x1, x2, x3 etc. • Parameters: numerical values that are included in the objective functions and constraints E.g. 40 hrs

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

  8. Linear Programming: Maximization Problem Model Formulation Step 1: Define the decision variables • x1 – number of bowls • x2– 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 Step 1: Define the decision variables x1 – number of bowls x2 – number of mugs Step 2: Define the objective function Z = 40x1 + 50x2 Step 3: Define the constraints x1 + 2x2 ≤ 40 4x1 + 3x2 ≤ 120 Non-negativity constraints x1, x2 ≥ 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: MaximizeZ = 40x1+ 50x2 Where x1+ 2x2 ≤ 40 4x1 + 3x2 ≤ 120 x1, x2≥ 0

  11. Linear Programming: Maximization Problem The solution of this model will result in numeric values for x1 and x2 that will maximize total profit, Z, but should not be infeasible • A feasible solution does not violate any of the constraints. E.g. x1= 5, x2= 10 • An infeasible problem violates at least one of the constraints. E.g. x1= 10, x2= 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 1st 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 x1 = 0 and solving for x2: • A second point can be found by letting x2 = 0 and solving for x1:

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

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

  18. Linear Programming: Graphical Solution 3rd Step - Identify the Optimal Solution Point To find point B, we place a straightedge parallel to the objective function line $800 = 40x1 + 50x2 in Figure 2.10and 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.

  19. Linear Programming: Graphical Solution 4th Step - Solve for the values of x1 and x2

  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 x1: • Now let x1 in the 1st eq. equal x1 in the 2nd eq. 40 - 2x2 = 30 - (3x2/4)

  22. Linear Programming: Graphical Solution • And solve for x2: • Substituting x2 = 8 in one the original equations:

  23. Linear Programming: Graphical Solution The optimal solution is point B Where x1 = 24 and x2 = 8.

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

  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:

  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 Step 1: Define the decision variables x1 = bags of Super-gro x2 = bags of Crop-quick Step 2: Define the objective function Minimize Z = $6x1 + 3x2 Step 3: Define the constraints 2x1 + 4x2 ≥ 16 lb. 4x1 + 3x2 ≥ 24 lb Minimum Requirement Non-negativity constraints x1, x2 ≥ 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 = $6x1 + 3x2 Where 2x1 + 4x2 ≥ 16 lb. 4x1 + 3x2 ≥ 24 lb x1, x2 ≥ 0

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

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

  32. Linear Programming: Graphical Solution 3rd 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 x2: • Given that the optimal solution is x1 = 0, x2 = 8, the minimum cost, Z, is

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