Introduction to Linear Programming and Graphical Solutions for Profit Maximization
This guide introduces linear programming, a mathematical technique used by firms to optimize objectives such as profit maximization and cost minimization while considering constraints like resource limitations. It outlines the process for formulating linear programming models, including defining decision variables, objective functions, and constraints. Further, it explains graphical solutions for maximizing profits using constrained resources, demonstrating how to plot constraint lines and identify feasible and optimal solution points. The systematic approach helps in making informed decisions in business environments.
Introduction to Linear Programming and Graphical Solutions for Profit Maximization
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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 • Linear Programming is a model that consists of linear relationships representing a firm’s decision(s), given an objective and resource constraints.
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.
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).
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
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
Linear Programming: Model Formulation Beaver Creek Pottery Company Objective of Firm: Maximize Profits
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
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
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
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
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
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
Linear Programming: Graphical Solution 1st Step - Plot constraint lines as equations
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:
Linear Programming: Graphical Solution 2nd - Identify Feasible Solution
Linear Programming: Graphical Solution 3rd Step - Identify the Optimal Solution Point
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.
Linear Programming: Graphical Solution 4th Step - Solve for the values of x1 and x2
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
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)
Linear Programming: Graphical Solution • And solve for x2: • Substituting x2 = 8 in one the original equations:
Linear Programming: Graphical Solution The optimal solution is point B Where x1 = 24 and x2 = 8.
Linear Programming: Minimization Model A Famer’s Field Objective of Firm: Minimization of Cost
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:
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.
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
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.
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
Linear Programming: Graphical Solution 1st Step - Plot constraint lines as equations
Linear Programming: Graphical Solution 2nd Step – Identify the feasible solution to reflect the inequalities in the constraints
Linear Programming: Graphical Solution 3rd Step - Locate the optimal point.
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.
Linear Programming: Graphical Solution • And solve for x2: • Given that the optimal solution is x1 = 0, x2 = 8, the minimum cost, Z, is