Management Science – MNG221 Linear Programming: Graphical Solution.

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.


• Steps in applying the linear programming technique1. Problem must be solvable by linear

programming2. The unstructured problem must be

formulated as a mathematical model.3. Problem must be solved using established

mathematical techniques.


• 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).

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


• 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


Objective of Firm: Maximize Profits

Beaver Creek Pottery Company

Linear Programming: Maximization Problem

Model Formulation

Step 1: Define the decision variablesx1 – number of bowlsx2 – number of mugs

Step 2: Define the objective functionMaximize profit

Step 3: Define the constraintsClay - A total 120LbsLabour – A total 40hrs


Resource Requirements

Product Labour(Hr/Unit)

Clay(Lb/Unit)

Profit$/Unit

Bowl 1 4 40

Mug 2 3 50

Step 1: Define the decision variablesx1 – number of bowlsx2 – number of mugs

Step 2: Define the objective functionZ = 40x1 + 50x2

Step 3: Define the constraintsx1 + 2x2 ≤ 404x1 + 3x2 ≤ 120

Non-negativity constraintsx1, x2 ≥ 0

Step 4: Solve the problem

There are 40 labour hours and 120 pounds of clay available


• The complete linear programming model for this problem can now be summarized as follows:

Maximize Z = 40x1 + 50x2

Wherex1 + 2x2 ≤ 40

4x1 + 3x2 ≤ 120

x1, x2 ≥ 0


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


• 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


• 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


1st Step - Plot constraint lines as equations


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:


2nd - Identify Feasible Solution


3rd Step - Identify the Optimal Solution Point


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 2.10 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.

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4th Step - Solve for the values of x1 and x2


• 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


• 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)


And solve for x2:

Substituting x2 = 8 in one the original equations:


The optimal solution is point B Where x1 = 24 and x2 = 8.

Linear Programming: Minimization Model

Objective of Firm: Minimization of Cost

A Famer’s Field


• 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 ContributionBrand NITROGEN

(LB./BAG)PHOSPHATE

(LB./BAG)

Super-gro 2 4

Crop-quick 4 3


• 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 FunctionMinimize cost

Step 3: Define the ConstraintsThe field requirements for nitrogen and

phosphate

Linear Programming: Minimization Problem

Chemical Contribution

Brand NITROGEN (LB./BAG)

PHOSPHATE (LB./BAG)

Super-gro 2 4

Crop-quick 4 3

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.


• 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


1st Step - Plot constraint lines as equations


2nd Step – Identify the feasible solution to reflect the inequalities in the constraints


3rd Step - Locate the optimal point.


• 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.


And solve for x2:

Given that the optimal solution is x1 = 0, x2 = 8, the minimum cost, Z, is

Management Science – MNG221 Linear Programming: Graphical Solution.

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