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
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.
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).
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
Resource Requirements
Product Labour
Clay
Profit
(Hr/Unit)
(Lb/Unit)
$/Unit
Bowl
1
4
40
Mug
2
3
50
There are 40 labour hours and
120 pounds of clay available
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
x 1, x 2 ≥ 0
Step 4: Solve the problem
Linear Programming: Maximization Problem
• The complete linear programming model for
this problem can now be summarized as
follows:
Maximize Z = 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 2.10
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.
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:
Chemical Contribution
Brand
NITROGEN
PHOSPHATE
(LB./BAG)
(LB./BAG)
Super-gro
2
4
Crop-quick
4
3
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
Chemical Contribution
Brand
NITROGEN
(LB./BAG)
PHOSPHATE
(LB./BAG)
Super-gro
2
4
Crop-quick
4
3
The field requires at
least 16 pounds of
nitrogen and 24 pounds
of phosphate.
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
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
x 1, x 2 ≥ 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