Linear Programming: Model Formulation and Graphical Solution

LINEAR PROGRAMMING:
MODEL FORMULATION
AND GRAPHICAL SOLUTION
Why use linear There should be an existing PROBLEM
programming?
(maximization of profit and minimization cost)
Solve by developing a Mathematical Model/ Problem
Linear Programming is a model that consists of linear

relationships representing a firm’s decision(s), given an
objective and resource constraints
A Maximization Model There are three steps in applying the linear programming
Example technique
• the problem must be identified as being solvable by
Linear Programming
linear programming.
technique derives its
name from the fact that • unstructured problem must be formulated as a
the functional mathematical model.
relationship in the • the model must be solved by using established
mathematical model are mathematical techniques
linear.
Katutubo, Inc is a cooperative primarily producing two kinds of products (Bowl and Mug) using their raw
materials (clay). The following are the given data of the said coop:
Labor Clay (Lbs) Profit
Bowl (X) 1 4 40
Mug (Y) 2 3 50
Max Capacity/
Availability ~
Constraints 40 120 lbs
Model Formulation includes:
MODEL – Abstract representation of an existing problem

Product cost = 5
Selling Price = 20
P = 20x – 5x
P and x are variables or a symbol to represent an item that can take on any value
Parameters – 20 and 5. Are known, constant values that are often coefficients of variables in equations. Usually remains constant
during the process of solving a specific problem. Derived from data (or pieces of information) from the problem environment. – linear
in variables
Equation as a whole is a functional relationship that includes variables, parameters, and equations
Decision variables – are mathematical symbols that represent levels of activity by the firm.
X for Bowl and y for Mug
The objective Function – linear mathematical relationship that describes the objective of the firm in terms of the decision variables.
Model Constraint – are also linear relationships of the decision variables ~ restrictions in the operating environment – limited
resources or restrictive guidelines.
Parameters – are numerical values that are included in the objective functions and constraints
Decision variables – how Define the Constraints – resources (clay and labor)
many bowls and mugs to available
produce Labor: 1x + 2y <= 40 hrs*
x&y Clay: 4x+3y <= 120 lbs
Objective Function - *Why less than or equal to? Because that is the limitation
maximize profit that can be used and not an amount that must be used ~
Maximize P = $40x + 50y allows some flexibility
40x = Profit from bowls FINAL RESTRICTION

50y = Profit from mugs X >= 0
Y >=0
~as it is impossible to produce negative items. A.k.a. non-

negativity constraints
COMPLETE LINEAR
POUNDS OF CLAYS
PROGRAMMING MODEL 4(5) + 3(10) <= 120
50 <= 120
Maximize P = 40x + 50y
Therefore, those number of units are feasible/ does meet all of the
Subject to: constraints
1x+2y <=40
4x + 3y <=120 Z = 40(5) + 50(10)
= 700
X,y >= 0
SOLUTION TO THIS MODEL What if the management decided to produce 10 bowls and 20 mugs
P= 40(10) + 50(20)
= 400 + 1000
Ex. 5 units for bowl & 10 units = 1,400
for Mugs
LABOR HOURS The foregoing is infeasible as it violates one of the constraints
1(10) + 2(20) … 40
1(5) + 2(10) <=40 50 …/ 40
25<=40
The solution that achieves the objectives is x=24 bowls and y = 8
mugs
Or a profit of 1,360
1. Graphical Solutions of Linear
Programming Models
Graphical solutions are limited to Y

linear programming problems with Graph of the labor
only two decision variables constraint line
The graphical method provides a
picture of how a solution is obtained
for a linear programming problem
Maximize Z = 40x + 50y

Subject to:
1x+2y <=40
4x + 3y <=120
X,y >= 0
x + 2y = 40
x – number of bowls produced

y – number of mugs produced
Y Y
Labor Constraint Area Constraint Area for Clay
4x + 3y <= 120
x + 2y <= 40
Y Y
Graph of both The feasible
model constraints solution area
constraints
4x + 3y = 120
x + 2y = 40
2. The Optimal Solution Point
The second step in the graphical
solution method is to locate the Y
point in the feasible solution Objective
area that will result in the greatest function line
total profit for Z $800
Assuming Profit is = 800
Objective Function:
800 = 40x1 + 50x2
800 = 40x +50y

Observations:
Alternative objective function lines for profits,
Z, of $800, $1,200, and $1,600
1. Profit increases as the objective function line
moves away from the origin
Y
2. Given this characteristic, the maximum profit
will be attained at the point where the objective
function line is farthest from the origin and is
800 = 40x +50y
still touching a point in the feasible solution
1,200 = 40x +50y area.
1,600 = 40x +50y
3. The Solution Values
The third step in the graphical Y Optimal solution
solution approach is to solve for the coordinates
values of and once the optimal
solution point has been found.
4x + 3y = 120
Y Identification of
optimal solution
point
x + 2y =40
800 = 40x +50y

Linear Programming: Model Formulation and Graphical Solution

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LINEAR PROGRAMMING:

Solve by developing a Mathematical Model/ Problem

Linear Programming is a model that consists of linear

Labor Clay (Lbs) Profit

MODEL – Abstract representation of an existing problem

40x = Profit from bowls FINAL RESTRICTION

~as it is impossible to produce negative items. A.k.a. non-

Graphical solutions are limited to Y

Maximize Z = 40x + 50y

x – number of bowls produced

Assuming Profit is = 800

800 = 40x +50y

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