Linear Programming Problem present.pptx

Topic : Application of Linear Inequalities in LP
Formulations
Course Leader:
Dr. Sudeshna Rath

Linear Programming: Model Formulation and
Graphical Solution

 Objectives of business decisions frequently involve maximizing
profit or minimizing costs
 Linear programming uses linear algebraic relationships to
represent a firm’s decisions, given a business objective, and
resource constraints
 Steps in application:
1. Identify problem as solvable by linear programming
2. Formulate a mathematical model of the unstructured
problem
3. Solve the model
4. Implementation
Linear Programming: An Overview

Linear Programming: An Overview(Contd..)
• Linear Programming is the most widely applied mathematical technique that helps in decision
making and planning for the optimal allocation of limited resources.
• It deals with the optimization (maximization or minimization) of a function of variables known as
objective function, subject to a set of linear equations and/or inequalities known as constraints.
• The objective function may be profit, cost, production capacity or any other measure of
effectiveness, which is to be obtained in the best possible or optimal manner.
• The constraints may be imposed by different resources such as raw material availability, market
demand, production process and equipment, storage capacity etc.
• It helps in attaining the optimum use of productive factors. Linear programming indicates how a
manager can utilize his productive factors most effectively by a better selection and distribution of
these elements. For example, more efficient use of manpower and machines can be obtained by
the use of linear programming.

• Decision variables - mathematical symbols representing levels
of activity of a firm
• Objective function - a linear mathematical relationship
describing an objective of the firm, in terms of decision
variables - this function is to be maximized or minimized
• Constraints – requirements or restrictions placed on the firm
by the operating environment, stated in linear relationships of
the decision variables
• Parameters - numerical coefficients and constants used in the
objective function and constraints
Model Components

Summary of Model Formulation Steps
Step 1 : Clearly define the decision variables
Step 2 : Construct the objective function
Step 3 : Formulate the constraints

LP Model Formulation A Maximization Example
Beaver Creek Pottery Company is a small crafts operation run by a Native American
tribal council. The company employs skilled artisans to produce clay bowls and
mugs with authentic Native American designs and colors. The two primary
resources used by the company are special pottery clay and skilled labor. Given
these limited resources, the company desires to know how many bowls and mugs
to produce each day in order to maximize profit. The two products have the
following resource requirements for production and profit per item produced (i.e.,
the model parameters):
This scenario is illustrated in Figure 1.8

LP Model Formulation
A Maximization Example (1 of 4)
• Product mix problem - Beaver Creek Pottery Company
• How many bowls and mugs should be produced to maximize
profits given labor and materials constraints?
• Product resource requirements and unit profit:
Resource Requirements
Product
Labor
(Hr./Unit)
Clay
(Lb./Unit)
Profit
(Rs./Unit)
Bowl 1 4 40
Mug 2 3 50
Resource Availability: 40 hrs of labor per day
120 lbs of clay

Figure 1.8 Process in A Beaver Creek Pottery Company

Resource 40 hrs of labor per day
Availability: 120 lbs of clay
Decision x1 = number of bowls to produce per day
Variables: x2 = number of mugs to produce per day
Objective Maximize Z = 40x1 + 50x2
Function: Where Z = profit per day in Rs.
Resource 1x1 + 2x2  40 hours of labor
Constraints: 4x1 + 3x2  120 pounds of clay
Non-Negativity x1  0; x2  0
Constraints:

Complete Linear Programming Model:
Maximize Z = 40x1 + 50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0

A feasible solution does not violate any of the constraints:
Example: x1 = 5 bowls
x2 = 10 mugs
Z = 40x1 + 50x2 = 700
Labor constraint check: 1(5) + 2(10) = 25 < 40 hours
Clay constraint check: 4(5) + 3(10) = 70 < 120 pounds
Feasible Solutions

An infeasible solution violates at least one of the constraints:
Example: x1 = 10 bowls
x2 = 20 mugs
Z = 40x1 + 50x2 = 1400
(Z is in Rs.)
Labor constraint check: 1(10) + 2(20) = 50 > 40 hours
Infeasible Solutions

• Graphical solution is limited to linear programming
models containing only two decision variables (can
be used with three variables but only with great
difficulty)
• Graphical methods provide visualization of how a
solution for a linear programming problem is
obtained
Graphical Solution of LP Models

Coordinate Axes
Graphical Solution of Maximization Model (1 of 12)
Figure 1.9: Coordinates for Graphical Analysis
4x1 + 3x2  120
x1, x2  0
X1 is bowls
X2 is mugs

Labor Constraint
Figure 1.10: Graph of Labor Constraint
Z is Profit in Rs.
4x1 + 3x2  120
x1, x2  0

Labor Constraint Area
Figure 1.11: Labor Constraint Area
Z is Profit in Rs.
4x1 + 3x2  120
x1, x2  0

Clay Constraint Area
Figure 1.12 : Clay Constraint Area
4x1 + 3x2  120
x1, x2  0

Both Constraints
Figure1.13 : Graph of Both Model Constraints
Z is in Rs.
4x1 + 3x2  120
x1, x2  0

Feasible Solution Area
Figure 1.14: Feasible Solution Area
4x1 + 3x2  120
x1, x2  0

Objective Function Solution = Rs.800
Figure 1.15 : Objective Function Line for Z = Rs.800
Z is Profit in Rs.
4x1 + 3x2  120
x1, x2  0

Alternative Objective Function Solution Lines
Figure1.16: Alternative Objective Function Lines
Z is in Rs.
4x1 + 3x2  120
x1, x2  0

Optimal Solution
Figure 1.17: Identification of Optimal Solution Point
4x1 + 3x2  120
x1, x2  0

Optimal Solution Coordinates
Figure 1.18 : Optimal Solution Coordinates
Z is Profit in Rs.
4x1 + 3x2  120
x1, x2  0

Extreme (Corner) Point Solutions
Figure1.19: Solutions at All Corner Points
Z is Profit in Rs.
4x1 + 3x2  120
x1, x2  0

Optimal Solution for New Objective Function
Z is Profit in Rs.
4x1+ 3x2  120
x1, x2  0
Figure 1.20: Optimal Solution with Z = 70x1 + 20x2

 Standard form requires that all constraints be in the form of
equations (equalities)
 A slack variable is added to a  constraint (weak inequality) to
convert it to an equation (=)
 A slack variable typically represents an unused resource
 A slack variable contributes nothing to the objective function value
Slack Variables

Linear Programming Model: Standard Form
Max Z = 40x1 + 50x2 + 0s1 + 0s2
subject to:1x1 + 2x2 + s1 = 40
4x1 + 3x2 + s2 = 120
x1, x2, s1, s2  0
Where:
x1 = number of bowls
x2 = number of mugs
s1, s2 are slack variables
Figure 1.21: Solution Points A, B, and C with Slack

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-grow and Crop-quick. Each brand yields a specific amount of
nitrogen and phosphate per bag, as follows:
LP Model Formulation – Minimization
Chemical Contribution
Brand
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 at
least 24 pounds of phosphate. Super-grow 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.

LP Model Formulation – Minimization (1 of 8)
Chemical Contribution
Brand
Nitrogen
(lb/bag)
Phosphate
(lb/bag)
Super-gro 2 4
Crop-quick 4 3
 Two brands of fertilizer available - Super-grow, Crop-quick
 Field requires at least 16 pounds of nitrogen and 24 pounds of
phosphate
 Super-grow costs Rs.6 per bag, Crop-quick Rs.3 per bag
 Problem: How much of each brand to purchase to minimize total
cost of fertilizer given following data ?
This scenario is illustrated in Figure.1.22

Figure1.22: Fertilizing farmer’s field

Decision Variables:
x1 = bags of Super-gro
x2 = bags of Crop-quick
The Objective Function:
Minimize Z = 6x1 + 3x2
Where: Rs.6x1 = cost of bags of Super- Gro
Rs.3x2 = cost of bags of Crop-Quick
Z is the Total Cost in Rs.
Model Constraints:
2x1 + 4x2  16 lb (nitrogen constraint)
4x1 + 3x2  24 lb (phosphate constraint)
x1, x2  0 (non-negativity constraint)

Z is Total Cost in Rs.
subject to: 2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
Figure 1.23: Graph of Both Model Constraints
Constraint Graph – Minimization (4 of 8)

Figure 1.24: Feasible Solution Area
Feasible Region– Minimization (5 of 8)
Z is Total Cost in Rs.
4x1 + 3x2  24
x1, x2  0

Figure 1.25: Optimum Solution Point
Optimal Solution Point – Minimization (6 of 8)
4x1 + 3x2  24
x1, x2  0

 A surplus variable is subtracted from a  constraint to
convert it to an equation (=)
 A surplus variable represents an excess above a
constraint requirement level
 A surplus variable contributes nothing to the calculated
value of the objective function
 Subtracting surplus variables in the farmer problem
constraints:
2x1 + 4x2 - s1 = 16 (nitrogen)
4x1 + 3x2 - s2 = 24 (phosphate)
Surplus Variables – Minimization (7 of 8)

Figure 1.26: Graph of Fertilizer Example
Graphical Solutions – Minimization (8 of 8)
Minimize Z = 6x1 + 3x2 + 0s1 + 0s2
subject to: 2x1 + 4x2 – s1 = 16
4x1 + 3x2 – s2 = 24
x1, x2, s1, s2  0

For some linear programming models, the general rules do
not apply.
• Special types of problems include those with:
 Multiple optimal solutions
 Infeasible solutions
 Unbounded solutions
Irregular Types of Linear Programming Problems

Figure 1.27: Example with Multiple Optimal Solutions
Multiple Optimal Solutions Beaver Creek Pottery
The objective function is
parallel to a constraint line.
Maximize Z=40x1 + 30x2
4x1 + 3x2  120
x1, x2  0
Where:
x1 = number of bowls
x2 = number of mugs

An Infeasible Problem
Figure 1.28: Graph of an Infeasible Problem
Every possible solution
violates at least one constraint:
x1  4
x2  6
x1, x2  0

An Unbounded Problem
Figure 1.29: Graph of an Unbounded Problem
Value of the objective
function increases indefinitely:
subject to: x1  4
x2  2
x1, x2  0

Characteristics of Linear Programming Problems
• A decision amongst alternative courses of action is required
• Decision is represented in the model by decision variables
• Problem encompasses a goal, expressed as an objective
function, that the decision maker wants to achieve
• Restrictions (represented by constraints) exist that limit the
extent of achievement of the objective
• The objective and constraints must be definable by linear
mathematical functional relationships

• Proportionality - The rate of change (slope) of the objective
function and constraint equations is constant
• Additivity - Terms in the objective function and constraint
equations must be additive
• Divisibility -Decision variables can take on any fractional value
and are therefore continuous as opposed to integer in nature
• Certainty - Values of all the model parameters are assumed to
be known with certainty (non-probabilistic)
Properties of Linear Programming Models

Moore’s Company produces a food mixture in 1,000-pound batches. The mixture
contains two ingredients— A and B . The cost per pound of each of these
ingredients is as follows:
Problem Statement
Example Problem No. 1
Each batch has the following recipe requirements:
a. At least 500 pounds of A
b. At least 200 pounds of B
The ratio of A to B must be at least 2 to 1. The company wants to know the
optimal mixture of ingredients that will minimize cost. Formulate a linear
programming model for this problem.
Ingredient Cost/lb.
A ₹ 3
B ₹ 5

Problem Statement
Example Problem No. 1 (1 of 3)
■ Food mixture in 1000-pound batches
■ Two ingredients, A (Rs.3/lb) and B (Rs.5/lb)
■ Recipe requirements:
at least 500 pounds of “A”
at least 200 pounds of “B”
■ Ratio of A to B must be at least 2 to 1
■ Determine optimal mixture of ingredients that will minimize
costs

Step 1:
Identify decision variables.
x1 = lb of A in mixture
x2 = lb of B in mixture
Step 2:
Formulate the objective function.
where Z = cost per 1,000-lb batch in Rs.
3x1 = cost of A
5x2 = cost of B
Solution

Step 3:
Establish Model Constraints
x1 + x2 = 1,000 lb
x1  500 lb of A
x2  200 lb of B
x1/x2  2/1 or x1 - 2x2  0
x1, x2  0
The Model: Minimize Z = 3x1 + 5x2
subject to: x1 + x2 = 1,000 lb
x1  50
x2  200
x1 - 2x2  0
Solution

Solve the following model
graphically:
subject to: x1 + 2x2  10
6x1 + 6x2  36
x1  4
x1, x2  0
Step 1: Plot the constraints
as equations
Figure 1.30: Constraint Equations

6x1 + 6x2  36
x1  4
x1, x2  0
Step 2: Determine the feasible
solution space
Figure 1.31: Feasible Solution Space and Extreme Points

6x1 + 6x2  36
x1  4
x1, x2  0
Step 3 and 4: Determine the
solution points and optimal
solution
Figure 1.32: Optimal Solution Point

Practice Problem 2
A firm uses lathes, milling machines, and grinding machines to produce two machine parts, Part I and Part II. The
machining times required for each part on different machines and the maximum time available per week are
provided in the table below:
Determine the number of parts I and II to be manufactured per week
in order to maximize the total profit.
1.Formulate the problem as a Linear Programming (LP) model by
defining:
1. The objective function to maximize profit
2. All constraints based on machining times and the maximum
time available for each machine
2.Ensure to include non-negativity constraints for the solution.

Practice Problem 4
A company manufactures two products, A and B, with profits earned per unit of ₹3 and ₹4, respectively.
Each product is processed on two machines, M₁ and M₂.
•Product A requires one minute of processing time on M₁ and two minutes on M₂.
•Product B requires one minute on M₁ and one minute on M₂.
•Machine M₁ is available for a maximum of 7 hours and 30 minutes (450 minutes) per working day.
•Machine M₂ is available for a maximum of 10 hours (600 minutes) per working day.
Task:
Determine the number of units of products A and B to be produced per day in order to maximize the total
profit.
1.Formulate the Linear Programming (LP) model by:
1. Defining the objective function for profit maximization.
2. Writing the constraints based on the availability of machine time on M₁ and M₂.
2.Ensure to include non-negativity constraints for the solution.

Summary
• Introduction to the concepts of Management Science/Operations
Research has been discussed with reference to origin, history,
application
• Several illustrations have been used to formulate the Linear
Programming Problem in relation to objective function and
constraints
• The meaning of Feasible and Infeasible solutions has been discussed
graphically
• Demonstration of Maximisation and Minimisation problems has
been done graphically
• Examples of Bounded and Unbounded solutions have been
graphically analysed

References
• Heaussler E F , Paul RW (2017), Introductory Mathematical Analysis, Pearson
Education
• Trivedi K and Trivedi C (2011) Business Mathematics, Pearson Education
• Dowling, Edward (2011), Schaum's Outline of Introduction to Mathematical
Economics, 3rd edition, McGraw-Hill Education

Disclaimer
• All data and content provided in this presentation are
taken from the reference books, internet – websites
and links, for informational purposes only.

Linear Programming Problem present.pptx

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