Course
Year
: D0744 - Deterministic Optimization
: 2009
Introduction to Linear Programming
and Formulation
Meeting 2
Introduction
• A model consisting of linear relationships
representing a firm’s objective and resource
constraints
• LP is a mathematical modeling technique used to
determine a level of operational activity in order to
achieve an objective, subject to restrictions called
constraints
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Linear Programming
Linear Programming is a mathematical technique
for optimum allocation of limited or scarce
resources, such as labor, material, machine,
money, energy and so on , to several competing
activities such as products, services, jobs and so
on, on the basis of a given criteria of optimality.
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Linear Programming (LP) Problem
• A mathematical programming problem is one that seeks to
maximize or minimize an objective function subject to
constraints.
• If both the objective function and the constraints are linear, the
problem is referred to as a linear programming problem.
• Linear functions are functions in which each variable appears in
a separate term raised to the first power and is multiplied by a
constant (which could be 0).
• Linear constraints are linear functions that are restricted to be
"less than or equal to", "equal to", or "greater than or equal to" a
constant.
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Building Linear Programming Models
1. What are you trying to decide - Identify the decision variable to
solve the problem and define appropriate variables that
represent them.
For instance, in a simple maximization
problem, RMC, Inc. interested in producing two products: fuel
additive and a solvent base. The decision variables will be X1 =
tons of fuel additive to produce, and X2 = tons of solvent base
to produce.
2. What is the objective to be maximized or minimized?
Determine the objective and express it as a linear function.
When building a linear programming model, only relevant costs
should be included, sunk costs are not included. In our
example, the objective function is: = 40X1 + 30X2; where 40
and 30 are the objective function coefficients
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3. What limitations or requirements restrict the values of the
decision variables? Identify and write the constraints as
linear functions of the decision variables.
Constraints
generally fall into one of the following categories:
a. Limitations - The amount of material used in the production
process cannot exceed the amount available in inventory. In our
example, the limitations are:
Material 1 = 20 tons
Material 2 = 5 tons
Material 3 = 21 tons available.
The material used in the production of X1 and X2 are also known.
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To produce one ton of fuel additive uses .4 ton of material 1, and .60 ton
of material 3. To produce one ton of solvent base it takes .50 ton of
material 1, .20 ton of material 2, and .30 ton of material 3. Therefore, we
can set the constraints as follows:
.4X1 + .50 X2 <= 20
.20X2 <= 5
.6X1 + .3X2 <=21, where .4, .50, .20, .6, and .3 are called
constraint coefficients. The limitations (20, 5, and 21) are called Right
Hand Side (RHS).
b. Requirements - specifying a minimum levels of performance.
For instance, production must be sufficient to satisfy customers’
demand
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Linear Progamming Application
Model Type
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Application
9
LP Model Formulation
• Decision variables
mathematical symbols representing levels of activity of an
operation
• Objective function
• a linear relationship reflecting the objective of an operation
• most frequent objective of business firms is to maximize
profit
• most frequent objective of individual operational units (such
as a production or packaging department) is to minimize cost
• Constraint
a linear relationship representing a restriction on decision
making
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Max/min
z = c1x1 + c2x2 + ... + cnxn
subject to:
a11x1 + a12x2 + ... + a1nxn (≤, =, ≥) b1
a21x1 + a22x2 + ... + a2nxn (≤, =, ≥) b2
am1x1 + am2x2 + ... + amnxn (≤, =, ≥) bm
xj = decision variables
bi = constraint levels
cj = objective function coefficients
aij = constraint coefficients
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Formulation Modeling : Example
RESOURCE REQUIREMENTS
PRODUCT
Bowl
Mug
Labor
(hr/unit)
1
2
Clay
(lb/unit)
4
3
Revenue
($/unit)
40
50
There are 40 hours of labor and 120 pounds of clay available
each day
Decision variables
x1 = number of bowls to produce
x2 = number of mugs to produce
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Maximize Z = $40 x1 + 50 x2
Subject to
x1 + 2x2 ≤ 40 hr (labor constraint)
4x1 + 3x2 ≤ 120 lb (clay constraint)
x1 , x2 ≥ 0
Solution is x1 = 24 bowls
Revenue = $1,360
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x2 = 8 mugs
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