Chapter TWO OR Final - Teacher Note

4/22/24, 8:14 AM Chapter TWO OR final - teacher note
UNIT TWO: THE LINEAR PROGRAMMING (LP)

2.1 Introduction to LP
The purpose of this section is to introduce you with linear programming. Emphasis is given on
familiarization with terminologies, problem recognition, model formulation, and applications of LP.
Constrained Optimization
A commonly encountered form of decision making involves situations in which the set of acceptable
solutions is somehow restricted. The restrictions may be imposed internally or externally. For example, an
internal restriction might be the amount of raw materials that a department has available to produce its
products. This would impose a limit on the amount of product that could be produced. Other internal
restrictions can include availability of labor time, machine time, technical requirements, and budgets. An
external restriction might be labor regulations (safety equipment, training requirements, and
overtime) that limit the options open to decision makers. The restrictions are called constraints for
purposes of linear programming. The goal in LP is to find the best possible solution given the constraints
imposed by the problem; hence the term constrained optimization. We are trying to optimize (find the
maximum or minimum values) a given objective function subject to constraints or limiting conditions.
Linear Programming
In 1947, George Dantzig developed the use of linear algebra for determining solutions to problems that
involved the optimal allocation of scarce resources such as labor, finance, time, etc. In spite of numerous
potential applications in business, response to this new technique was low, primarily because of the
substantial computational burden it required. Subsequent advances in computer technology and related
computer software have removed the computational burden and this has led to widespread use of linear
programming in business.
The programming aspect of LP refers to the use of algorithms. An algorithm is a well defined sequence of
steps that will lead to an optimal solution. The term linear refers to straight line relationships. LP models
are based on linear relationships. Taken as a whole, the term linear programming refers to a family of
mathematical techniques that can be used to find solutions to constrained optimizations problems.
Linear programming is a problem solving approach that has been developed to help managers in making
decisions. The following decisions described some typical applications of LP:
a) A manufacturer wants to develop a production schedule and an inventory policy that will satisfy sales
demand in future periods. Ideally, the schedule and policy will enable the company to satisfy demand
and at the same time minimize the total production and inventory costs. The objective here is to
minimize cost subject to demand and production capacity constraints.
b) A financial analyst must select an investment portfolio from a variety of stock and bond investment
alternatives. The analyst would like to establish the portfolio that maximize the return on investment
subject to total amount of investment funds available and the maximum amounts that can be invested
in each stock and bond.
c) A marketing manager wants to determine how best to allocate a fixed advertising budget among
alternative advertising medias such as radio, TV, newspaper, and magazine. The manager would like
to determine the media mix that maximizes the advertising effectiveness subject to a fixed advertising
budget and the availability of the various media.
Linear programming models

Linear programming models are mathematical representations of LP problems. LP models have certain
characteristics in common. The characteristics can be grouped into two categories:
Components and assumptions: The components relate to the structure of a model, whereas the
assumptions reveal the conditions under which the model is valid.
Components of LP models: The components are the building blocks of an LP model. The following are
the components of LP models:
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a) Objective: The objective in problem solving is the criterion by which all decisions are evaluated. As
such, it provides the focus for problem solving. In linear programming models, a single, quantifiable
objective must be specified by the decision maker. For example, the objective might relate to profits, or
costs or market share, but to only one of these. Moreover, because we are dealing with optimization, the
objective will be either maximization or minimization. Hence, every LP problem will be either a
maximization problem or a minimization problem. Maximization problems often involve profits,
revenue, market share, or return on investment, where as minimization problems often involve costs,
time, or risk. Once the objective is specified, it becomes the measure of effectiveness against which
alternate solutions are judged. The optimal solution will be the one that yields the highest profit or the
one that provides the lowest cost. If a problem concerns maximizing profits, the profit per unit for each
variable either may be given or it may have to be derived by finding the difference between revenue per
unit and cost per unit for each decision variable. For instance, if the revenue is Birr 50 per unit for a
product and the cost is Birr 30 per unit, the profit per unit is Birr 20 (Birr 50-Birr 30). Note that the
units of all the coefficients in the objective function must be the same (e.g., all in Birrs, all in hours,
etc). In addition, all terms in the objective function must include a variable, and all decision variables
must be represented in the objective function.
b) Decision variables: The decision maker can control the value of the objective, which is achieved
through the choices in the level of decision variables. For example, a problem may involve maximizing
the profit obtained from the sale of three products, each with its own unit profit. If we assume that all
the output can be sold, the problem facing the decision maker is: How much of each product should be
produced to obtain the maximum profit? In terms of the LP model, the decision variables represent
unknown quantities to be solved for. The decision variables will be defined in terms of a quantity of an
amount which is usually represented by a variable and a subscript number like x 1, x2, x3... If for example,
our case is to maximize profit through the production and sell of two products, x 1 represents the quantity
of the first product and x2 the quantity of the second product. The decision variables are the variables
whose specification describes the solution for the problem. It typically comprises the set of decisions to
be made.
c) Constraints: The ability of a decision maker to select values of the decision variables in an LP
problem is subject to certain restrictions or limitations. These restrictions are called constraints. These
can come from a variety of sources. The restriction may reflect availabilities of resources (e.g., raw
materials, labor time, machine time, work space, storage space, etc), legal or contractual requirements
(e.g., product standards, work standards, and so on ), technological requirements (e.g., tensile strength),
or they may reflect other limits based on forecasts, customer orders, company policies, and so on.
Constraints may be in terms of an upper limit, lower limit or equality. So when you formulate the
model, carefully differentiate which one is called for. If for example one constraint says “at least 100
unit of the first product must be made”, the constraint will be expressed as: x1  100. If it were “to a
maximum of 50 units of the first product must be made”, the constraint will be: x 1 50. And had the
constraint been explained like “exactly 60 units of the first product is needed”, the constraint could have
been x1 = 60. The usual way to write constraints is by putting the variables in the left side of the
expression while a constant alone is placed on the right side. This constant is usually referred to as the
right hand side value (RHSV). Generally speaking, a constraint has four elements:
a) A right hand side (RHS) quantity that specifies the limit for that constraint. It must be a constant,
not a variable.
b) An algebraic sign that indicates whether the limit is an upper bound ( ) that cannot be exceeded,
a lower bound ( ) that is the lowest acceptable amount, or an equality (=) that must be met
exactly.
c) The decision variables to which the constraint applies.
d) The impact that one unit of each decision variable will have on the right hand side quantity of the
constraint.
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d) Parameters: The objective and the constraints in the model are mathematical expressions. Each
expression consists of numerical values and symbols. The symbols represent decision variables,
whereas the numerical values are called parameters. The parameters are fixed values that specify the
impact that one unit of each decision variables will have on the objective function and on any constraint
it pertains to as well.
Assumptions of LP Model
The major assumptions of LP model are linearity, divisibility, certainty, and non-negativity.
i Linearity: The linearity requirement is that each decision variable has a linear impact on the objective
function and in each constraint in which it appears. This means, for example, that if the profit of x 1 is
Birr 4 per unit, that same figure holds regardless of the quantity of x 1: It is true over the entire range of
possible values of x1. The same applies to each of the constraints: It is required that the same coefficient
(for example, 2 kg per unit) applies over the entire range of possible values of the decision variable. In
terms of a mathematical model, a function is linear when the variables included are all to the power 1
(that is not squared, cubed, square root, etc) and no products (e.g., x 1x2) appear. Thus, 2x1 + 3x2=40 is
linear constraint: where as the following constraints are not linear.
2x12 + 3x2=40 (x1 is squared)
2x1 + x1x2=40 (the second term is a product of variables)
ii Divisibility: The divisibility requirement pertains to potential values of decision variables. It is assumed
that non-integer values for decision variables are acceptable. For example, an optimal solution to a
decision variable can take values like 3.5, 1.2, 2.5, etc… regardless of the nature of the item represented
by the decision variable. For instance, suppose 3.5 is determined to be the optimal number of television
sets to be produced per hour. This would results in 7 sets every two hours, which would be acceptable.
iii Certainty: The certainty requirement involves two aspects in LP models. One aspect relates to the
model parameters (i.e., the numerical values). It is assumed that these values are known and constant.
For instance, in the above example 2x1 appears in a labor constraint with hours as the unit of measure, it
is assumed that each unit of x1 will require exactly 2 hours of labor. If the right side of the constraint is
300 labor hours, it is assumed that this value will remain 300. The second aspect is that all relevant
constraints are identified and included in the model.
iv Non-negativity: The non-negativity requirement is that negative values of variables are unrealistic and,
therefore, will not be considered in any potential solutions; only positive values and zero will be
allowed. The non-negativity constraints typically apply in an LP model, whether they are explicitly
stated or not.
Formulating LP Models
Once a problem has been defined, the attention of the analyst shifts to formulating a model. Just as it is
important to carefully define a problem, it is important to carefully formulate the model that will
be used to solve the problem.
Formulation of LP models is the process of translating the verbal statement of a problem into a
mathematical statement. Formulation of the model is the same as assembling the components of the LP
models that are mentioned above. Each component will be depicted as a mathematical expression in the
linear programming model. It is worth acquiring this knowledge since the model is an abstraction of the
real problem at hand. That is, if the model is ill-formulated, it will fail to represent the problem properly
and it can easily lead to poor decisions.
Formulating linear programming models involve the following steps:
✓ Identify the decision variables.
✓ Determine the objective functions.
✓ Identifying the constraints.
✓ Determining appropriate values for parameters and decide whether an upper limit, lower
limit, or equality is called for.
✓ Use this information to build a model.
✓ Validate the model.
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In many cases, the decision variables are obvious; in others, a brief discussion with the appropriate
manager is necessary. However, identifying the constraints and then determining appropriate value for the
parameters can require considerable time and effort on the part of the analyst. Potential sources of
information include historical records, interviews with managers and staff, and data collection.
Accounting records or personnel can generally provide information about the values of parameters. Once
the information about constraints and parameters has been obtained, constructing an appropriate model
becomes the focus. Validating the model will involve a critical review of the output, perhaps under a
variety of inputs, in order to decide if the results are reasonable
Example: The Microcomputer Problem
A firm that assembles computer and computer equipment is about to start production of two new
microcomputers. Each type of micro-computer will require assembly time, inspection time, and storage
space. The amount of each of these resources that can be devoted to the production of microcomputers
is limited. The manager of the firm would like to determine the quantity of each microcomputer to
produce in order to maximize the profit generated by sales of these microcomputers. In order to
develop a suitable model of the problem, the manager has met with design and manufacturing
personnel. As a result of those meetings, the manager has obtained the following information:
Type-1 Type-2
Profit per unit Birr 60 Birr 50
Assembly time per unit 4 hours 10 hours
Inspection time per unit 2 hours 1 hour
Storage space time per unit 3 cubic feet 3 cubic feet
The manager also has acquired information on the availability of company resources. These (daily)
amounts are:
Resource Amount available
Assembly time 100 hours
Inspection time 22 hours
Storage space 39 cubic feet
The manager also met with the firm’s marketing manager and learned that demand for the microcomputers
was such that whatever combination of these two types of microcomputer is produced, all of the output can
be sold.
Required: Formulate the linear programming model of this problem.
Solution:
a. First, we must define the decision variables. Based on the statement, “The manager…..would like to
determine the quantity of each microcomputer to produce” the decision variables are the quantities of
each type of computer. Thus,
x = quantity of type 1 to produce
y = quantity of type 2 to produce
b. Next, we formulate the objective function. The profit per unit of type1 is Birr 60 and the profit
per unit of type 2 is Birr 50, so the appropriate objective function is:
Maximize Z = 60x + 50y, where Z is the value of the objective function given the values of the
decision variables (x and y).
c. As for the constraints, there are three resources with limited availability: assembly time, inspection
time, and storage space. The fact that availability is limited means that these constraints will all be 
constraints. The type 1 microcomputer requires 4 hours of assembly time per unit, where as the type 2
microcomputer requires 10 hours of assembly time per unit. Therefore, with a limit of 100 hours
available, the assembly constraint is 4x + 10y  100 hours
Similarly, each unit of type 1 requires 2 hours of inspection time, and each unit of type 2 requires 1
hour of inspection time. With 22 hours available, the inspection constraint is:
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2x + y  22 hours
The storage constraint is determined in a similar manner. It is 3x + 3y  39
There are no other systems or individual constraints.
The non-negativity constraints are: x, y 0 (or x  0, y 0).
In summary, the mathematical model of the microcomputer problem is:
Maximize profit: 60x + 50y
Subject to:
4x + 10y  100 hours (Assembly time constraint)
2x + y  22 hours (Inspection time constraint)
3x + 3y  39 cubic feet( Storage space constraint)
x, y 0
Linear Programming Applications
Product mix
Organizations often produce similar products or offer similar services that use the same resources (for
example, labor, time, materials, etc). Because of the limits in the amounts of these resources that are
available during any time period, a decision must be made concerning how much of each product or
service to produce or make available during the time period that will be consistent with the goal of the
organization. The basic question that can be answered using linear programming is: What mix of output
(or service) will maximize profit (revenue, etc) given the availability of scarce resources? The
microcomputer problem described above is an example of a product mix problem.
Additional Example:
A toy manufacturer makes three versions of a toy robot. The first version requires 10 minutes each for
fabrication and packaging and 2 pounds of plastic, the second version requires 12 minutes each for
fabrication and packaging and 3 pounds of plastic, and the third version requires 15 minutes each for
fabrication and packaging and 4 pounds of plastic. There are 8 hours of fabrication and packaging
time available and 200 pounds of plastic available for the next production cycle. The unit profits are
Birr 1, Birr 5, and Birr 6 for version 1, 2 and 3 respectively. A minimum of 10 units of each must be
produced to fill backorders. Formulate an LP model that will determine the optimal production
quantities for profit maximization.
Solution
a) Identify the decision variables:
x1=number of version 1 robots
b) Identify the constraints by name:
Fabrication and packaging time
Quantity of plastic
Quantity of version 1 robots (minimum of 10)
c) Write the objective function:
Maximize profit: x1 + 5x2 + 6x3
d) Write the constraints:
Note that fabrication and packaging times are given in minutes per unit, but that available time is given
in hours. It is necessary that both sides of a constraint have the same units. This can be accomplished by
converting 8 hours to 480 minutes.
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The constraints are:

Fabrication and packaging 10x1 + 12x2 + 15x3480
minutes Plastic 2x1 + 3x2 + 4x3 200 pounds
Version 1 x1 10 robots
x1, x2, x3  0
In sum , the model is:
Maximize profit: x1 + 5x2 + 6x3
Subject to:
Fabrication and packaging 10x1 + 12x2 + 15x3480 minutes
Plastic 2x1 + 3x2 + 4x3 200 pounds
x1, x2, x3  0
Diet Problems
This type of problem usually involves the mixing of raw materials or other ingredients to obtain an end
product that has certain characteristics. For instance, food processors and dieticians generally are
concerned with meeting dietary needs in food products. There may be specific requirements pertaining to
nutrients, calories, sodium contents, and so on. The general question to be answered by linear
programming is: What mix of inputs (e.g., different food types) will achieve the desired results for the
least cost? The following example illustrates this type of problem.
Example: A cereal manufacturer is investigating the possibility of introducing a new cereal. It would
be composed of wheat, rice, and cornflakes. The cost per kg and dietary requirements are shown in the
following table:
Nutrients Wheat Rice Cornflakes Amount
Protein 6 4 4 at least 280 grams
Carbohydrate 25 35 25 at least 260 grams
Calories 80 120 130 not more than 1500 calories
Cost per gram Birr 0.05 Birr 0.08 Birr 0.04
Formulate an LP model for this problem that will determine the optimal quantities of wheat, rice, and corn
flakes that will achieve the requirements at minimum cost.
x1=the amount of wheat
x2 =the amount of rice
x3 =the amount of cornflakes
Minimize cost: 0.05x1 + 0.08x2 + 0.04x3
Subject to:
Protien 6x1 + 4x2 + 4x3  280 grams
Carbohydrate 25x1 + 35x2 + 25x3  260 grams
Calories 80x1 + 120x2 + 130x3 1500
calories
x1, x2, x3  0
Portfolio Selection
These problems generally involve allocating a fixed Birr amount (e.g., Birr 100,000) among a variety of
investments, such as bonds, stocks, and so on. Usually the goal is to maximize income or total return.
Example: A conservative investor has Birr 100,000 to invest. The investor has decided to use three
vehicles for generating income: municipal bonds, a certificate of deposit (CD), and a money market
account. After reading a financial newsletter, the investor also has identified several additional
restrictions on the investments:
a) No more than 40 percent of the investment should be in bonds.
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b) The proportion allocated to the money market account should be at least double the amount in the
CD.
The annual return will be 8 percent for bonds, 9 percent for the CD, and 7 percent for the money
market account. Assume the entire amount will be invested.
Formulate the LP model for this problem assuming that the investor wants to maximize the total
annual return.
Solution
Let
x1=amount invested in bonds
x2 = amount invested in the CD
x3 = amount invested in the money market account
The annual return from each investment is the product of the rate of return and the amount of investment:
Maximize: 0.08x1 + 0.09x2 + 0.07x3
The requirement that no more than 40 percent be in bonds produces this constraint:
x1 0.4(100,000), whcih becomes x1 40,000.
The requirement that the proportion invested in the money market account be at least double the amount
in the CD leads to: x3 2x2. Then, subtracting 2x2 from both sides gives us:
x3-2x20, which can be rearranged so that x2 comes before x3: -2x2 +x3 0
Finally, the investor must recognize that the amounts invested in bonds, the CD, and the money market
account must equal Birr 100,000. This gives us:
x1 + x2 + x3=Birr 100,000
In sum , the model is:
Maximize: 0.08x1 + 0.09x2 + 0.07x3
Subject to
Amount in bonds x1  40,000
Money/CD -2x2 +x3 0
Total amount x1 + x2 + x3 =Birr 100,000
x1, x2, x3  0
Take about 25
minutes.
A small firm produces a variety of chemical products. In a particular production process, three raw
materials are blended (mixed together) to produce two products: a fuel additive and a solvent base.
Each tonu of fuel additive is a mixture of 0.4 ton of material 1 and 0.6 of material 3. A ton of solvent
base is a mixture of 0.5 ton of material 1, and 0.2 of material 2, and 0.3 ton of material 3. The firm’s
production is constrained by a limited availability of the three raw materials. For the current
production period, the firm has the following quantities of each raw material:
Raw Material Amount available for
production Materials 1 20 tons
Materials 2 5 tons
Materials 3 21 tons
The accounting department has determined, after deducting all relevant costs, that the firm
will make a Birr 40 for every ton of the fuel additive and Birr 30 for every ton of solvent base
products. Assuming that the firm is interested in maximizing the total profit contribution,
formulate the linear programming model.
2.2 Graphic Solutions to LP Problems
In the previous section, the basic concepts of LP such as definitions of LP, component and assumptions of
LP models, formulation of LP model and application of LP have been discussed. Accordingly, a well
formulated LP model should be solved and the graphic method is one of the solution procedures used in
solving linear programming models. Thus, this section presents the graphic solution approach to LP
problems.
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Introduction
Graphical linear programming is a relatively straight forward method for determining the optimal
solution to certain LP problems. In practical terms, this method can be used only to solve problems that
involve two decision variables. This is because we cannot plot an equation having more than two
variables in a two dimensional coordinate plane.
In order to demonstrate the graphical method, the microcomputer problem that was formulated in the
preceding section will be solved. The model is
Maximize profit: 60x + 50y
Subject to:
Assembly time constraint: 4x + 10y  100 hours
Inspection timer constraint: 2x +  22 hours
y
Storage space constraint: 3x + 3y  39 cubic feet
x, y  0
Graphing the model
Graphing an LP model consists of the following distinct steps.
1) Plot each of the constraints
2) Determine the region or area that contains all of the points that satisfy the entire set
of constraints.
3) Determine the optimal solution.
Fundamental theorem of LP: If there is a solution to an LP problem, at least one optimal solution will
occur at a corner point of the feasible solution space. A corner point is a point in the feasible solution
region that is the intersection of two boundary lines (constraints
The extreme point theorem states that for problems that have optimal solutions, a solution will occur at an
extreme or corner point. As a result, in searching for an optimal solution to a problem, we need only
consider the extreme points because one of those must be optimal. Further, we could identify the optimal
solution by selecting the corner point that has the best value (i.e., maximum or minimum, depending on
which type of problem was involved) of the objective function. Hence the necessary steps for the extreme
point approach are:
a) Graph the problems.
b) Determine the values of the decision variables at each corner point. Sometimes this can be done
by inspection (observation); usually it is done using simultaneous equations.
c) Substitute the values of the decision variables at each corner point into the objective function
to obtain its value at each corner point.
d) After all corner points have been so evaluated, select the one with the highest value of the objective
function (for a maximum problem) or lowest value (for a minimum problem) as the optimal solution.
Slack
After we have got the optimal solution, we have to substitute the value of the decision variables into the
constraints and check whether all the resources available are used or not. The amount of unused resources
is known as slack. If there is an unused resource we can use it for any other purpose. Slack can range
from zero, for a case in which all of a particular resource is used, to the original amount of the resource
that was available (i.e., none of it is used). Slack can potentially exist in a  constraint.
Let us see what slack remains for the optimal solution to the above LP problem. The amount of unused
resource can be computed by substituting the values of the decision variables into each constraint and
Take about 30 minutes.

Given the LP model:
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Maximize profit: 10x1 + 16x2

Subject to:
10x1 + 20x2 110
25x1 + 20x2  200
x1, x2  0
a) Find the optimal solution using the graphic method.
b) Determine the amount of slack for each constraint.
c) What is the value of the objective function at the optimal solution?
d) What is the maximum available resource for each constraint?
Minimization example
Graphical solutions to minimization problems are very similar to solutions to maximization problems.
The main differences are that the constraints, in general, are the greater than or equal to variety, which
causes the feasible solution space to be restricted to an area away from the origin, instead of close to the
origin, as in a maximization problem, and the optimum being the point with the smallest possible value of
the objective function, instead of the largest.
Example: Solve the following LP model using the graphic approach:
Minimize cost:
0.1x+0.07y Subject to:
6x + 2y  18
8x+ 10y 40
y 1
x, y  0
Solution
Graph the constraints using this procedure:
1. Treat each constraint as equality.
2. For the first two constraints, find the x intercept by setting y equal to 0 and solving for x,
and then find y intercept by setting x equal to 0 and solving for y. For the third constraint,
y is simply equal to 1.
3. Plot these constraints and shade in the feasible solution space as shown below:
Dear learner, find the x and y intercepts by your own.
Solution
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4/22/24, 8:14 Chapter TWO OR final - teacher note
The extreme points can be determined either by inspection or from simultaneous equations. The results
are summarized as follows:
Extreme points How Value of the objective function
x Y determined?
0 9 Observation 0.10(0)+0.07(9)=Birr 0.63
5 0 Observation 0.10(5)+0.07(0)=Birr 0.50
2.27 2.19 Simultaneous 0.10(2.27)+0.07(2.19)=Birr 0.38 (smallest)
equation
The minimum value of the objective function is Birr 0.38, which occurs when x=2.27 and y=2.19.
Surplus
If there is a difference between the minimum required amount and the amount used, the difference is
called surplus. Surplus is the amount by which the optimal solution causes a  constraint to exceed the
required minimum amount. It can be determined in the same way that slack is calculated. Substitute the
optimal values of the decision variables into the constraint and solve. The difference between the resulting
value and the original right-hand side amount is the amount of surplus. Surplus can potentially occur in a
 constraint.
Computing the amount of surplus

Constraints Minimum requirement Amount used Surplus=Amount used minus
x = 2.27, y = minimum requirement
2.19
1 18 6*2.27+2*2.19=18 18-18=0
2 40 8*2.27+10*2.19=40 40-40=0
3 1 2.19 2.19-1=1.19
Interpretation: The value of x is 2.27 units and the value of y is 2.19 units to attain a minimum cost of
Birr 0.38. There is a surplus of 1.19 units in the third constraint.
Some Special Issues on graphic method

The following are some of the special issues that may arise in solving of linear programming problems:
a) Problems with no feasible solutions
b) Unbounded problems
c) Redundant constraints
d) Problems with multiple optimum solutions
a) No feasible solutions: It is possible to formulate a problem for which it is impossible to satisfy the set of
constraints. This situation sometimes occurs in problems that have a mix of greater than constraints and
less than constraints, where in order to satisfy one of the constraints, another constraint must be violated.
Infeasibility means that a feasible region doesn’t exist; that is, no points satisfy all the constraints. It
looks like the following graph.
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4/22/24, 8:14 AM Chapter TWO OR final - teacher note

UNIT TWO: THE LINEAR PROGRAMMING (LP)

Linear programming models

The constraints are:

Take about 30 minutes.

Maximize profit: 10x1 + 16x2

Computing the amount of surplus

Some Special Issues on graphic method

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