QA CHAPTER II.pptx

CHAPTER TWO
LINEAR PROGRAMMING
MODELS(LPM)
1

Chapter objectives
• After studying this lesson, you should be able to:
– Define the basic concepts in LP;
– Formulate Linear Programming Problem
– Identify the characteristics of linear programming problem
– Make a graphical analysis of the linear programming problem
– Solve the problem graphically
– Understand the basics of simplex method
– Explain the simplex calculations
– Describe various solutions of Simplex Method
– Understand Big-M method
– Realize the special cases of LP model;
– Perform sensitivity analysis of an LP model
– Understand the Dual Linear programming Problem
– Formulate and solve the Dual Problem
2

2.1. INTRODUCTION
• In 1947, George Danzig developed the use of
algebra for determining solutions to problems that
involved the optimal allocation of scarce
resources.
• In spite of numerous potential applications in
business, response to this new technique was low
due to substantial computational burden, which is
now removed with subsequent advances in
computer technology and related software during
the last three decades.
3

Cont’d
• The term linear implies that all the mathematical
relations used in the problem are linear or straight-line
relations.
• The term programming refers to the method of
determining a particular program or plan of action, i.e.,
the use of algorithms that is a well defined sequence of
steps that will lead to an optimal solution.
• Taken as a whole, the term linear programming refers
to a family of mathematical techniques for
determining the optimum allocation of resources and
obtaining a particular objective when there are
alternative uses of the limited or constrained
resources.
4

2.2. CHARACTERISTICS OF LPM
• Linear programming models are mathematical
representations of LP problems.
• Linear programming models have certain characteristics in
common.
• Knowledge of these characteristics enables us to recognize
problems that are amenable to a solution using LP models,
and to be able to correctly formulate an LP model.
• These characteristics can be grouped as components and
assumptions.
• The components relate to the structure of a model, where as
the assumptions reveal the conditions under which the
model is valid.
5

2.1.1. COMPONENTS OF LP MODELS
• There are four major components of LP models
including: Objective function, decision variables,
constraints and parameters.
1. Objective and Objective Function-The objective in
problem solving is the criterion by which all decisions
are evaluated.
• Because we are dealing with optimization, the objective
will be either maximization or minimization.
• Hence, every LP problem will be either maximization or
a minimization problem.
• An LP model consists of a mathematical statement of
the objective called the objective function.
6

Cont’d
2. Decision variables- They represent unknown quantities to be
solved for.. For example, how much of each product should
be produced in order to obtain the greatest profit?
3. Constraints- However, the ability of a decision maker to
select values of the decision variables in an LP problem is
subject to certain restrictions or limits coming from a variety
of sources.
– The restrictions may reflect availabilities of resources
(e.g., raw materials, labor time, etc.), legal or contractual
requirements (e.g., product standards, work standards,
etc.), technological requirements (e.g., necessary
compressive strength or tensile strength) or they may
reflect other limits based on forecasts, customer orders,
company policies, and so on.
7

Cont’d
In LP model, the restrictions are referred to as constraints.
Only solutions that satisfy all constraints in a model are
acceptable and are referred to as feasible solutions.
The optimal solution will be the one that provides the best
value for the objective function.
• Constraints can be arranged into three groups:
– System constraints – involve more than one decision
variable,
– Individual constraints – involve only one variable,
and
– Non-negativity constraints – specify that no variable
will be allowed to take on a negative value. The non-
negativity constraints typically apply in an LP model,
whether they are explicitly stated or not.
8

Cont’d
4. Parameters- The objective function and the
constraints consist of symbols that represent the
decision variables (e.g., X1, X2, etc.) and
numerical values called parameters.
• The parameters are fixed values that specify the
impact that one unit of each decision variable will
have on the objective and on any constraint it
pertains to as well as the numerical value of each
constraint.
• The following simple example illustrates the
components of LP models:
9

2.1.2 ASSUMPTIONS OF LP MODELS
• A linear programming model is valid under the following
assumptions:
• 1. Linearity (proportionality)
• The linearity requirement is that each decision variable has a linear
impact on the objective function and in each constraint in which it
appears.
• In terms of a mathematical model, a function or equation is linear
when the variables included are all to the power 1 (not squared,
cubed, square root, etc.) and no products (e.g., x1x2) appear.
• On the other hand, the amount of each resource used (supplied) and
its contribution to the profit (or cost) in the objective function must
be proportional to the value of each decision variable.
• For example, if production of one unit requires 5 hours of a
particular resource, then making 3 units of that product requires 15
hours (3x5) of that resource.
11

2. Divisibility (Continuity)
• The divisibility requirement pertains to potential values of
decision variables.
• It is assumed that non-integer values are acceptable.
However, if the problem concerns, for example, the optimal
number of houses to construct, 3.5 do not appear to be
acceptable.
• Instead, that type of problem would seem to require strictly
integer solutions. In such cases, integer-programming
methods should be used. It should be noted, however, that
some obvious integer type situations could be handled under
the assumption of divisibility.
• For instance, suppose 3.5 to be the optimal number of
television sets to produce per hour, which is unacceptable,
but it would result in 7 sets per two hours, which would
then be acceptable.
12

3. Certainty
• -This requirement involves two aspects of LP models.
• One aspect relates to the model parameters, i.e., the
numerical values. It is assumed that these values are
known and constant.
• In practice, production times and other parameters may
not be truly constant. Therefore, the model builder must
make an assessment as to the degree to which the
certainty requirement is met.
• Large departures almost surely will have a significant
effect on the model.
• The other aspect is the assumption that all relevant
constraints have been identified and represented in the
model.
13

5. Non-negativity
• -It assumes 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
and the non-negativity assumption is inherent
in LP models.
• For example X=-12, can’t be a solution as it
doesn’t give any meaning as negative number
of units can’t be produced
14

2.3. SOLVING LP MODELS
• Linear programming model can be solved by
either of:
• Graphical method or
• Simplex or algebraic method
15

2.3.1.GRAPHICAL METHODS
• Graphical linear programming is a relatively straightforward for
determining the optimal solution to certain linear programming
problems involving only two decision variables.
• Although graphic method is limited as a solution approach, it is very
useful in the presentation of LP, in that it gives a “picture” of how a
solution is derived thus a better understanding of the solution.
• Moreover, graphical methods provide a visual portrayal of many
important concepts.
• In this method, the two decision variables are considered as ordered
pairs (X1, X2), which represent a point in a plane, i.e, X1 is
represented on X-axis and X2 on Y-axis.
• Graphical method has the following advantages:
– It is simple
– It is easy to understand, and
– It saves time.
16

STEPS IN GRAPHICAL SOLUTION METHODS
Step 1. Formulate the LPM of the problem
Steps in formulating LP models:
a)Identify the decision variables.
b)Determine the objective function.
c)Identify the constraints.
d)Build and validate the model.
Step 2. Plot each of the constraints on the graph
Step 3. Identify the feasible region -The feasible region is the
solution space that satisfies all the constraints simultaneously.
Step 4. Locate the optimal solution- Graphic approach to find
a solution for LP model consisting of only two decision
variables:
a) The extreme point enumeration method
b) The objective (Iso-profit or cost) function line approach
17

Formulating the LPM of the Problem
• Example1
• ABC Company is to start a production of two spare parts. The
production manager would like to produce so many parts to
earn per unit profit of Birr 60 from the sale of type I and Birr 50
from the sale of type II parts. However, the total available time
to assemble the items, to inspect them and the space for their
storage is limited to 100 hours, 22 hours and 39 square feet
respectively. Four hours of assembly time, two hours of
inspection time and three square feet of storage space is
required to produce each unit of type I spare part. On the other
hand 10 hours of assembly time, 1 hour of inspection time and
3 square feet of storage space is needed to produce each unit of
type II spare parts. How many of each item should the
Company produce and sale to maximize its profit?
• Required: Formulate the LPM of the Problem
18

Example 2
• An individual investor has Birr 70,000 to divide among several investments.
The alternative investments are municipal bonds with an 8.5% return,
certificates of deposits with a 10% return, Treasury bill with a 6.5% return, and
income bonds with a 13% return. The amount of time until maturity is the same
for each alternative. However, each investment alternative has a different
perceived risk to the investor; thus it is advisable to diversify. The investor
wants to know how much to invest in each alternative in order to maximize the
return. The following guidelines have been established for diversifying the
investment and lessening the risk perceived by the investor.
– No more than 20% of the total investment should be in an income bonds.
– The amount invested in certificates of deposit should not exceed the amount
invested in other three alternatives.
– At least 30% of the investment should be in treasury bills and certificates of
deposits.
– The ratio of the amount invested in municipal bonds to the amount invested in
treasury bills should not exceed one to three.
– The investor wants to invest the entire Birr 70,000.
• Required: Formulate a LP model for the problem.
19

Example 3:
• A chemical corporation produces a chemical mixture for the
customer in 1000- pound batches. The mixture contains three
ingredients- Zinc, mercury and potassium. The mixture must
conform to formula specifications (i.e., a recipe) supplied by a
customer. The company wants to know the amount of each
ingredient to put in the mixture that will meet all the requirements of
the mix and minimize total cost.
• The customer has supplied the following formula specifications for
each batch of mixture.
1. The mixture must contain at least 200 lb of mercury
2. The mixture must contain at least 300 lb of zinc
3. The mixture must contain at least 100 lb of potassium
• The cost per pound of mixture is Birr4, of zinc Birr 8 and of
potassium Birr 9.
• Required: Formulate LPM for the problem
20

1.MAXIMIZATION CASE
• Example 1 of the previous example 1
• After formulating the LPM of the Problem,
the next steps will be:
• Plotting each constraints on the graph, then
• Identifying the feasible solution and finally
• Locating optimal solution
21

Figure 2-1 Feasible Region Based on a Plot of the First Constraint
(assembly time) and the Non-negativity Constraint
23

Figure 2–2 A Completed Graph of the Server Problem
Showing the Assembly and Inspection Constraints and the
Feasible Solution Space
24

Figure 2–3 Completed Graph of the Server Problem
Showing All of the Constraints and the Feasible Solution Space
25

Finding the Optimal Solution
• The extreme point approach
– Involves finding the coordinates of each
corner point that borders the feasible
solution space and then determining
which corner point provides the best value
of the objective function.
– The extreme point theorem
– If a problem has an optimal solution at
least one optimal solution will occur at a
corner point of the feasible solution
space.
26

The Extreme Point Approach
1. Graph the problem and identify the feasible
solution space.
2. Determine the values of the decision variables
at each corner point of the feasible solution
space.
3. Substitute the values of the decision variables at
each corner point into the objective function
to obtain its value at each corner point.
4. After all corner points have been evaluated in a
similar fashion, select the one with the highest
value of the objective function (for a
maximization problem) or lowest value (for a
minimization problem) as the optimal solution.
27

Figure 2–5 Graph of Server Problem with Extreme
Points of the Feasible Solution Space Indicated
28

Table 2–3 Extreme Point Solutions for the Server
Problem
29

SLACK VERSUS SURPLUS
• Slack is the amount of a scarce resource that is unused by a given
solution. Slack can potentially exist in a < constraint.
– Slack variables are considered in the objective function by using a
coefficient of zero for each of them. When all the constraints are
written as equalities after adding a slack variable to each of them, the
linear program is said to be in standard form.
• Surplus on the other hand 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 can, i.e.,
substitute the optimal values of the decision variables into the left
side of the constraint and solve.
– The difference between the resulting value and the original right hand
side amount is the amount of surplus.
– Surplus should also be accounted for in the objective function by using
coefficients of zero likewise.
30

Table 3–5 Computing the Amount of Slack for the
Optimal Solution to the Server Problem
31

2.MINIMIZATION CASE
• Minimize Z with inequalities of constraints in > form
• Example:
• Suppose that a machine shop has two different types of
machines; machine 1 and machine 2, which can be used to
make a single product .These machines vary in the amount
of product produced per hr., in the amount of labor used
and in the cost of operation.
• Assume that at least a certain amount of product must be
produced and that we would like to utilize at least the
regular labor force. How much should we utilize each
machine in order to utilize total costs and still meets the
requirement?
32

Figure 2–8 A Comparison of Maximization and
Minimization Problems
36

3. MIXED CONSTRAINT CASE
• It is only in some cases that the maximization problems consist of constraints
connected to the RHS value with only <. By the same token, constraints of
minimization problems are not always connected to their RHS with>. This is to
mean that both maximization and minimization problems may consist of constraints
connected to RHS with a mix of algebraic signs ( <, >, =).
•
Example
Minimize Z=1500x+2400y
Subjected to:
4x+Y>24 …I
2x+3y>42 …II
X+4y>36 …III
X<14 …IV
y<14 …V
x, y>0
37

Exercise
1. Formulate the LP
ABC Furniture manufacturer produces two products: Beds and
Chairs. Each unit of Bed requires 3 hrs in molding unit, 4hrs in
painting unit, and 1 hr in finishing. On the other hand, each unit of
Chair requires 3 hrs in molding unit, 2 hrs in the paint shop and 2
hours in finishing. Each week, there are 210 hrs available in molding,
200hrs in painting, and 120 hrs in finishing unit. The demand for
Beds cannot exceed 40 units per week. Each unit of Bed contributes
Birr 20 to profit, while each unit of chair contributes Birr 30.
2. Maximize Z 3x + 4y
S.T.
5x + 4y ≤ 200
3x + 5y ≤ 150
5x + 4y ≤ 100
8x + 4y ≤ 80
X,y ≥ 0 40
3. Minimize Z: 10x + 4y
S.T.
3x + 6y ≥ 250
15x + y ≥ 150
2x + 9y ≥ 200
X,y ≥ 0

2.3.2.Some Special Issues
a) No Feasible/Infeasible Solutions
– Occurs in problems where to satisfy one of the
constraints, another constraint must be violated.
41
In the graph, there is no
common point in the shaded
area.
All constraints cannot be
satisfied simultaneously and
there is no feasible solution to
the problem.

b) Unbounded Problems
– Exists when the value of the objective
function can be increased without limit.
42
Note
Here that the two corners of the region are
A(0,3) and .B(2,1). The value of Max Z(A)=6
and Max Z(B)=8. But there exist number of
points in the shaded region for which the
value of the objective function is more than 8.
For example, the point (10, 12) lies in the
region and the function value at this point is
70 which is more than 8.
Remark:
An unbounded solution does not
mean that there is no solution to the
given LPP, but implies that there exits
an infinite number of solutions.

c)Redundant Constraints
– A constraint that does not form a unique boundary of the
feasible solution space; its removal would not alter the
feasible solution space.
– If a constraint when plotted on a graph doesn’t form part
of the boundary making the feasible region of the
problem that constraint is said to be redundant.
43
Note:
The packaging hour’s constraint does
not form part of the boundary making
the feasible region.
Thus, this constraint is of no
consequence and is therefore, redundant.
The inclusion or exclusion of a
redundant constraint does not affect the
optimal solution of the problem.

d) Multiple Optimal Solutions
–Problems in which different combinations
of values of the decision variables yield the
same optimal value.
44
 Both C and D are optimal
solutions. Any point on the line
segment CD will also lead to the
same optimal solution.
 Multiple optimal solutions
provide more choices for
management to reach their
objectives.

2.3.2.THE SIMPLEX APPROACH
• The graphical approach is useful in solving linear
programming models having only two variables.
• When the model has more than two variables, the
appropriate approach is the Simplex procedure.
• It begins with a feasible solution which is not
optimal, and the solution is improved through
continuous algebraic manipulations (iterations) until
the optimal solution is determined.
45

Cont’d
• The graphical method to solving LPPs provides fundamental
concepts for fully understanding the LP process. However,
the graphical method can handle problems involving only
two decision variables (say X1 and X2).
• Simplex Method which is an efficient approach to solve
applied problems containing numerous constraints and
involving many variables that cannot be solved by the
graphical method.
• The simplex method is an ITERATIVE or “step by step”
method or repetitive algebraic approach that moves
automatically from one basic feasible solution to another
basic feasible solution improving the situation each time until
the optimal solution is reached at.
46

STEPS IN SIMPLEX APPROACH
• Step1. Formulate the linear programming model of the
real world problem
• Step2. Express the mathematical model of L.P. problem
in the standard form
• Step 3. Set up the initial simplex tableau
• Step 4.Check if the current solution is optimum or not.
• Step 5. Further iterate towards an optimum solution.
• Step 6. Repeat Step3-5 until optimal solution is reached
47

A.MAXIMIZATION CASE
Let us take the example of a spare part production
problem
– Step 1 Formulate LPP Model
– Step 2 Standardize the problem
i.e. Convert constraint inequality into equality
form by introducing a variable called Slack variable.
• Slack Variables:
–A slack variable(s) is added to the left hand
side of a < constraint to covert the constraint
inequality in to equality.
48

Cont’d
– The value of the slack variable shows unused resource.
– Slack variables represent unused resource or idle capacity.
– Thus, they don’t produce any product and their contribution
to profit is zero.
– A slack variable emerges when the LPP is a maximization
problem.
– Slack variables are added to the objective function with zero
coefficients.
– Let that s1, s2, and s3 are unused labor, machine and
marketing hrs. respectively.
49

50
Example: Taking the microcomputer problem its
standard form is as follows:
Solve the problem using the simplex approach
Zmax = 60X1 + 50X2
4X1 + 10X2  100
2X1 + X2  22
3X1 + 3X2  39
X1, X2  0
The standard form is:

Standardized Form
Zmax = 60X1 + 50X2 + 0S1 + S2 + 0S3
4X1 + 10X2 + S1 = 100
2X1 + X2 + S2 = 22
3X1 + 3X2 + S3 = 39
X1, X2, S1, S2, S3  0
51

Cont’d
– To represent the data, the simplex method uses a
table called the simplex table or the simplex
matrix.
– In constructing the initial simplex tableau, the
search for of the optimal solution begins at the
origin. Indicating that nothing can be produced;
– Thus, first assumption, No production implies
that x1 =0 and x2=0
52

Cont’d
 4X1 + 10X2 + S1 +0 s2+ 0 s3= 100
4(0) +0 + s1 +0 s2+ 0 s3= 100
s1= 100 – Unused assembly time.
 2x1+1x2 +0 s1 + s2+ 0 s3= 22
2(0) +1(0) + 0s1 + s2+ 0 s3= 22
s2= 22 – Unused inspection time.
 3x1+3x2+ 0s1 +0s2+ s3= 39
3(0 )+3(0)+ 0s1 +0 s2+ s3= 39
s3= 39 – Unused storage space.
53
Therefore,
Max.Z=60x1 +50x2 + 0 s1 +0 s2+ 0 s3
=60(0) +50(0) + 0(100) +0(22) + 0(39)
= 0

Step 3. Construct the initial tableau
The initial tableau always represents the “Do Nothing”
strategy, so that the decision variables are initially non-
basic.
• List the variables across the top of the table and write the
objective function coefficient of each variable jut above
it.
• There should be one row in the body of the table for each
constraint.
• List the slack variables in the basis column, one per raw.
• In the Cj column, enter the objective function coefficient
of zero for each slack variable. (Cj - coefficient of
variable in the objective function)
• Compute values for row Zj
• Computer values for Cj – Zj.
54

Interpretation
• Cj = the coefficient of the variables in the
objective function in the standard form.
• Cb = the coefficient of basic variables in the obj.
fun.
• X1, X2, ..Xn = decision variables .
• S1, s2, Sn = basic variables (slack).
• a11, a12,… coefficient of DV in the constraint set.
• 1, 0, 1 are coefficient of basic variables in the
constraints set.
• RHSV = the right hand side value of the
constraints.
56 56

Step 3 cont’d
Basic
variable
Cj 60
X1
50
X2
0
S1
0
S2
0
S3
RHSV Ratio
S1 0 4 10 1 0 0 100 100/4 = 25
S2 0 2* 1 0 1 0 22 22/2 = 11
S3 0 3 3 0 0 1 39 39/3 = 13
Zj 0 0 0 0 0 0
Cj-
Zj
60 50 0 0 0 0
57

Evaluate, the initial Tableau, does it optimal or
not?
58
Based on the above explanation it is not optimal:
Current Solution is x1= X2 =0, and S1= 100, S2= 22
and S3 =39
Decision: S2 departing and X1 is entering variable

Step 4.
• Choose the “incoming” or “entering” variables
• Note:
The entering variable is the variable that has the
most positive value in the Cj - Zj row also called as
indicator row.
Or the entering variable is the variable that has the
highest contribution to profit per unit.
a) X1 in our case is the entering variable
b) The column associated with the entering
variable is called key or pivot column ( X1
column in our case )
59

Step 5
• Choose the “leaving “or “outgoing” variable
• In this step, we determine the variable that will leave the solution for X1 (or
entering variable)
Note:
– The row with the minimum or lowest positive (non-negative)
replacement ratio shows the variable to leave the solution.
Replacement Ratio (RR) = Solution Quantity (Q)
Corresponding values in pivot column
Note: RR>0
– The variable leaving the solution is called leaving variable or outgoing
variable.
– The row associated with the leaving variable is called key or pivot row
(s3 column in our case)
– The element that lies at the intersection of the pivot column and pivot
row is called pivot element(No 2 in our case)
60

Step 6: Develop subsequent tableaus
• Repeat step 3-5 till optimum basic feasible solution is
obtained.
Or: repeat step 3-5 till no positive value occurs in the Cj - Zj
row.
• Note:
o Divide each element of the pivot row by the pivot element
to find new values in the key or pivot row.
o Perform row operations to make all other entries for the
pivot column equal to zero.
Raw operation= old Raw Value – Key column Value (New raw
value
61

Cont’d
62
Basic
variable
Cj 60
X1
50
X2
0
S1
0
S2
0
S3
RHS
V
Ratio
S1 0 0 8 1 -2 0 56 56/8 = 7
X1 60 1 1/2 0 1/2 0 11 11/. 5 =
22
S3 0 0 3/2 0 -
3/2
1 6 6/1.5 = 4
Zj 60 30 0 30 0 660
Cj-Zj 0 20 0 -30 0 0

Optimal table
63
Basic
variable
Cj 60
X1
50
X2
0
S1
0
S2
0
S3
RHS
V
Ratio
S1 0 0 0 1 6 -16/3 24
X1 60 1 0 0 1 -1/3 9
X2 50 0 1 0 -1 2/3 4
Zj 60 50 0 10 40/3 740
Cj-Zj 0 0 0 -10 -40/3

Cont’d
• Since all the Cj - Zj < 0 optimal solution is
reached at
X1 = 9
X2 = 4
S1 = 24 hrs
Z = Birr 740
• “A simplex solution is a maximization problem is
optimal if the Cj – Zj row consists entirely of
zeros and negative numbers (i.e., there are no
positive values in the bottom row).”
64

B.MINIMIZATION PROBLEMS
• There are two methods to solve minimization LP problems:
1. Direct method/Big M-method/
–Using artificial variables
2. Conversion method
–Minimization by maximizing the dual
• Surplus Variable (-s):
 A variable inserted in a greater than or equal to constraint
to create equality. It represents the amount of resource
usage above the minimum required usage.
 Surplus variable is subtracted from a > constraint in the
process of converting the constraint to standard form.
 Neither the slack nor the surplus is negative value. They
must be positive or zero. 65

Cont’d
• Example:
• 2x1+x2 < 40 ==>is a constraint inequality
x1= 12 and x2= 11==> 2x1+x2+s = 40 ==>2(12)+11+s= 40
==> s=5 unused resource
• 5x1+3x2 < 45
x1= 12 and x2= 11==> 5x1+3x2+s= 45 ==>5(12)+3(11)+s = 45
==> s=0 unused resource (No idle resource)
• 5x1+2x2 >20
x1= 4.5 and x2= 2==> 5x1+2x2- s = 20 ==>5(4.5)+2(2)-s = 20
==> s=6 unused resource
• 2x1+x2 >40
x1= 0 and x2= 0(No production)==> 5x1+2x2- s = 20 ==>5(4.5)+2(2)-s= 20
==> s=-6 (This is mathematically unaccepted)
66

Artificial variable
• Thus, in order to avoid the mathematical contradiction,
we have to add artificial variable (A)
• Artificial variable is a variable that has no
meaning in a physical sense but acts as a tool to
create an initial feasible LP solution.
• Note:
Type of constraint To put into
standard form
< --------------------------------------------- Add a
slack variable
= ---------------------------------------------Add an artificial
variable
> ---------------------- Subtract a surplus variable and add
artificial variable 67

Big M-Method
• The Big-M Method is a method which is used in
removing artificial variables from the basis.
• In this method; we assign coefficients to artificial
variables, undesirable from the objective function
point of view.
• If objective function Z is to be minimized, then a
very large positive price (called penalty) is
assigned to each artificial variable.
• Similarly, if Z is to be maximized, then a very
large negative price (also called penalty) is
assigned to each of these variables.
68

Characteristics of Big-M Method:
A. High penalty cost (or profit) is assumed as M
B. M is assigned to artificial variable A in the objective function Z.
C. Big-M method can be applied to minimization as well as maximization
problems with the following distinctions:
– Minimization problems
-Assign +M as coefficient of artificial variable A in the objective function
Z
– Maximization problems:
-Here –M is assigned as coefficient of artificial variable A in the objective
function Z
D. Coefficient of S (slack/surplus) takes zero values in the objective function
Z
E. For minimization problem, the incoming variable corresponds to the
highest negative value of Cj-Zj.
F. Solution is optimal when there is no negative value of Cj-Zj.(For
minimization case)
69

Example:
• 1. Minimize Z=25x1 +30x2
Subject to:
20x1+15x2 > 100
2x1+ 3x2 > 15
x1, x2 > 0
70

Solution
Step 1: Standardize the problem
Minimize Z=25x1 +30x2 +0s1+0s2 +MA1+MA2
Subject to:
20x1+15x2- s1+A1 = 100
2x1+ 3x2 –s2+A2 = 15
x1, x2 , s1, s2 ,A1 ,A2 > 0
Step 2: Initial simplex tableau
The initial basic feasible solution is obtained by setting x1= x2=
s1= s2=0
No production, x1= x2= s1=0==>20(0) +15(0) - 0+A1 = 100 ==>
A1 = 100
x1= x2= s2=0==>0(0)+3(0) -
0+A2 =15==> A2 = 15
71

Cont’d
Initial simplex tableau
72
Note:
Once an artificial variable has left the basis, it has served its purpose and can therefore
be removed from the simplex tableau. An artificial variable is never considered for re-

Cont’d
2nd Simplex Tableau
73

Cont’d
3rd Simplex Tableau
74

Cont’d
Cj - Zj > 0==>Optimal solution is reached
X1=5/2
X2=10/3 and
MinZ=162.5
Note:
As long as an “A” variable is available in the solution
variable column, the solution is infeasible.
75

C.MAXIMIZATION WITH MIXED
CONSTRAINTS
• The simplex method requires that all the
constraints be in standard form.
• Constraints that are ≤ can be put in to standard
form by adding a slack variable in the constraint.
• Constraints with ≥or = sign are handled a bit
differently.
76

Cont’d
• To change equality constraints to standard form
add artificial variables
• To convert this inequality to standard form
subtract surplus variable first and add artificial
variable
77

Solve using simplex method
Maximize Z= 6X + 8Y
Subject to:
Y ≤4
X+ Y =9
6X+2Y ≥ 24
X, Y ≥ 0
78

In standard form
Max! 6X + 8Y + 0S1 + 0S3 – MA2 – MA3
Subject to: Y+ S1 = 4
X+Y + A2 = 9
6X+ 2Y –S3 + A3 = 24
All variables ≥ 0
79

Initial table
BV CBV X Y S1 S3 A2 A3
6 8 0 0 -M -M
Quantity
S1 0
A2 -M
A3 -M
0 1 1 0 0 0
1 1 0 0 1 0
6 2 0 -1 0 1
4
9
24
Z
C-Z
-7M -3M 0 M -M -M
6+ 7M 8+3M 0 -M 0 0
-33M
80

Second table
BV CBV X Y S1 S3 A2 A3
6 8 0 0 -M -M
Quantity
S1 0
A2 -M
X 6
0 1 1 0 0 0
0 2/3 0 1/6 1 -1/6
1 1/3 0 -1/6 0 1/6
4
5
4
Z
C-Z
6 2-2M/3 0 -1M/6 -1 -M 1+M/6
0 6+2M/3 0 1+ M/6 0 -1-m/6
24 - 5M
81

Third table
BV CBV X Y S1 S3 A2
6 8 0 0 -M
Quantity
Y 0
A2 -M
X 6
0 1 1 0 0
0 0 -2/3 1/6 1
1 0 -1/3 -1/6 0
4
7/3
8/3
Z
C-Z
6 8 6 + 2M/3 -M/6 -1 -M
0 0 -6-2M/3 1+ M/6 0
48 -7M/3
82

Fourth table
BV CBV X Y S1 S3
6 8 0 0
Quantity
Y 8
S3 0
X 6
0 1 1 0
0 0 -4 1
1 0 -1 0
4
14
5
Z
C-Z
6 8 2 0
0 0 -2 0
62
83

Special case in simplex method
a) Unbounded solution
– Zmax
84
Cj
6 9 0 0
Basic X1 X2 S1 S2 Q
X2 5 -1 1 2 0 30
S2 0 -2 0 -1 1 10
Zj -9 9 18 0 270
Cj - Zj
15 0 -18 0 -

b) Multiple optimal solution
• Zmax
85
Cj
2 4 0 0
Basic X1 X2 S1 S2 Q
X2 4 1/2 1 1/2 0 5/2
S2 0 1/2 0 -1/2 1 3/2
Zj 2 4 3/2 0 10
Cj - Zj
0 0 -3/2 0 -

c) Infeasible solution
86
Cj
5 8 0 0 M
Basic X1 X2 S1 S2
A2
Q
X1 5 1 0 1 3 0 200
X2 8 0 1 -1 -2 0 100
A2 M 0 0 0 -1 1 20
Zj 5 8 -3 -1-M M 1800+20M
Cj - Zj
0 0 3 1+M 0 -

d) Degeneracy
87
Cj 2 3 0 0 0
Basic X1 X2 S1 S2
S3
Q
0 S1 2 5 1 0 0 20
0 S2 1 1 0 1 0 5
0 S3 1 2 0 0 1 8
Zj 0 0 0 0 0 0
Cj - Zj
2 3 0 0 0 -

2.4. POST OPTIMALITY ANALYSIS
• Carried out after the optimal solution is
found
• Is begins with the final simplex tableau
• There are two scenarios in this case
–Duality and
–Sensitivity analysis
88

2.4.1. Sensitivity Analysis
• Sensitivity analysis carries the LP analysis beyond the
determination of the optimal solution and begins with
the final simplex tableau.
• Its purpose is to explore how changes in any of the
parameters of a problem, such as the coefficients of the
constraints, coefficients of the objective function, or the
right hand side values; would affect the solution.
• For this, instead of resolving the entire problem as a
new problem with new parameters, we may consider
the original optimal solution as an initial solution for
the purpose of knowing the ranges, both lower and
upper, within which a parameter may assume a value.
89

a) A change in the RHS of a constraints
• Simultaneous changes are not contemplated
• Change in RHS or Q of one constraint is considered at a time
• The range of feasibility is the range over which the RHS value of a
constraint can be changed and still have the same shadow prices.
• Starts with Considering shadow price
• Shadow price: is a marginal value; it indicates the impact that a
one unit change in the value of the constraint would have on the
value of the objective function.
• Shadow prices are the values in the Zj-row of slack columns
90

Determining Range of Feasibility: How?
Feasibility Ratio (FR) = quantity ÷ respective slack variable coefficient:
in the final optimal solution tableau
The determine the range of feasibility according to the following rule
The effect of a RHS change on the objective function depends on whether the change
leads to relaxation or tightening of the constraint
≤ ≥
Allowable RHS increase The negative FR closest to zero Smallest positive FR
Allowable RHS decrease Smallest positive FR The negative FR closest to
zero
91

Coun’t…
• Relaxation of a constraint has favorable impact on the objective function.
• However, tightening has unfavorable impact on the objective function
92

Example
The LPM of the micro computer problem is:
Max Z: 60X1 + 50X2
Subject: Assembly time: 4X1 + 10X2 ≤100
Inspection time: 2X1 + X2≤22
Storage space: 3X1 + 3X2 ≤39
X1 and X2 ≥ 0
If the amount of assembly time was increased by one hour, there would
be no effect on profit; if the inspection time was increased by one hour,
the profit would have increased by $10, and if storage space was increased
by one cubic foot, profit would increase by $40/3. The reverse holds true.
one –unit decrease in inspection time would decrease profit by $10 since
its shadow price is $ 10.
93

Consider the previous optimal solution and compute
1. Compute the range of feasibility for the constraints in the above
illustration.
2. The manager in the above problem is contemplating one of two
possible changes in the level of the storage constraint. One change
would be an increase of 3 cubic feet in its level and another change
will be an increase of 8 cubic feet. Determine the revised optimal
solution for each possible change.
94

Example
3. Suppose the manager in the above problem is
contemplating a decrease in the storage space
due to an emergency situation. There are two
possibilities being considered, a decrease of 6
cubic feet and a decrease of 9 cubic feet; what
will be the impact on the optimal solution?
95

Determine the FR and range of feasibility for each CONS.
1. Recall the original value of the resources
Original value constraints S1 S2 S3
100 S1 1 6 -16/3
22 S2 0 -1 -1/3
39 S3 0 -1 2/3
2. Ratio = Q/respective slack values
S1= 24/1= 24 S2= 24/6= 4 S3= 24/-16/3= -4.5
9/0=undefined 9/-1= -9 9/-1/3= -27
4/0= undefined 4/-1= -4 4/2/3= 6
NB: If there is no suitable negative/positive FR, then the allowable
increase/decrease is ∞ (undefined)
96

3. Find the range of feasibility
Constraints Original
value
Lower limit Upper limit Range of
feasibility
S1 100 100-24= 76 100+∞ 76-∞
S2 22 22-4= 18 22+4= 26 18-26
S3 39 39-6 = 33 39+4.5= 43.5 33-43.5
Therefore:
Constraint one (assembly line): 100-24 up to 100+∞= (76-∞)
Constraint two (inspection time): 22-4 up to 22-4= (18-26)
Constraint three (storage space): 39-6 up to 39+4.5= (33-43.5)
97

b) A change of coefficient of objective function
• Two cases
Case 1: Range of insignificance :
The range over which the non basic variables objective function
coefficient can change without making these variables entering in the
solution
Case 2. Range of optimality
The range over which the objective function coefficient of basic variables
can change without changing the optimal values i.e. without changing basic
and non basic variables but change the optimal function value.
98

Range of optimality
• Range of optimality is calculated for only original
decision variables because slack/surplus variables
and artificial variables do not have any contribution
for the objective function.
99

Cont’d
Determine the range of optimality for the following final simplex tableaue
Min Z = 10X1 + 3X2
Subject to:
2X1 + X2 ≥80
X1 + 4X2 ≥ 200
X1 and X2 ≥ 0
100

Range of Insignificance (RoI)
–The range over which a non-basic variable’s objective function
coefficient can change without causing that variable to enter in to
the solution mix is called its range of insignificance
–In a maximization problem if a variable is not currently in solution, its
objective function coefficient would have to increase by an amount
that exceeds the Cj –Zj value in order for it to enter as a basic
variable in the optimal solution. Any lesser value of its objective
function coefficient would keep it out of solution
–In a minimization problem if a variable is not in solution, its
objective function coefficient would have to decrease by an amount
that exceeds the Cj –Zj value for that variable in the final tableau in
order for it to enter as a basic variable in the optimal solution.
101

The final tableau of the above problem is:
Basic variable
10
X1
3
X2
0
S1
0
S2
r.h.s.
S2---0 0 0 -4 1 120
X2---3 2 1 -1 0 80
Zj 6 3 -3 0 160
Cj- Zj 4 0 3 0 240
102

Cont’d
• X1 is non-basic variable in this optimal table, so,
determine RoI for X1
Since the problem is minimization the objective
function coefficient of x1 must decrease by an
amount that exceeds 4 (i.e the Cj-Zj row
coefficient of x1 in the optimal table.)
• Therefore the range of insignificance for X1 is:
• Its current OF coefficient-4 up to its current of
coefficient + ∞
• 10 – 4 up to 10+ ∞: 6 - ∞
• Interpretation: The cost per unit of Product A
must be less than 6 before the company starts
manufacturing product A
103

2.4.2.Duality
• Every LPP has another LPP associated with it, which is called its dual.
• The given problem is called the primal.
• The dual of a dual is the primal
104
Primal Dual
Objective is minimization Objective is maximization and vice
versa
> type constraints < type constraints
No of columns No of rows
No of rows No of columns
No of decision variables No of constraints
No of constraints No of decision variables
Coefficient of Object function RHS value
RHS value Coefficient of Object function

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