CHAPTER TWO - OPERATIONS RESEARCH (2).pptx

Chapter Two
Linear Programming
By Asmamaw T. (PhD)
1

2.1 Concepts of Linear Programming
• Linear programming- is an optimization method which
shows how to allocate scarce resources in the best
possible way subject to more than one limiting condition
expressed in the form of inequalities and /or equations.
• It enables users to find optional solution to certain
problems in which the solution must satisfy a given set
of requirements or constraints.
• Optimization in linear programming implies
maximization (max) Profit, revenue, sales, market share
or minimization (min) Cost, time, distance, or a certain
objective function.
• The goal in linear programming is to find the best
solution given the constraints imposed by the problem,
hence the term constrained optimization.
2

Formulation of Linear Programming Model
•
• LP models are mathematical representation of LP
problems. Some models have a specialized format
where as others have a more generalized format.
• Despite this, LPMs 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 correctly formulate an
LP model.
• The characteristics can be grouped into two categories:
Components and assumptions. The components
relate to the structure of a model, where as the
assumptions describe the conditions under which the
model is valid.
3

Components of LP Model
1. The Objective function: is the mathematical/
quantitative expression of the objective of the
company/ model. The objective in problem solving
is the criterion by which all decisions are evaluated.
• In LPMs a single quantifiable objective must be
specified by the decision maker.
• For example, the objective might relate to profits,
or costs or market share, best to only one of these.
Moreover, because we are dealing with
optimization, the objective will be either
maximization or minimization, but not both at a
time.
5

2. The Decision Variables: represent unknown quantities
to be resolved for. These decision variables may represent
such things as the:
• Number of units of different products to be sold
• The number of dollars to invest in various projects
• The number of ads to place with different media
• Since the decision maker has freedom of choice among
actions, these decision variables are controllable variables.
6

3. The constraints: are restrictions which define or limit the
attainability (achievability) feasibility of a proposed course
of action. They limit the degree to which the objective can
be pursued.
• A typical restriction embodies scarce resources (such as
labor supply, RMs, production capacity, machine time,
storage space), legal or contractual requirements (leg.
Product standards, work standards), or they may reflect
other limits based on forecasts, customer orders, company
policies etc.
7

4. 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 to
the numerical value of each constraint.
• The components are the building blocks of an LP model.
We can better understand their meaning by examining a
simple LP model as follows.
8

• System constraints- involve more than one
decision variables.
• None-negativity constrains- 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.
11

Assumption of LP models
1. 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.
• Taking the above example, producing one more unit of
products add birr 4 to the total profit. This is true over
the entire range of possible values of x1. The same
applies (true) to each of the constraints.
2. Divisibility: The divisibility requirement pertains to
potential values of decision variables. It is assumed that
non-integer values are acceptable. For example: 3.5 TV
sets/ hr would be acceptable 7 TV sets/ 2hr.
12

3. Certainty: The parameters are known and constant. It is
assumed that the decision maker here is completely certain
(i.e., deterministic conditions) regarding all aspects of the
situation, i.e., availability of resources, profit contribution of the
products, technology, courses of action and their
consequences etc.
4. Non-negativity: The non-negativity constraint 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.
5. We assume here fixed technology. Fixed technology refers to
the fact that the production requirements are fixed during the
planning period and will not change in the period.
13

Formulating Linear Programming Models
The formulation of LPM involves the following steps:
1. Define the problem: This involves the determination of specific objectives.
The word expression of the specific objective.
Example: to determine the number of units of P1 and P2 to be produced per
month so as to maximize profit given the restrictions (constraints).
2. Determine the decision variables: This involves the representation of the
unknown quantities by letters
Example: let x1 and x2 represent the number of units of P1 and P2 to be
produced per month respectively.
14

3. Formulate the objective function: in developing the
objective function make sure that:
• All the decision variables are represented in the objective function
• The unit of measurement of all the coefficients in the objective
function must be the same.
 Example, if we use kg for x1, we should use kg for x2, too.
• All the terms in the objective must include a variable
15

4. Formulate the constraints: decide on the constraints and
formulate the mathematical relationship that is used to express
the limitations
5. Determine the non-negativity assumptions
Set all decision variables greater or equal to zero
16

• The general linear programming problem can be represented in the
following mathematical forms
Maximization Example
Max Z= C1X1 + C2X2 + C3X3 + C4X14… CnXn
Subject to:
A11 x1 + A12x2+ A13x3+ A14x4+ A15x5+ A16x6+ A17x7+ …. + A1nxn <R1
A21 x1 + A22x2+ A23x3+ A24x4+ A25x5+ A26x6+ A27x7+ …. + A2nxn <R2
Ai1 x1 + Ai2x2+ Ai3x3+ Ai4x4+ Ai5x5+ Ai6x6+ Ai7x7+ …. + Ainxn < Rn
Xij > o for J= 1, 2,3,4… n
17

Minimization Example
Min Z= C1X1 + C2X2 + C3X3 + C4X14… CnXn
Subject to:
A11 x1 + A12x2+ A13x3+ A14x4+ A15x5+ A16x6+ A17x7+ …. + A1nxn >R1
A21 x1 + A22x2+ A23x3+ A24x4+ A25x5+ A26x6+ A27x7+ …. + A2nxn >R2
Ai1 x1 + Ai2x2+ Ai3x3+ Ai4x4+ Ai5x5+ Ai6x6+ Ai7x7+ …. + Ainxn > Rn
Xij > o for J= 1, 2,3,4… n
18

Application Area of LP
 Production management (product mix, blending problems, production
planning, Assembly line balancing.),
 Marketing management (media selection, revenue determination,
physical distribution, etc)
 Personnel management (staffing problem, determination of equitable
salary and etc)
 Financial management (portfolio selection, profit planning)
 Agricultural application
 Military applications
19

Example 1
Given: A firm that assembles computer equipments is about to start production
of two new micro computers (type 1 and type2). 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 the micro
computers is limited. The manager of the firm would like to determine the
quantity of each micro computer to be produce in order to maximize the profit
generated by the sale of the micro computers.
20

Further information
Type1 Type 2
Unit profit $60 $50
Assembly time/unit 4 10
Inspection time/unit 2 1
Storage space/unit 3 3
The manager has also information on the availability of company resources:
Resources Amount available
Assembly time 100hrs
Inspection time 22hrs
Storage space 39cubic feet
Required: Formulate LP Model
21

Solution
Step1: Define the problem
 To determine the quantity of type I and type II micro computers to
be produced per day so as to maximize profit, given the restrictions
Step2: Identify the decision variables
 Let x1 = the quantity of Type I micro computers to be produced
& sold per day.
 x2 = the quantity of Type II micro computers to be produced&
sold per day.
 Z = daily profit
22

Solution
Step 3: Develop the Objective Function
Max Z= 60x1 + 50x2
Step 4: Formulate the Constraints
Assembly time: 4x1 + 10x2 < 100
Inspection time: 2x1 + x2 <22
Storage Space: 3x1 + 3x2 <39
Step 5: Non-negativity
X1, X2 > 0
23

Solution
The complete LPM is:
Max Z = 60x1 + 50x2
Subject to:
4x1 + 10x2 < 100
2x1 + x2 <22
3x1 + 3x2 < 39
x1, x2 > 0
24

Example 2
Two machines C1 and C2 produce two grades of tyres A and B. In one hour of
operation, machine C1 produces 50 units of grade A and 30 units of grade B
tyres while machine C2 produces 30 units of grade A and 40 units of grade B
tyres. The machines are required to meet a production schedule of at least 1400
units of grade A and 1200 units of grade B tyres. The cost of operating
machine C1 is $50 per hour and the cost of operating machine C2 is $80 per
hour.
Required
• Formulate the LPM if the objective is to minimize the cost of operating the
machines.
25

Solution
Step 1: Define the Problem
Determine the number of hours the two machines operate in order to
minimize the cost, meeting the production requirement.
Step 2: Determine the decision variables
Let x1= be the number of hours machine C1 should operate
x2= be the number of hours machine C2 should operate
Z= be the total cost
Step 3: Formulate the objective function
Min Z = 50x1 + 80x2
26

Solution
Step 4: List the constraints
Grade A: 50x1 + 30x2 > 1400
Grade B: 30x1 + 40x2 > 1200
Step 5: Non-negativity
x1, x2 > 0
Therefore ,The complete LPM is
Min Z = 50x1 + 80x2
Subject to:
50x1 + 30x2 > 1400
30x1 + 40x2 > 1200
x1, x2 > 0
27

Exercise – 1- A retail store stocks two types of shirts A and B. These are
packed in attractive cardboard boxes. During a week the store can sell a
maximum of 400 shirts of type A and a maximum of 300 shirts of
type B. The storage capacity, however, is limited to a maximum of 600 of both
types combined. Type A shirt fetches a profit of birr 2/- per unit and type B a
profit of birr 5/- per unit. How many of each type the store should
stock per week to maximize the total profit? Formulate a mathematical
model of the problem.
Exercise – 2- An aviation fuel manufacturer sells two types of fuel A and
B. Type A fuel is 25 % grade 1 gasoline, 25 % of grade 2 gasoline and
50 % of grade 3 gasoline. Type B fuel is 50 % of grade 2 gasoline and
50 % of grade 3 gasoline. Available for production are 500 liters per
hour grade 1 and 200 liters per hour of grade 2 and grade 3 each. Costs
are 60 paise per liter for grade 1, 120 paise for grade 2 and100 paise for
grade 3. Type A can be sold at Rs. 7.50 per liter and B can be sold at Rs.
9.00 per liter. How much of each fuel should be made and sold to
maximize the profit.
28

Minimization Problems
• Example: the agricultural research institute suggested to a
farmer to spread out at least 4800 kg of a special phosphate
fertilizer and not less than 7200 kg of a special nitrogen
fertilizer to raise productivity of crops in his fields. There are
two sources for obtaining these mixtures A and B. Both of
these are available in bags weighing 100 kg each and they
cost $40 and $24 respectively. Mixture A contains phosphate
and nitrogen equivalent of 20 kg and 80 kg respectively,
while mixture B contains these ingredients equivalent of 50
kg each.
Write this as a linear programming problem and determine
how many bags of each type the farmer should buy in order
to obtain the required fertilizer at minimum cost.
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1. Problem Definition
Determining the number of bags of mixtures A and
B that must be purchased, so as to satisfy the
requirements set and minimize total cost, given the
limiting factors.
2. Variable representation
Let X1 be the number of bags of mixture A that
must be purchased
X2 be the amount of bags of mixture B that must
be purchased
Z be the minimum cost
30

Solving LP Problems (LPP)
• There are two methods to solve LP problems
1.Graphical Method
2.Simplex Method
32

Graphical Method
The graphic method is applicable when we have two decision
variables and the steps include the following:
Step 1: Formulate the mathematical model of the problem
Step 2: Plot the constraints in the graph
Step 3: Identify the area that satisfies the entire set of constraints,
determine corners and their coordinate either from graphing procedure
or by the elimination procedure
33

Graphical Method
Step 4: Evaluate the objective function at each corner
 The largest value is the maximum and the smallest value is the
minimum
 If two corners have the same optimal value, then the optimum
occurs at every point on the line segment joining the respective
corners
34

Maximization Problem
Example: A furniture manufacturing company plans to make two
products namely, Chairs and Tables, from its available
resources which consist of 400 cubic feet of mahogany timber and
450 man hours of labor. It is known that to make a chair it
requires 5 cubic feet of timber and 10 man hours and yields a
profit of Birr 45/chair. To manufacture a table, it requires 20
cubic feet of timber and 15 man hours and yields a profit of Birr
80/table. The problem is to determine how many chairs and tables
the company can make keeping within the resources constraints so
that is maximizes the profit.
Required: Formulate LP Model and solve the LP problem using a
graphical method.
35

Solution
Formulate the LPM (Step wise)
Step 1: Define the problem
to determine the number of chairs and tables to be produced in order to maximize profit under the
limited resources/constraints/.
Step 2: Represent the decision variables
Let x1 =be the number of chairs to be produced
x2= be the number of tables to be produced
Z= be the total be the total profit
Max Z = 45x1 + 80x2
M.Timber: 5x1 + 20x2 < 400
Man hr Labor: 10x1 + 15x2 <450
Step 5: Non-negativity x1, x2 > 0
36

Solution
Max z = 45x1 + 80x2
Subject to: 5X1 + 20X2 < 400
10X1 + 15X2 < 450
x1, x2 > 0
37

Solution
Step 2: Plot the constraints on the graph by taking two coordinates
satisfying the equation
5X1 + 20X2 < 400 i.e. X1 + 4X2 < 80 - - - (a)
10X1 + 15X2 < 450 i.e. 2X1 + 3X2 < 90 - - - (b)
X1, X2 > 0
Converting (a) and (b) to equality, we get
X1 + 4X2 = 80 . . . . (1)
2X1 + 3X2 = 90 . . . (2)
38

Solution
Step 3: Identify the area that satisfies the entire set of constraints
(Feasible region), determine corners and their coordinate either from
graphing procedure or by the elimination procedure.
Step 4: Evaluate the objective function at each corner.
 The largest value is the maximum and the smallest value is the
minimum.
 If two corners have the same optimal value, then the optimum occurs
at every point on the line segment joining the respective corners.
39

Solution
Coordinate of corner point Objective Function
Max Z=45X1 + 80X2
Value
(0, 0) 45 ×0 + 80×0 = 0 0
(0, 20) 0 + 1600 1600
(24, 14) 45 × 24 + 80 × 14 2200*
(45, 0) 45 × 45 + 80 × 0 2025
Hence, the maximum profit (2200) is obtained when the
company produces 24 chairs and 14 tables
40

Minimization Problem
• Graphical solutions to minimization problems are very similar to
solutions to maximization problems
• The difference in minimization problems are:
 The constraints are in (>=)greater than or equal to sign,
the feasible solution area is away from the origin, instead of close to
the origin
 Optimum solution is the smallest possible value, instead
of largest value
41

Minimization Problem
Example : Two machines C1 and C2 produce two grades of tyres A
and B. In one hour of operation, machine C1 produces 50 units of
grade A and 30 units of grade B tyres while machine C2 produces 30
units of grade A and 40 units of grade B tyres. The machines are
required to meet a production schedule of at least 1400 units of
grade A and 1200 units of grade B tyres. The cost of operating
machine C1 is $50 per hour and the cost of operating machine C2 is
$80 per hour.
Required: How many hours should each machine operate so that the
cost of production is minimized?
42

Solution
Formulate the LPM (Step wise)
Step 1: to determine the number of hours the two machines operate in order to minimize cost,
meeting the production requirement.
Step 2: Represent the decision variables
Let x1 be the number of hours machine C1 should operate
x2 be the number of hours machine C2 should operates
Z be the total cost
Min Z = 50x1 + 80x2
Grade A: 50x1 + 30x2 > 1400
Grade B: 30x1 + 40x2 > 1200
Step 5: Non-negativity x1, x2 > 0
43

Solution
Min Z= 50x1 + 80x2
Subject to:
50x1 + 30x2 > 1400
30x1 + 40x2 > 1200
x1, x2 > 0
44

Solution
Exercise the next steps
Step 2: Plot the constraints on the graph
Step 3: Identify the feasible region
Step 4: Evaluate the objective function at each corner points
45

Solution
Coordinate of corner point Objective Function
Min Z= 50x1 + 80x2
Value
(46.7, 0) 50 × 46.7 + 80× 0 2335
(16.40, 18.18 ) 50 ×18.18+ 80× 16.40 2221*
(0,40) 50× 0+ 80 × 30 2400
Therefore, the company can minimize its cost to $2221 by
operating machine 1 for 18.18 hours and Machine 2 for
16.40 hours
46

Graphical Solutions for the Special Cases of
LP problems
1. Unbounded problems
2. Problems with no feasible solutions
3. Redundancy in constraints
4. Problems with Multiple optimal solutions
47

Un bounded Problems
An unbounded problem occurs when the value of decision variable increased
indefinitely without violating any of the constraints. The reason for it may be
concluded to be wrong formulation of the problem such as incorrectly
maximizing instead of minimizing and/or errors in the given
problem.
Example : Max Z = 10X1 + 20X2
Subject to:
2X1 + 4X2 > 16
X1 + 5X2 > 15
X1, X2 > 0
48

Problems with no feasible solutions
Infeasibility is a condition that arises when there is no solution to a LP
problem that satisfies all the constraints. This problem occurs when the problem has
a mix of greater and less than constraints ( LPP formulated with conflicting
constraints).
Example: Max Z = 3X1+2X2
Subject to:
2X1 + X2 < 2
3X1 + 4X2 > 12
X1, X2 > 0
49

Redundancy in constraints
• In some cases, a constraint does not form a unique boundary of the feasible
solution space.
• Such a constraint is called a redundant constraint. A constraint is redundant if its
removal would not alter the feasible solution space. Redundancy of any
constraint does not cause any difficulty in solving an LP problems graphically.
Constraints appear redundant when it may be more binding (restrictive) than others.
Example: Max Z= x1 + x2
Subject to:
2x1+3x2<60
2X1+x2<20
X2<25
X1, x2>0
50

Problems with Multiple optimal solutions
• Recall the optimum solution is that extreme point for which the
objective function has the largest value.
• It is of course possible that in a given problem there may be more
than one optimal solution.
Example : Max Z = 8X1+16X2
Subject to:
X1 + X2 < 200
3X1 + 6X2 < 900
X2 < 125
X1, X2 > 0
51

EXERCISE
Use graphical method to solve the following LPP.
a. Max.Z = 7/4X1+3/2X2
St:
8X1+5X2 < 320
4X1+5X2 < 20
X1 > 15
X2 > 10
X1, X2 > 0
52

b. Max.Z = 3X1+2X2
St:
-2X1+3X2 < 9
X1-5X2 > -20
X1, X2 > 0
c. Max.Z=3X1+2X2
St:
X1-X2 < 1
X1+X2> 3
X1, X2> 0
53

The Simplex Method
• Graphical method of solving LPP can be used when we have
two decision variables in the problem.
• If we have more than two decision variables, we can’t use
graphical method
• There fore, SIMPLEX Method is useful to solve LPP with
two and more decision variables
55

The Simplex Method
• Simplex method deals with iterative process, which consists of
first designing a Basic Feasible Solution and proceed towards
the Optimal Solution and testing each feasible solution for
Optimality to know whether the solution on hand is optimal or
not
56

Comparison b/n graphical and Simplex
1. The graphical method is used when we have two decision
variables in the problem. Whereas in simplex, the problem may
have any number of decision variables
2. In graphical method, the inequalities are assumed to be
equations, so as to enable to draw straight lines. Whereas in
simplex method the inequalities are converted to equations by
adding Slack variable (in maximization) and Surplus
variable (in minimization)
57

The Simplex Method: Maximization Example
There are 8 steps to be followed:
Step1:Formulate the linear programming model of the real world
problem, i.e., obtain a mathematical representation of the problem's
objective function and constraints.
Step2: Express the mathematical model of LP problem in the
STANDARD FORM by adding slack variables in the left-
hand side of the constraints and assign a zero coefficient to
these in the objective function.
58

Example : Maximize Z = C1X1+ C2X2 + ... +CnXn + OS1 + OS2 +... +0Sm
Subject to
a11X1+a12X2+... + a1nxn+s1=b1
a2lX1+al22X2+... + a2nXn+S2 = b2
amlXl + am2 X2 +... + amnxn + Sm = bm
where X1, X2... Xn and S1, S2 ... Sm are non-negative.
Step 3: Design the initial feasible solution
 An initial basic feasible solution is obtained by setting the decision
variables equal to zero
X1= X2 = ... = Xn = 0. Thus, we get S1 = b1, S2 = b2 ... Sm = bm
59

Step 4: Set up the initial simplex Tableau
• For computational efficiency and simplicity, the initial basic
feasible solution, the constraints of the standard LPM as well as
the objective function can be displayed in a tabular form, called the
simplex tableau.
60

Cj C1 C2 Cn,…. 00..0 Quantity Ratio
Basic
variable
CB X1, X2, ... Xn S1 S2 ... Sm
S1 CB1 a11 a12 ... a1n 1 0 ... 0 b1
S2 CB2 a21 a22 ... a2n 0 1 ... 0 b2
S3 CB3 am1 am2.amn 0 0 ... 1 b3
Zj=ΣCBiXj 0 0 ... 0 0 0 ... 0 ΣCBiXj
Cj-Zj C1-Z1 C2-Z2 ... Cn-Zn
Simplex Tableau
61

 Cj--coefficients of the variables in the objective function
 X1, X2. Xn----Non-basic variables
 S1, S2... Sm-------basic variables
 Zj entries represent the decrease in the value of objective function
that would result if one of the variables not included in the solution
were brought into the solution.
 Cj-Zj--index row or net evaluation row, is used to
determine whether or not the current solution is optimal or not
62

Step 5: Test if the current solution is optimum or not.
 If all the elements or entries in the Cj- Zj row (i.e., index row) are
negative or zero, then the current solution is optimum.
 If there exists some positive number, the current solution can be
further improved by removing one basic variable from the basis and
replacing it by some non-basic one.
 So, start trying to improve the current solution in line with the
following steps.
63

Step 6: Iterate, further, towards an optimum solution
 To improve the current feasible solution, we need to replace one current
basic variable (called the leaving variable) by a new non-basic
variable (called the entering variable)
Step 7:Evaluate the new solution by constructing the next
simplex tableau
 After identifying the entering and leaving/departing variable, all that
remains is to find the new basic feasible solution by constructing a new
simplex tableau from the current one.
64

Step 8: If any of the numbers in Cj - Zj row are positive, repeat the
steps (6-7) again until an optimum solution has been obtained.
Therefore : A simplex solution in a maximization problem is
optimal if the Cj-Zj row consists of entirely zeros and negative
numbers, i.e. there are no positive values in the row.
65

Maximization Example
Max Z = 60X1+50X2
Subject to:
4X1+10X2 < 100
2X1+ X2 < 22
3X1+ 3X2 <39
X1, X2 > 0
66

Solution
Step 1: Formulate the linear programming model of the real world
problem i.e mathematical representation of LPP
Step 2: Put in Standard form
Max Z = 60X1+50X2+0S1+0S2+0S3
Subject to:
4X1+10X2+S1 = 100
2X1+ X2+ S2 = 22
3X1+ 3X2+ S3 = 39
X1, X2, S1, S2, S3 > 0
67

Solution
Step 3: Design the initial feasible solution.
X1= X2=0 so S1= 100 S2= 22 S3=39
Step 4: Set up the initial simplex tableau
Cj
Basic Variables
60
X1
50
X2
0
S1
0
S2
0
S3
Quantity
S1 0 4 10 1 0 0 100
S2 0 2 1 0 1 0 22
S3 0 3 3 0 0 1 39
Zj
Cj-Zj
0 0 0 0 0 0
60 50 0 0 0
68

Second Tableau
Cj 60 50 0 0 0 Quantity
BVs X1 X2 S1 S2 S3
S1 0 4 10 1 0 0 100 100/4= 25
S2 0 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 0
Cj-Zj 60 50 0 0 0
Pivot Row Pivot Column Pivot Element
2
69
Smallest non-negative
quotient

Second tableau
X1 X2 S1 S2 S3
S1 0 0 8 1 -2 0 56
X1 60 1 1/2 0 1/2 0 11
S3 0 0 3/2 0 -3/2 1 6
Zj 60 30 0 30 0 660
Cj-Zj 0 20 0 -30 0
70

Third tableau
X1 X2 S1 S2 S3
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
Therefore, S1 = 24, X1 = 9, and X2 = 4 producing a
maximum profit of $740.
71

The various steps involved in using simplex method for minimization
problems are:
Step1: Formulate the linear programming model, and
express the mathematical model of LPP in the standard form by
introducing surplus and artificial variables in the left hand side
of the constraints.
 Assign a 0 (zero) and +M as coefficient for surplus and
artificial variables respectively in the objective function.
 M is considered to be a very large number so as to finally drive
out the artificial variables out of basic solution.
73

• Artificial variables have no physical interpretation, they merely
serve as a device to enable us to use simplex process.
• During simplex process, any artificial variables are quickly
eliminated from the solution and hence the optimal solution should
never contains an artificial variable with a non-zero value
• Artificial variables are required only for manual solution;
computer codes do not require to input artificial variables
74

Step 2: Set up an initial solution
• Just to start the solution procedure, the initial basic feasible
solution is obtained by assigning zero value to decision
variables.
• This solution is summarized in the initial simplex tableau.
• Complete the initial simplex tableau by adding two final
rows Zj, and Cj- Zj. These two rows help us to know
whether the current solution is optimum or not.
75

Step 3: Test for optimality of the solution
 If all the entries of Cj - Zj, row are positive and zero,
then the solution is optimum.
 However, this situation may come after a number of iterations.
But if at least one of the Cj - Zj values is less than zero, the
current solution can be further improved by removing one basic
variable from the basis and replacing it by some non-basic one.
76

Step 4: (i) Determine the entering variable into the
basic solution. To do this, we identify the column with the
largest negative value in the Cj - Zj row of the
tableau (the variable in this column is entering variable)
(ii) Next, determine the departing variable from the
basic solution. If an artificial variable goes out of solution,
then we discard it totally and even this variable may not
form part of further iterations. Same procedure, as in
maximization case, is employed to determine the departing
and entering variable.
77

Step 5: Update the new solution
• We evaluate the entries for next simplex tableau in exactly
the same manner as was discussed earlier in the
maximization example.
78

Step 6: Step (3-5) are repeated until an optimum solution is
obtained
• So the following are the essential things to observe in solving for minimization
problems:
• The entering variable is the one with the largest negative value in
the Cj-Zj row while the leaving variable is the one with the smallest
non-negative ratio.
• The optimal solution is obtained when the Cj-Zj row contains entirely
zeros and positive values.
79

Example
Min Z = 7X1+9X2
Subject to
3X1+6X2 > 36
8X1+4X2 > 64
X1, X2 > 0
80

Solution
First, put the LPM in standard form
Min Z = 7X1+9X2+0S1+0S2+MA1+MA2
Subject to
3X1+6X2-S1+A1 = 36
8X1+4X2-S2+ A2 = 64
X1, X2 , S1,S2,A1, A2> 0
81

Solution
Then, develop the Initial Simplex Tableau
Cj 7 9 0 0 M M
BV X1 X2 S1 S2 A1 A2
Quantity
A1 M 3 6 -1 0 1 0 36
A2 M 8 4 0 -1 0 1 64
Zj 11M 10M -M -M M M 100M
Cj-Zj 7-11M 9-10M M M 0 0
82

Solution
Next, develop the second Simplex Tableau
Cj 7 9 0 0 M
X1 X2 S1 S2 A1
Quantity
A1 M 0 9/2 -1 3/8 1 12
X1 7 1 ½ 0 -1/8 0 8
Zj 7 7/2+9/2M -M 3/8M-7/8 M 56+12M
Cj-Zj 0 11/2-9/2M M 7/8-3/8M 0
83

Solution
Next, develop the third Simplex Tableau
Cj 7 9 0 0
X1 X2 S1 S2 Quantity
X2 9 0 1 -2/9 1/12 8/3
X1 7 1 0 1/9 -1/6 20/3
Zj 7 9 -11/9 -5/12 212/3
Cj-Zj 0 0 11/9 5/12
Therefore, the optimal solution is: X1 = 20/3 and X2 = 8/3
and value of objective function is 212/3.
84

Post optimality Analysis
1. Duality in LPP
2. Sensitivity Analysis
Sensitivity analysis involves an examination of the potential
impact of changes in any of the parameters (numerical
values) of a problem where as the dual involves setting up and
solving LP problem that is almost a “mirror image” of LP
problem that has been formulated.
85

Duality in LPP
• Every linear programming problem can have two forms: the original
formulation of the problem is referred to as its primal form. The
other form is referred to as its dual form.
• The dual is just a mirror image of the primal and hence it is
called an alternate form of linear programming problem
• The solution to the primal problem contains the solution to the dual
problem, and vice versa.
• If so, why are we interested in dual form?
86

Duality in LPP
• We need to have the dual LPP for an economic analysis.
• Analysis of the dual help the manager to evaluate the potential
impact of a new product, and it can be used to determine the
marginal value of resources (Constraints).
• Relative to the new product, a manager would want to know what
impact adding a new product would have on the solution quantities
and the profit; relative to resources, a manager can refer to dual
solution to determine how much profit one unit of each resource is
equivalent to.
87

Formulating the dual
• A LPP formulated directly from the description of the
problem is a primal problem.
• The dual, on the other hand, is an alternate formulation of
the problem that requires the existence of the primal.
• Let’s take an example how the dual is formulated from the
primal problem
88

Formulating the dual from the primal
Max Z= 60x1 + 50x2
Subject to:
4x1 + 10x2 < 100
2x1 + x2 < 22
3x1 + 3x2 < 39
x1, x2 > 0
Primal problem
89
Min Z = 100y1 + 22y2+ 39y3
Subject to:
4y1 + 2y2 + 3y3 > 60
10y1 + y2 +3y3> 50
y1, y2 ,y3 > 0
Dual problem

Sensitivity Analysis
• Sensitivity analysis (A Post optimality analysis) is made
beyond the determination of optimal solution
• It starts from the final simplex tableau.
• The purpose is to explore how changes in any of the
parameters of a problem, such as the coefficient of the
objective function, or the right-hand side values, would affect
the solution
• This analysis help the manager in many ways
90

• The parameters are assumed to be constant,
but in real life there is nothing constant.
Hence, the optimal values of the decision variable may be
wrong.
• Therefore, we need to check how sensitive is the solution
to the alternate values of parameters.
• Hence, the decision maker usually would want to perform
sensitivity analysis before implementing the solution
91

• If the additional analysis satisfies the decision maker that
the solution is reasonable, the decision maker can proceed
with greater confidence in using the solution
• Conversely, if sensitivity analysis indicates that the solution
is sensitive to the changes that are within the realm
of possibility, the decision maker will be warned that more
precise information on parameters are needed.
92

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