Chapter 4 Simplex Method ppt

Linear Optimization(MATH 2062)
Dereje Tigabu(MSc.)
Department of Mathematics
Debark University
October 7, 2020
(Department of Mathematics Debark University )Linear Optimization(MATH 2062) October 7, 2020 1 / 107

Chapter 4
Simplex Method1
1
Checking the results of a decision against its expectations shows executives
what their strengths are, where they need to improve, and where they lack
knowledge or information. Peter Drucker

Outline of the Chapter
1 Introduction
2 Linear Programs in Standard Form
3 Basic Feasible Solutions
4 Fundamental Theorem of Linear Programming
5 Algebra of the Simplex Method
6 The simplex Algorithm
7 Degeneracy and Finiteness of Simplex Algorithm
8 Finding a Starting Basic Feasible Solution
Two -phase method
Big - M method
9 Some Complications and Their Resolution
Unrestricted Variables
Tie for Entering Basic Variable (Key Column)
Tie for Leaving Basic Variable (Key Row) Degeneracy
10 Types of Linear Programming Solutions
Alternative (Multiple) Optimal Solutions
Unbounded Solution
Infeasible Solution

1. Introduction
We shall discuss a procedure called the simplex method for solving
an LP model of such problems.
This method was developed by G B Dantzig in 1947.
For LP problems with several variables, we may not be able to graph
the feasible region, but the optimal solution will still lie at an extreme
point of the many-sided, multidimensional figure (called an
n-dimensional polyhedron) that represents the feasible solution space.
The simplex method examines these extreme points in a systematic
manner, repeating the same set of steps of the algorithm until an
optimal solution is found.
It is also called the iterative method.
Since the number of extreme points of the feasible solution space are
finite, the method assures an improvement in the value of objective
function as we move from one iteration (extreme point) to another
and achieve the optimal solution in a finite number of steps.
The method also indicates when an unbounded solution is reached.

Slack variables : If the constraints are in the form of ”≤”, then we
add variable to the left hand side of inequality to make equality. This
variables are called slack variables.
Remark: A slack variable represents unused resource, either in the
form of time on a machine, labour hours, money, warehouse space
or any number of such resources in various business problems. Since
these variables yield no profit, therefore such variables are added to
the original objective function with zero coefficients.
Surplus variables : If the constraints are in the form of ” ≥ ” ,
then we subtract variables to the left hand side of inequality to make
equality. This variables are called the surplus variables.
Remark: A surplus variable represents amount by which solution values
exceed a resource. These variables are also called negative slack variables.
Surplus variables like slack variables carry a zero coefficients in the objective
function.(Department of Mathematics Debark University )Linear Optimization(MATH 2062) October 7, 2020 5 / 107

Conditions for Application of Simplex Method
1 The right hand side of each of the constraint bi should be
non-negative. If LPP has a negative constraint, we should
convert it to positive by multiplying both sides by -1.
2 Each of the decision variables of the problem should be
non-negative. If one of the choice variables is not feasible we
can’t apply the Simplex method. Therefore feasibility is the
necessity condition for application of the Simplex method.
3 The inequality constraint of resources or any other activities
must be converted in to equality equations by adding slack
variables (≤ type inequality) or by subtracting surplus variables
(≥ type inequality) to the left of the inequality.

2. Linear Programs in Standard Form
The standard form have the following characteristics.
(i) All the constraints should be expressed as equations by adding
slack or surplus and/ or artiﬁcial variables.
(ii) The right hand side of each constraint should be made of
non-negative, if it is not, this should be done by multiplying
both sides of the resulting constraint by -1.
Minimization and Maximization Problems
Max(Z) = −Min(−Z).
After the optimization of the new problem is completed, the
objective value of the old problem is -1 times the optimal
objective value of the new problem.
Let Z = −Z. After the Z value is found, replace Z = −Z .

Consider an LP model,
Max/Min Z = cT
X
S.t : Ax(≤, ≥)b
xi ≥ 0
The standard form of the Linear programming problem is expressed as
Optimize(Max or Min)Z = CT X + 0S
Subject to AX ± S = b.
and X, S ≥ 0.
where CT = (c1, c2, ..., cn) is the row vector, X = (x1, x2, ..., xn)T
b = (b1, b2, ..., bm)T and S = (s1, s2, ..., sm)T are column vectors.
and A =








a11 a12 . . . a1n
a21 a22 . . . a2n
. . .
. . . . . .
. . .
am1 am2 . . . amn








is the m × n matrix of coeﬃcients of variables of x1, x2, ..., xn in the
constraints.

Example 1
Transform the following LPP into standard form of LPP.
Maximize Z = x1 − x2 + x3
Subject to x1 + x2 − 3x3 ≥ 4
2x1 − 4x2 + x3 ≥ −5
x1 + 2x2 − 2x3 ≤ 3
x1 ≥ 0, x2 ≥ 0, x3 ≥ 0
Solution:
Since the second constraint have a negative value on the right we can
multiply both side by (-1) so,
Maximize Z = x1 − x2 + x3 + 0x4 + 0x5 + 0x6
subject to x1 + x2 − 3x3 − x4 = 4
−2x1 + 4x2 − x3 + x5 = 5
x1 + 2x2 − 2x3 + x6 = 3
x1, x2, x3, x4, x5, x6 ≥ 0
Therefore, the above problem is standard LPP.

Variables Unrestricted In Sign
The diﬀerence of two non-negative variables is a variable unrestricted
in sign.
Let x1 and x2 be two non-negative variables. The diﬀerence of these
two variables is a variable x3 i.e x3 = x1 − x2 which is unrestricted in
sign.
If x1 > x2, then x3 > 0.
If x1 < x2, then x3 < 0.
If x1 = x2 , then x3 = 0.

Example 2.
Write down the following LPP, where the variables are non-negative in
standard LPP form
Maximize Z = 2x1 + 3x2 − x3
Subject to 4x1 + x2 + x3 ≥ 4
7x1 + 4x2 − x3 ≤ 25,
x1, x3 ≥ 0, x2 unrestricted sign.
Solution: Since x2 is unrestricted sign, we will convert x2 by x2 = x2 − x2 ,
where x2, x2 ≥ 0.
Maximize Z = 2x1 + 3x2 − 3x2 − x3 + 0x4 + 0x5
Subject to 4x1 + x2 − x2 + x3 − x4 = 4
7x1 + 4x2 − 4x2 − x3 + x5 = 25,
x1, x3, x4, x5, x2, x2 ≥ 0,

3. Basic Feasible Solutions
Consider the system Ax = b and x ≥ 0, where A is an m × n matrix
and b is any m vector.
Suppose that the rank(A, b) = rank(A) = m. After possibly
rearrangement of the columns of A, let A = [B, N] where B is an
m × m invertible matrix and N is m × (n − m) matrix.
The point X =
XB
XN
, where XB = B−1b and XN = 0 is called a
basic solution of the system.
If XB ≥ 0, then X is called a basic feasible solution of the system.
Here B is called the basic matrix (or simply the basis ) and N is
called the non basic matrix.
The components of XB are called basic variables, and the
components XN are called non basic variables.
If XB > 0, then X is called a non degenerate basic feasible
solution and if at least one component of XB is zero, then X is called
a degenerate basic feasible solution.

Example 1
Consider the polyhedral set deﬁned by the following inequalities. Find
the basic solution.
(a)
x1 + x2 ≤ 6
x2 ≤ 3
x1, x2 ≥ 0
(b)
x + y ≤ 3
2x − y ≤ 4
x, y ≥ 0

Solution
(a) By introducing the slack variables x3 and x4 , the problem is put in the
following standard format:
x1 + x2 + x3 = 6
x2 + x4 = 3
x1, x2, x3, x4 ≥ 0
Note that, the constraint matrix
A = [a1 a2 a3 a4] =
1 1 1 0
0 1 0 1
From forgoing deﬁnition, basic feasible solutions corresponding to ﬁnding
a 2 × 2 basic matrix B with non-negative B−1b. The following are possible
ways of extracting B out of A.

B = [a1, a2] =
1 1
0 1
XB =
x1
x2
= B−1b =
1 −1
0 1
6
3
=
3
3
XN =
x3
x4
=
0
0
B = [a1, a4] =
1 0
0 1
XB =
x1
x4
= B−1b =
1 0
0 1
6
3
=
6
3
XN =
x2
x3
=
0
0
B = [a2, a3] =
1 1
1 0
XB =
x2
x3
= B−1b =
0 1
1 −1
6
3
=
3
3
XN =
x1
x4
=
0
0

B = [a2, a4] =
1 0
1 1
XB =
x2
x4
= B−1b =
1 0
−1 1
6
3
=
6
−3
XN =
x1
x3
=
0
0
B = [a3, a4] =
1 0
0 1
XB =
x3
x4
= B−1b =
1 0
0 1
6
3
=
6
3
XN =
x1
x2
=
0
0
We have four basic feasible solutions.
Namely X1 =




3
3
0
0



 , X2 =




6
0
0
3



 , X3 =




0
3
3
0



 , X4 =




0
0
6
3





These basic feasible solutions, projected in R2 that is in the (x1, x2) space
give rise to the following four points.
3
3
,
6
0
,
0
3
,
0
0
these points are precisely the extreme points of the feasible region.

In this example, the possible number of basic feasible solutions is bounded
by the number of ways of extracting two columns out of four columns to
form the basis. therefore, the number of basic feasible solutions is less
than or equal to
4
2
= 4!
2!(4−2)! = 6.
Out of these six possibilities, one point violates the non-negativity of B−1b.
furthermore, a1 and a3 could not have been used to form a basic since
a1 = a3 =
1
0
are linearly dependent, and hence the matrix
1 1
0 0
does not qualify as a basis. This leaves four basic feasible solutions.
(b)
x + y + z = 3
2x − y + w = 4
x, y, z, w ≥ 0
1 1 1 0
2 −1 0 1




x
y
z
w



 =
3
4

Here, 4
2 = 6 possible square matrix obtained from the given system.
If we select column 3 and 4 and assign zero value to the variable
associated with column 1and 2, z = 3, w = 4. So, (0, 0, 3, 4) is basic
solution and x, y is non-basic variable and z and w is basic variable.
B =
1 0
0 1
. XB =
z
w
=
3
4
and XN =
x
y
=
0
0
Again take column 1 and 2 and assign zero value to the variable associated
with column 3 and 4, B =
1 1
2 −1
and XB = B−1b and follow the same
fashion as above.
In general, the number of basic feasible solutions is less than or
equal to
n
m
= n!
m!(n−m)!

Example 2: Degenerate basic feasible solutions
Consider the following system of inequalities
x1 + x2 ≤ 6
x2 ≤ 3
x1 + 2x2 ≤ 9
x1, x2 ≥ 0
Solution: After adding the slack variables x3, x4 and x5, we get
x1 + x2 + x3 = 6
x2 + x4 = 3
x1 + 2x2 + x5 = 9
x1, x2, x3, x4, x5 ≥ 0
A = [a1, a2, a3, a4, a5] =


1 1 1 0 0
0 1 0 1 0
1 2 0 0 1



Let us consider the basic feasible solution for B = [a1, a2, a3]
XB =


x1
x2
x3

 =


1 1 1
0 1 0
1 2 0


−1 

6
3
9

 =


0 −2 1
0 1 0
1 1 −1




6
3
9

 =


3
3
0

, XN =
x4
x5
=
0
0
Note that this basic feasible solution is degenerate since the basic variable
x3 = 0.
Now consider the basic feasible solution with B = [a1, a2, a4]
XB =


x1
x2
x4

 =


1 1 0
0 1 1
1 2 0


−1 

6
3
9

 =


2 0 −1
−1 0 1
1 1 −1




6
3
9

 =


3
3
0

, XN =
x3
x5
=
0
0

Note that these basic feasible solution give rise to the same point obtained
by B = [a1, a2, a3]. It can be also checked the other basic feasible solution
with basis B = [a1, a2, a5] is give by
XB =


x1
x2
x5

 =


3
3
0

 , XN =
x3
x4
=
0
0
Note that all the three foregoing bases represent the single extreme point
or basic feasible solution (x1, x2, x3, x4, x5) = (3, 3, 0, 0, 0). This basic
feasible solution is degenerate since each associated basis involves a basic
variable at level zero.

4. Fundamental Theorem of Linear Programming
Theorem: For an arbitrary linear programming in standard form of
the following are true.
i . If there is no optimal solution, then the problem is either
infeasible or unbounded
ii . If a feasible solution exists, then a basic feasible solution exists
iii . If an optimal solution exists, then a basic optimal solution
exists

Key to the Simplex Method
The key to the simplex method lies in recognizing the optimality of a given
extreme point solution based on local considerations without having to
(globally) enumerate all extreme points or basic feasible solutions.
Consider the following linear programming problem :
LP: Minimize CT X
Subject to AX = b
X ≥ 0
where A is an m × n matrix with rank m, suppose that we have a basic
feasible solution
B−1b
0
whose objective value Z0 is given by
Z0 = CT B−1b
0
= (CT
B , CT
N )
B−1b
0
= CT
B B−1
b (3.1)

Now, let XB and XN denote the set of basic and non basic variables for
the given basis and X =
XB
XN
be an arbitrary feasible solution. Then
feasibility requires that XB ≥ 0, XN ≥ 0 and that b = AX = BXB + NXN.
Multiplying by B−1 and rearranging the terms, we get
B−1
(b = BXB + NXN)
B−1
b = XB + B−1
NXN
XB = B−1
b − B−1
NXN
XB = B−1
b −
j∈J
B−1
aj xj
XB = B−1
b −
j∈J
yj xj
XB = b −
j∈J
yj xj (3.2)
Where b = B−1
b, yj = B−1
aj and J is the current set of the indices of the non
basic variables.

Letting Z denote the objective function value. we get
Z = CT X
= (CT
B , CT
N )
XB
XN
= CT
B XB + CT
N XN
= CT
B (B−1b − j∈J B−1aj xj ) + j∈J cj xj
= CT
B (B−1b) − j∈J CT
B B−1aj xj + j∈J cj xj
= Z0 − j∈J zj xj + j∈J cj xj
= Z0 − j∈J(zj − cj )xj (3.3)
Where zj = CT
B B−1aj for each non basic variable
Using the foregoing transformation, the linear programming problem may
be written as
Minimize Z = Z0 −
j∈J
(zj − cj )xj
subject to
j∈J
yj xj + XB = B−1
b
xj ≥ 0, j ∈ J, and XB ≥ 0 (3.4)

5. Algebra of the Simplex Method
Consider the representation of the linear programming LP in the non
basic variable space written in equality form as in equation (3.4).
If (zj − cj ) ≤ 0 for j ∈ J, then xj = 0 for j ∈ J and XB = B−1b is
optimal for the LP. otherwise, while holding (p-1) non basic variable
ﬁxed at zero, the simplex method considers increasing the remaining
variables, say xk.
Naturally we would like zk − ck to be positive and perhaps the most
positive of all the zj − cj , j ∈ J.
Now ﬁxing xj = 0 for j ∈ J − {k} we obtain from equation (3.4)that
z = z0 − (zk − ck)xk (3.5)
and















xB1
xB2
.
.
.
xBr
.
.
.
xBm















=















b1
b2
.
.
.
br
.
.
.
bm















−















y1k
y2k
.
.
.
yrk
.
.
.
ymk















xk (3.6)

If yik ≤ 0, then xBi
increases as xk increase and so xBi
continues to be
non-negative.
If yik ≥ 0 then xBi
will decrease as xk increases.
In order to satisfy non-negativity, xk is increase until the ﬁrst point at
which some basic variable xBr drops to zero.
Examining equation (3.6), it is then clear that the ﬁrst basic variable
dropping to zero corresponds to the minimum of bi
yik
for positive yik.
more precisely, we can increase xk until
xk = br
yrk
= minimum1≤i≤m{ bi
yik
: yik > 0} (3.7)
In the absence of degeneracy, br > 0 and hence xk = br
yrk
> 0, from
equation (3.5) and the fact that zk − ck > 0, it then follows z < z0
and the objective function strictly improves.
As xk increases from level 0 to br
yrk
, a new feasible solution obtained.

Substituting xk = br
yrk
in equation (3.6) gives the following point:
xBi
= bi − yik
yrk
br , i = 1, 2, ..., m
xk = br
yrk
(3.8)
and all other xj - variables are zero.
From equation (3.8), xBr = 0 and hence, at most m variables are
positive.
The corresponding columns in A are
aB1 , aB2 , ..., aBr−1 , ak, aBr+1 , ..., aBm .
Note that these columns are linearly independent. since yrk = 0.
(Recall that if aB1 , aB1 , ..., aBm are linearly independent, and if ak
replaces aBr , then the new columns are linearly independent if and
only if yrk
= 0.
Therefore, the point given by equation (3.8) is a basic feasible
solution.

Recall that a basic variable xB that first drops to zero is called a
blocking variable because it blocks the further increase of xk. Thus,
xk enters the basis and xB leaves the basis.
To summarize, we have algebraically described an iteration, that is,
the process of transforming from one basis to an adjacent basis.
This is done by increasing the value of a nonbasic variable xk with
positive zk − ck and adjusting the current basic variables.
In the process, the variable xB drops to zero.
The variable xk hence enters the basis and xB leaves the basis.
In the absence of degeneracy the objective function value strictly
decreases, and hence the basic feasible solutions generated are
distinct.
Because there exists only a finite number of basic feasible solutions,
the procedure would terminate in a finite number of steps.

Example 1.
Minimize x1 + x2
subject to x1 + 2x2 ≤ 4
x2 ≤ 1
x1, x2 ≥ 0
Solution: Introducing the slack variables x3 and x4 to put the problem in
a standard form.
This leads to the following constraint matrix A :
A = [a1, a2, a3, a4] =
1 2 1 0
0 1 0 1
Consider the basic feasible solution corresponding to B = [a1, a2].
In other words, x1 and x2 are the basic variables, while x3 and x4 are
the nonbasic variables.
The representation of the problem in this nonbasic variable space as
in Equation (3.4) with J= 3, 4 may be obtained as follows.

First, compute: B−1 =
1 2
0 1
−1
=
1 −2
0 1
, CBB−1 =
(1, 1)
1 −2
1 1
= (1, −1)
Hence y3 = B−1a3 =
1 −2
0 1
1
0
=
1
0
y4 = B−1a4 =
1 −2
0 1
0
1
=
−2
1
and XB = b = B−1b =
1 −2
0 1
4
1
=
2
1
Also, z0 = CBB−1b = (1, −1)
4
1
= 3
z3 − c3 = CBB−1a3 − c3 = (1, −1)
1
0
− 0 = 1
z4 − c4 = CBB−1a4 − c4 = (1, −1)
0
1
− 0 = −1

Hence, the required representation of the problem is
Minimize 3 − x3 + x4
subject to x3 − 2x4 + x1 = 2
x4 + x2 = 1
x1, x2, x3, x4 ≥ 0
Since z3 − c3 > 0, then the objective function improves by increasing
x3, the modiﬁed solution is given by
XB = B−1b − B−1a3x3
x1
x2
=
2
1
−
1
0
x3
The maximum value of x3 is 2 (any larger value of x3 will force x1 to
be negative).
Therefore the new basic feasible solution is (x1, x2, x3, x4) = (0, 1, 2, 0)
Here, x3 enters the basis and x1 leaves the basis.
Note that the new point has an objective value equal to 1, which is
an improvement over the previous objective value of 3. the
improvement is precisely (z3 − c3)x3 = 2.
Remark: CB is the coeﬃcient of the basic variable in the objective of the
LP.

Termination with an optimal solution
Consider the following problem, where A is an m × n matrix with rank m.
Minimize CT X
Subject to AX = b
X ≥ 0
Suppose that X∗ is a basic feasible solution with basis B; that is
X∗ =
B−1b
0
Let Z∗ denote the objective value at X∗, that is Z∗ = CT
B B−1b. Suppose
further that Zj − Cj ≤ 0, for all non basic variables, and hence there are no
non basic variables that are eligible to enter the basis. let x be any feasible
solution with objective function value z, then from equation (3.3) we have
Z∗ − Z = j∈J(Zj − Cj )xj (3.9)
Because Zj − Cj ≤ 0 and Xj ≥ 0 for all variables, then Z∗ ≤ Z, and so X∗
is an optimal basic feasible solution, since if (Zj − Cj ) ≤ 0 for all j ∈ J,
then the current basic feasible solution is optimal.

Example 2.
Minimize 2x1 − x2
subject to −x1 + x2 ≤ 2
2x1 + x2 ≤ 6
x1, x2 ≥ 0
Introducing the slack variables x3 and x4. This leads to the following
constraints
−x1 + x2 + x3 = 2
2x1 + x2 + x4 = 6
x1, x2, x3, x4 ≥ 0

Consider the basic feasible solution with basis
B = [a1, a2] =
−1 1
2 1
andB−1 =
−1/3 1/3
2/3 1/3
XB = B−1b − B−1NXN
xB1
xB2
=
x1
x2
=
−1/3 1/3
2/3 1/3
2
6
−
−1/3 1/3
2/3 1/3
1 0
0 1
x3
x4
=
4/3
10/3
−
−1/3
2/3
x3 −
1/3
1/3
x4 (3.10) Currently,
x3 = x4 = 0, x1 = 4/3 and x2 = 10/3, note that
z3 − c3 = CBB−1a3 − c3 =
(2, −1)
−1/3 1/3
2/3 1/3
1
0
− 0 = −4/3 0
1
0
− 0 = −4/3
z4 − c4 = CBB−1a4 − c4 = (2, −1)
−1/3 1/3
2/3 1/3
0
1
− 0 = 1/3

Hence, the objective improves by holding x3 non- basic and
introducing x4 in the basis.
Then x3 is kept at zero level, x4 is increased and x1 and x2 are
modiﬁed according to equation (3.10). we see that x4 can be
increased to 4, at which x1 drops to zero.
Any further increase of x4 results in violating the non-negativity of x1
and so x1 is the blocking variable.
With x4 = 4 and x3 = 0, the modiﬁed value of x1 and x2 are 0 and 2
respectively.
The new basic feasible solution is
(x1, x2, x3, x4) = (0, 2, 0, 4)
Note that a4 replaces a1, that is x1 drop from the basic and x4 enters
the basis.
The new set of basic and non basic variables and their values are
given as :
XB =
xB4
xB2
=
x4
x1
=
4
2
, XN =
x3
x1
=
0
0

Moving from old to the new basic feasible solution is illustrated in
ﬁgure given below, note that as x4 increases by one unit x1 decrease
by 1/3 unit and x2 decrease by 1/3 unit, that is we move in the
direction (-1/3, -1/3) in the (x1, x2) space.
This continues until we are blocked by the non negativity restriction
x1 ≥ 0.
At this point x1 drops to zero and leaves the basis.

The simplex Algorithm
The key solution concepts
The simplex method focuses on corner point feasible(CPF) solutions.
The simplex method is an iterative algorithm(a systematic solution
procedure that keeps repeating a ﬁxed series of steps, called an
iteration, until a desired result has been obtained ) with the following
structure.

Whenever possible, the initialization of the simplex method chooses
the origin point(all decision variables equal to zero) to be the initial
CPF solution
Given a CPF solution, it is much quicker computationally to gather
information about its adjacent CPF solutions than about other CPF
solutions. Therefore, each time the simplex method performs an
iteration to move from the current CPF solution to a better one, it
always chooses a CPF solution that is adjacent to the current one.
After the current CPF solution is identiﬁed, the simplex method
examines each of the edges of the feasible region that emanates from
this CPF solution. Each of these edges leads to an adjacent solution
at the other end, but the simplex method doesn’t even take the time
to solve for adjacent CPF solutions, instead, it simply identiﬁes the
rate of improvement in Z that would be obtained by moving along the
edge. and then chooses to move along the one with largest positive
rate of improvement.

A positive rate of improvement in Z implies that the adjacent CPF
solution is better than the current one, whereas a negative rate of
improvement in Z implies that the adjacent CPF solution is worse.
Therefore, the optimality test consists simply checking whether any of
the edges give a positive rate of improvement in Z. if none do, then
the current CPF solution is optimal.
The Simplex Method in Tabular Form
Steps
Initialization:
1 Convert (Transform) all the constraints to equality by introducing
slack, surplus, and artificial variables as follows
Constraint type Variable to be added
≤ +slack(s)
≥ -surplus(s)+artificial(A)
= + artificial(A)

The artiﬁcial variable refers to the kind of variable which is introduced in
the linear program model to obtain the initial basic feasible solution. It is
utilized for the equality constraints and for the greater than or equal
inequality constraints.
1 Construct the initial simplex tableau
Cj c1 . . . cn 0 . . . 0 0 . . . 0
CB BV X1 . . . Xn S1 . . . Sn A1 . . . An RHS
0 S b1
. . .
. . .
. . .
0 A bm
Z Z val
Zj − Cj
2 Test for optimality :
Case 1: In maximization problem the current bfs is optimal if every
element in the last row of the simplex tableau is nonnegative
Case 2 : In minimization problem the current bfs is optimal if every
element in the last row of the simplex tableau is non positive

Iteration :
Step 1 : Determine the entering basic variable by selecting the
variable (automatically a non basic variable ) with the most negative
value (in case of maximization)or the most positive(in case of
minimization) in the last row (Z-row). Put a box around the column
below this variable, and call it the ” pivot column”.
Step 2 : Determine the leaving basic variable by applying the
minimum ratio test as following:
1 Pick out each coeﬃcients in the pivot column that is strictly positive
2 Divide each of these coeﬃcients into the right hand side entry for the
same row
3 Identify the row that has the smallest of these ratios
4 The basic variable for that row is the leaving variable, so replace that
variable by the entering variable in the basic variable column of the next
simplex tableau. Put a box around this row and call it the ”pivot row”.
Step 3: Solve for the new basic feasible solution by using elementary
row operation(multiply or divide a row by a nonzero constant, add or
subtract a multiple of one row another row)to construct a new
simplex tableau, and then return the optimality test.

Example 1.
Solve the following problem using the simplex method
Maximize z = 3x1 + 5x2
Subject to x1 ≤ 4
2x2 ≤ 12
3x1 + 2x2 ≤ 18
x1, x2 ≥ 0
Solution: 1. Write the standard form of the above linear programming
problem.
Maximize z = 3x1 + 5x2
Subject to x1 + s1 = 4
2x2 + s2 = 12
3x1 + 2x2 + s3 = 18
x1, x2, s1, s2, s3 ≥ 0

2. Initial tableau
C 3 5 0 0 0
CB XB x1 x2 s1 s2 s3 b(RHS)
0 s1 1 0 1 0 0 4
0 s2 0 2 0 1 0 12
0 s3 3 2 0 0 1 18
Zj − Cj -3 -5 0 0 0
The basic feasible solution at the initial tableau is (0,0,4,12,18) where:
x1 = 0, x2 = 0, s1 = 4, s2 = 12, s3 = 18 and z = 0. Where s1, s2, s3 are
basic variable and x1 and x2 are non-basic variables. The solution at the
initial tableau is associated to the origin point at which all the decision
variables are zero.
Optimality test: By investigating the last row of the initial tableau, we
ﬁnd that there are some negative numbers. Therefore, the current solution
is not optimal.

Iteration
Step 1: Determine the entering variable by selecting the variable with the
most negative in the last row. From the initial tableau, in the last row
((Zj − Cj ) row), the coeﬃcient of X1 is -3 and the coeﬃcient of X2 is -5;
therefore, the most negative is -5. consequently, X2 is the entering
variable. X2 is surrounded by a box and it is called the pivot column.
Step 2: Determining the leaving variable by using the minimum ratio test
as follows:
C 3 5 0 0 0
CB XB x1 x2(Entering) s1 s2 s3 b(RHS)
0 s1 1 0 1 0 0 4
0 s2(Leaving) 0 2 0 1 0 12
0 s3 3 2 0 0 1 18
Zj − Cj -3 -5 0 0 0

Step 3: solving for the new BF solution by using the eliminatory row
operations as follows:
C 3 5 0 0 0
0 s1 1 0 1 0 0 4
5 x2 0 1 0 1/2 0 6
0 s3 3 0 0 -1 1 6
Zj − Cj -3 0 0 5/2 0
This solution is not optimal, since there is a negative numbers in the last
row. Apply the same rules we will obtain this solution:
C 3 5 0 0 0
0 s1 0 0 1 1/3 -1/3 2
5 x2 0 1 0 1/2 0 6
3 x1 1 0 0 -1/3 1/3 2
Zj − Cj 0 0 0 3/2 1 36
This solution is optimal; since there is no negative solution in the last row:
basic variables are x1 = 2, x2 = 6 and s1 = 2; the nonbasic variables are
s2 = s3 = 0, and the optimal value Z = 36.

Example 2.
Minimize x1 + x2 − 4x3
subject to x1 + x2 + 2x3 ≤ 9
x1 + x2 − x3 ≤ 2
−x1 + x2 + x3 ≤ 4
x1, x2, x3 ≥ 0
Solution : Introducing the non- negative slack variables s1, s2 and s3. The
problem becomes the following :
Minimize x1 + x2 − 4x3
subject to x1 + x2 + 2x3 + s1 = 9
x1 + x2 − x3 + s2 = 2
−x1 + x2 + x3 + s3 = 4
x1, x2, x3, s1, s2, s3 ≥ 0

Since b ≥ 0, then we can choose our initial basis as B = [a4, a5, a6] = I3,
and we indeed have B−1b = b ≥ 0. This gives the following initial tableau:
Iteration 1
C 1 1 -4 0 0 0 0
CB XB x1 x2 x3 s1 s2 s3 b(RHS)
0 s1 1 1 2 1 0 0 9
0 s2 1 1 -1 0 1 0 2
0 s3 -1 1 1 0 0 1 4
Zj − Cj -1 -1 4 0 0 0
The initial basic feasible solution in the above table is
x1 = 0, x2 = 0, x3 = 0, but it is not optimal, since there is positive element
the last row of the simplex table (Zj − Cj ). Identify the entering and
leaving variable

Iteration 2
C 1 1 -4 0 0 0 0
0 s1 1 1 2 1 0 0 9
0 s2 1 1 -1 0 1 0 2
0 s3 -1 1 1 0 0 1 4
Zj − Cj -1 -1 4 0 0 0
From the above simplex table s3 is the leaving variable and x3 is the
entering variable.
C 1 1 -4 0 0 0 0
0 s1 3 -1 0 1 0 -2 1
0 s2 0 2 0 0 1 1 6
-4 x3 -1 1 1 0 0 1 4
Zj − Cj 3 -5 0 0 0 -4
The initial basic feasible solution in the above table is x1 = 0, x2 = 0, x3 = 4, but
it is not optimal, since there is positive element in the last row of the simplex
table (Zj − Cj ).(Department of Mathematics Debark University )Linear Optimization(MATH 2062) October 7, 2020 50 / 107

Iteration 3
C 1 1 -4 0 0 0 0
0 s1 3 -1 0 1 0 -2 1
0 s2 0 2 0 0 1 1 6
-4 x3 -1 1 1 0 0 1 4
Zj − Cj 3 -5 0 0 0 -4
From the above simplex table s1 is the leaving variable and x1 is the
entering variable.
C 1 1 -4 0 0 0
1 x1 1 -1/3 0 1/3 0 -2/3 1/3
0 s2 0 2 0 0 1 1 6
-4 x3 0 2/3 1 1/3 0 1/3 13/3
Zj − Cj 0 -4 0 -1 0 -2
This is an optimal tableau since Zj − Cj ≤ 0 in the last row. The optimal solution
is given by x1 = 1/3, x2 = 0, x3 = 13/3, with z = -17.

Degeneracy and Finiteness of Simplex Algorithm
Degeneracy: 1. only one component of basic feasible solution
corresponding to the basic variable is zero.
2. at least two component of basic feasible solution corresponding to the
basic variables are zero.
Example 1:
Maximize 5x1 + 3x2
Subject to x1 + x2 ≤ 2
5x1 + 2x2 ≤ 10
x1 + 8x2 ≤ 12
x1, x2 ≥ 0
Solution :The standard linear programming problem for the above problem is
Maximize 5x1 + 3x2
Subject to x1 + x2 + s1 = 2
5x1 + 2x2 + s2 = 10
x1 + 8x2 + s3 = 12
x1, x2, s1, s2, s3 ≥ 0

Now we can construct the simplex tableau
C 5 3 0 0 0
0 s1 1 1 1 0 0 2 ←
0 s2 5 2 0 1 0 10
0 s3 3 8 0 0 1 12
Zj − Cj ↑-5 -3 0 0 0
The initial basic feasible solution of the LP problem is x1 = 0, x2 = 0, but
it can not be an optimal solution, since there is a negative element in the
last row of the simplex table. from the above table we can identify the
entering and leaving variables, thus x1 is the entering variable and s1 is the
leaving variable.

By applying elementary row operation we can get the following simplex
tableau
C 5 3 0 0 0
5 x1 1 1 1 0 0 2
0 s2 0 -3 -5 1 0 0
0 s3 0 5 -3 0 1 6
Zj − Cj 0 2 5 0 0
Thus the optimal solution is (x1, x2, s1, s2, s3) = (2, 0, 0, 0, 6) and the
optimal value is 10, since there exist a zero value of basic variable ,it is
degenerate case of ﬁniteness of simplex algorithm.

Inﬁnite number of solutions
The alternative optimal solution can be obtained by considering the zj − cj
row of the simplex table. That is, zj − cj = 0 for some non-basic variable
columns in the optimal simplex table.
Example
Maximize 10x1 + 20x2
subject to x1 ≤ 10
x2 ≤ 6
2x1 + 4x2 ≤ 36
x1, x2 ≥ 0
Solution: The standard linear programming problem of the above problem
is
Maximize 10x1 + 20x2
subject to x1 + s1 = 10
x2 + s2 = 6
2x1 + 4x2 + s3 = 36
x1, x2, s1, s2, s3 ≥ 0

Now construct the simplex tableau.
C 10 20 0 0 0
0 s1 1 0 1 0 0 10
0 s2 0 1 0 1 0 6 ←
0 s3 2 4 0 0 1 36
Zj − Cj -10 ↑ -20 0 0 0
C 10 20 0 0 0
0 s1 1 0 1 0 0 10
20 x2 0 1 0 1 0 6
0 s3 2 0 0 -4 1 12←
Zj − Cj ↑ -10 0 0 20 0

C 10 20 0 0 0
0 s1 0 0 1 2 -1/2 4
20 x2 0 1 0 1 0 6
10 x1 1 0 0 -2 1/2 6
Zj − Cj 0 0 0 0 5 180
We have basic feasible solution (x1, x2) = (6, 6) with optimal value z =
180. We apply a new simplex step with Zj − Cj = 0 for a non- basic
variable.
C 10 20 0 0 0
0 s1 0 0 1 2 -1/2 4
20 x2 0 1 0 1 0 6
10 x1 1 0 0 -2 1/2 6
Zj − Cj 0 0 0 0 5 180

C 10 20 0 0 0
0 s2 0 0 1/2 1 -1/4 2
20 x2 0 1 -1/2 0 1/4 4
10 x1 1 0 1 0 0 10
Zj − Cj 0 0 0 0 5 180
Now we have another basic solution (x1, x2) = (10, 4), but the optimal
values remains 180.
In our case, since (x1, x2) = (6, 6) and (x1, x2) = (10, 4) are solutions all
points of the segment (x1, x2)(x1, x2) = λ(6, 6) + (1 − λ)(10, 4), λ ∈ [0, 1]
are solutions.

Finding a Starting Basic Feasible Solution
In certain cases, it is diﬃcult to obtain an initial basic feasible solution of
the given LP problem. Such cases arise
1 when the constraints are of the ≤ type,
n
j=1
aij xj ≤ bi , xj ≥ 0
and value of few right-hand side constants is negative [i.e. bi < 0].
After adding the non-negative slack variable si (i = 1, 2, . . ., m), the
initial solution so obtained will be si = −bi for a particular resource, i.
This solution is not feasible because it does not satisfy non-negativity
conditions of slack variables (i.e. si ≥ 0).
2 when the constraints are of the ≥ type,
n
j=1
aij xj ≥ bi , xj ≥ 0

After adding surplus (negative slack) variable si , the initial solution so
obtained will be −si = bi or si = −bi
n
j=1
aij xj − si = bi , xj ≥ 0, si ≥ 0
This solution is not feasible because it does not satisfy non-negativity
conditions of surplus variables (i.e. si ≥ 0).
In such a case, artiﬁcial variables, Ai (i = 1, 2, . . ., m) are added to
get an initial basic feasible solution.
The resulting system of equations then becomes:
n
j=1
aij xj − si + Ai = bi
xj , si , Ai ≥ 0, i = 1, 2, ..., m

These are m simultaneous equations with (n + m + m) variables (n
decision variables, m artificial variables and m surplus variables).
An initial basic feasible solution of LP problem with such constraints
can be obtained by equating (n + 2m - m) = (n + m) variables equal
to zero.
Thus the new solution to the given LP problem is: Ai = bi (i = 1, 2,
. . . , m), which is not the solution to the original system of
equations because the two systems of equations are not equivalent.
Thus, to get back to the original problem, artificial variables must be
removed from the optimal solution.
There are two methods for removing artificial variables from the solution.
Two-Phase Method
Big-M Method or Method of Penalties
Remark Before the optimal solution is reached, all artificial variables must
be dropped out from the solution mix. This is done by assigning
appropriate coefficients to these variables in the objective function. These
variables are added to those constraints with equality (=) and greater than
or equal to (≥) sign.

Two -phase method
In the first phase of this method, the sum of the artificial variables is
minimized subject to the given constraints in order to get a basic
feasible solution of the LP problem.
The second phase minimizes the original objective function starting
with the basic feasible solution obtained at the end of the first phase.
Since the solution of the LP problem is completed in two phases, this
method is called the two-phase method.
Advantages of the method
No assumptions on the original system of constraints are made, i.e.
the system may be redundant, inconsistent or not solvable in
non-negative numbers.
It is easy to obtain an initial basic feasible solution for Phase I.
The basic feasible solution (if it exists) obtained at the end of phase I
is used as initial solution for Phase II

Steps of the Algorithm: Phase I
Step 1 If all the constraints in the given LP problem are less than or equal
to (≤) type, then Phase II can be directly used to solve the problem.
Otherwise, the necessary number of surplus and artificial variables are
added to convert constraints into equality constraints.
Step 2: Assign zero coefficient to each of the decision variables (xj ) and
to the surplus variables; and assign - 1 coefficient to each of the artificial
variables. This yields the following auxiliary LP problem.
Maximize Z∗ = m
i=1(−1)Ai
subject to the constraints
m
i=1
aij xj + Ai = bi , i = 1, 2, ..., m
and xj , Aj ≥ 0

Step 3: Apply the simplex algorithm to solve this auxiliary LP problem.
The following three cases may arise at optimality.
(i) Max Z∗ = 0 and at least one artificial variable is present in the basis
with positive value. This means that no feasible solution exists for
the original LP problem.
(ii) Max Z∗ = 0 and no artificial variable is present in the basis. This
means that only decision variables(xj ’s) are present in the basis and
hence proceed to Phase II to obtain an optimal basic feasible solution
on the original LP problem.
(iii) Max Z∗ = 0 and at least one artificial variable is present in the basis
at zero value. This means that a feasible solution to the auxiliary LP
problem is also a feasible solution to the original LP problem. In
order to arrive at the basic feasible solution, proceed directly to
Phase II or else eliminate the artificial basic variable and then proceed
to Phase II.

Remark Once an artificial variable has left the basis, it has served
its purpose and can, therefore, be removed from the simplex table.
An artificial variable is never considered for re-entry into the basis.
Phase II
Assign actual coefficients to the variables in the objective function
and zero coefficient to the artificial variables which appear at zero
value in the basis at the end of Phase I.
The last simplex table of Phase I can be used as the initial simplex
table for Phase II.
Then apply the usual simplex algorithm to the modified simplex table
in order to get the optimal solution to the original problem.
Artificial variables that do not appear in the basis may be removed.

Remark
1 All artificial vectors are not present in the bases which
indicates that all artificial vectors at zero level at the optimal
stage, thus the solution obtained is a basic feasible solution.
2 Some artificial vectors are present in the basis and some
artificial vectors are at positive level at the optimal stage. In
that case there are no feasible solution to the problem.
3 All artificial are at zero level but at least one artificial vector is
present in the basis at optimal stage. Here the solution under
test is an optimal solution. Here the converted equations are
consistent but some of the constraints may be redundant. By
redundancy means the system has more than enough
constraints.

Example 1.
Solve the following Linear programming problems by using two phase
method
Maximize Z = 3x1 − x2
subject to 2x1 + x2 ≥ 2
x1 + 3x2 ≤ 2
x2 ≤ 4
x1, x2 ≥ 0
Solution:Phase I Convert the above problem to Standard form of LPP
Maximize Z = 3x1 − x2/Z∗ = −A1
subject to 2x1 + x2 − s1 + A1 = 2
x1 + 3x2 + s2 = 2
x2 + s3 = 4
x1, x2, s1, s2, s3, A1 ≥ 0

Now, we will Max Z∗ = −A1 till zero in order to remove the variable A1,
where A1 is an artiﬁcial variable.
Maximize Z∗ = 0x1 − 0x2 + 0s1 + 0s2 + 0s3 − 1A1
subject to 2x1 + x2 − s1 + A1 = 2
x1 + 3x2 + s2 = 2
x2 + s3 = 4
x1, x2, s1, s2, s3, A1 ≥ 0
Cj 0 0 0 0 0 -1
CB XB x1 x2 s1 s2 s3 A1 b(RHS) Min ratio bi /yik
-1 A1 2 1 -1 0 0 1 2 1 ←
0 s2 1 3 0 1 0 0 2 2
0 s3 0 1 0 0 1 0 4 -
Zj − Cj ↑-2 -1 1 0 0 0 Z∗ = −2

Cj 0 0 0 0 0 -1
CB XB x1 x2 s1 s2 s3 A1 b(RHS)
0 x1 1 1/2 -1/2 0 0 × 1
0 s2 0 5/2 1/2 1 0 × 1
0 s3 0 1 0 0 1 × 4
Zj − Cj 0 0 0 0 0 × Z∗ = 0
Since all Zj − Cj = 0 and no artiﬁcial vector appears in the the basis, we
proceed to phase II.
Phase II
Cj 3 -1 0 0 0
3 x1 1 1/2 -1/2 0 0 1
0 s2 0 5/2 1/2 1 0 1 ←
0 s3 0 1 0 0 1 4
Zj − Cj 0 5/2 ↑ -3/2 0 0 Z = 3

Cj 3 -1 0 0 0
3 x1 1 3 0 1 0 2
0 s1 0 5 1 2 0 2
0 s3 0 1 0 0 1 4
Zj − Cj 0 10 0 3 0 Z = 6
Since all Zj − Cj ≥ 0, optimal basic feasible solution is obtained, therefore
the solution is Max Z = 6, x1 = 2, x2 = 0.

Example 2.
Maximize Z = 5x1 + 8x2
subject to 3x1 + 2x2 ≥ 3
x1 + 4x2 ≥ 4
x1 + x2 ≤ 5
x1, x2 ≥ 0
Solution: The standard linear programming problem
subject to 3x1 + 2x2 − s1 + A1 = 3
x1 + 4x2 − s2 + A2 = 4
x1 + x2 + s3 = 5
x1, x2, s1, s2, s3, A1, A2 ≥ 0

Auxiliary LPP
Maximize Z = 0x1 + 0x2 + 0s1 + 0s2 + 0s3 − A1 − A2
subject to 3x1 + 2x2 − s1 + A1 = 3
x1 + 4x2 − s2 + A2 = 4
x1 + x2 + s3 = 5
x1, x2, s1, s2, s3, A1, A2 ≥ 0
Phase I
Cj 0 0 0 0 0 -1 -1
CB XB x1 x2 s1 s2 s3 A1 A2 b(RHS) min ratio bi
-1 A1 3 2 -1 0 0 1 0 3 3/2
-1 A2 1 4 0 -1 0 0 1 4 1 →
0 s3 1 1 0 0 1 0 0 5 5
Zj − Cj -4 ↑ −6 1 1 0 0 0 Z∗ = −7

Cj 0 0 0 0 0 -1 -1
CB XB x1 x2 s1 s2 s3 A1 A2 b(RHS) min rat
-1 A1 5/2 0 -1 1/2 0 1 × 1 2/
0 x2 1/4 1 0 -1/4 0 0 × 1
0 s3 3/4 0 0 1/4 1 0 × 4 16
Zj − Cj ↑ −5/2 0 1 -1/2 0 0 × Z∗ = −1
Cj 0 0 0 0 0 -1 -1
CB XB x1 x2 s1 s2 s3 A1 A2 b(RHS) min ratio
0 x1 1 0 -2/5 1/5 0 × × 2/5
0 x2 0 1 1/10 -3/10 0 × × 9/10
0 s3 0 0 3/10 1/10 1 × × 37/10
Zj − Cj 0 0 0 0 0 × × Z∗ = 0
Since all Zj − Cj ≥ 0, Maz zx = 0, and no artiﬁcial vector appears in the
basis, we proceed to phase II.

Phase II
Cj 5 8 0 0 0
CB XB x1 x2 s1 s2 s3 b(RHS) min ratio bi /yik
5 x1 1 0 -2/5 1/5 0 2/5 2 →
8 x2 0 1 1/10 -3/10 0 9/10 1
0 s3 0 0 3/10 1/10 1 37/10 37
Zj − Cj 0 0 -6/5 ↑ −7/5 0 Z = 46/5
Cj 5 8 0 0 0
0 s2 5 0 -2 1 0 2 -
8 x2 3/2 1 -1/2 0 0 3/2 -
0 s3 -1/2 0 1/2 0 1 7/2 7→
Zj − Cj 7 0 ↑ −4 0 0 Z = 12

Cj 5 8 0 0 0
0 s2 3 0 0 1 2 16
8 x2 1 1 0 0 1/2 5
0 s1 -1 0 1 0 2 7
Zj − Cj 3 0 0 0 4 Z = 40
Since all Zj − Cj ≥ 0, optimal basic feasible solution is obtained. Therefore

Big - M method
Big-M method is another method of removing artificial variables from
the basis.
In this method, large undesirable (unacceptable penalty) coefficients
to artificial variables are assigned from the point of view of the
objective function.
If the 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.
The penalty is supposed to be designated by - M, for a maximization
problem, and + M, for a minimization problem, where M > 0.

The Big-M method for solving an LPP can be summarized in
the following steps:
Step 1: Express the LP problem in the standard form by adding slack
variables, surplus variables and/or artificial variables. Assign a zero
coefficient to both slack and surplus variables. Then assign a very large
coefficient + M (minimization case) and - M (maximization case) to
artificial variable in the objective function.
Step 2: The initial basic feasible solution is obtained by assigning zero
value to decision variables, x1, x2, ..., etc.

Step 3: Calculate the values of zj − cj in last row of the simplex table and
examine these values.
1 If all zj − cj ≤ 0, then the current basic feasible solution is optimal.
2 If for a column, k, zk − ck is most positive and all entries in this
column are negative, then the problem has an unbounded optimal
solution.
3 If one or more zj − cj > 0 (minimization case), then select the
variable to enter into the basis (solution mix) with the largest positive
zj − cj value (largest per unit increase in the objective function
value). This value also represents the opportunity cost of not having
one unit of the variable in the solution. That is,
zk − ck = Max{zj − cj : zj − cj > 0}
Step 4: Determine the key row and key element in the same manner as
discussed in the simplex algorithm.
Step 5: Continue with the procedure to update solution at each iteration
till optimal solution is obtained.

Remarks At any iteration of the simplex algorithm one of the fol-
lowing cases may arise:
1 If at least one artificial variable is a basic variable (i.e., variable
that is present in the basis) with zero value and the coefficient
it M in each zj − cj ( j = 1, 2, . . ., n) values is non-positive,
then the given LP problem has no solution. That is, the
current basic feasible solution is degenerate.
2 If at least one artificial variable is present in the basis with a
positive value and the coefficients M in each zj − cj ( j = 1, 2,
. . ., n) values is non-positive, then the given LP problem has
no optimum basic feasible solution. In this case, the given LP
problem has a pseudo optimum basic feasible solution.

Example 1.
Solve the following linear programming problem by using Big-M method.
Maximize Z = −2x1 − x2
subject to 3x1 + x2 = 3
4x1 + 3x2 ≥ 6
x1 + 2x2 ≤ 4
x1, x2 ≥ 0
Solution : The standard linear programming problem of the above LP
Maximize Z = −2x1 − x2 + 0s1 + 0s2 − MA1 − MA2
subject to 3x1 + x2 + A1 = 3
4x1 + 3x2 − s1 + A2 = 6
x1 + 2x2 + s2 = 4
x1, x2, s1, s2, A1, A2 ≥ 0

Cj -2 -1 0 0 -M -M
CB XB x1 x2 s1 s2 A1 A2 b(RHS) min ratio
-M A1 3 1 0 0 1 0 3 3/3 = 1 →
-M A2 4 3 -1 0 0 1 6 6/4 = 3/2
0 s2 1 2 0 1 0 0 4 4/1 = 4
Zj − Cj ↑ 2 − 7M 1-4M M 0 0 0 Z = -9M
Cj -2 -1 0 0 -M -M
-2 x1 1 1/3 0 0 × 0 1 1/1/3 = 3
-M A2 0 5/3 -1 0 × 1 2 2/5/3 = 6/5 →
0 s2 0 5/3 0 1 × 0 3 3/5/3 = 9/5
Zj − Cj 0 ↑ −5M+1
3 M 0 × 0 Z = -2-2m

Cj -2 -1 0 0 -M -M
-2 x1 1 0 1/5 0 × × 3/5
-1 x2 0 1 -3/5 0 × × 6/5
0 s2 0 0 1 1 × × 1
Zj − Cj 0 0 1/5 0 × × Z = -12/5
Since all Zj − Cj ≥ 0, optimal basic feasible is obtained. Therefore the solution is
Max Z = −12/5, x1 = 3/5, x2 = 6/5

Example 2.
Maximize Z = 3x1 − x2
Subject to 2x1 + x2 ≥ 2
x1 + 3x2 ≤ 3
x2 ≤ 4
x1, x2 ≥ 0
Solution: The standard linear programming problem
Maximize Z = 3x1 − x2 + 0s1 + 0s2 + 0s3 − MA1
Subject to 2x1 + x2 − s1 + A1 = 2
x1 + 3x2 + s2 = 3
x2 + s3 = 4
x1, x2, s1, s2, s3, A1 ≥ 0

Cj 3 -1 0 0 0 -M
CB XB x1 x2 s1 s2 s3 A1 b(RHS) min ratio
-M A1 2 1 -1 0 0 1 2 2/2 = 1 →
0 s2 1 3 0 1 0 0 3 3/1 = 3
0 s3 0 1 0 0 1 0 4 -
Zj − Cj ↑ −2M − 3 -M + 1 M 0 0 0 Z = -2M
Cj 3 -1 0 0 0 -M
CB XB x1 x2 s1 s2 s3 A1 b(RHS) min ratio bi /yik
3 x1 1 1/2 -1/2 0 0 × 1 -
0 s2 0 5/2 1/2 1 0 × 2 2/1/2 = 4 →
0 s3 0 1 0 0 1 × 4 -
Zj − Cj 0 5/2 ↑ −3/2 0 0 × Z = 3

Cj 3 -1 0 0 0 -M
CB XB x1 x2 s1 s2 s3 A1 b(RHS) min ratio bi /yik
3 x1 1 3 0 1/2 0 × 3
0 s1 0 5 1 2 0 × 4
0 s3 0 1 0 0 1 × 4
Zj − Cj 0 10 0 3/2 0 × Z = 9
Since all Zj − Cj ≥, optimal basic feasible solution is obtained. Therefore,

Example 3
Maximize Z = 3x1 + 2x2 + x3
subject to 2x1 + x2 + x3 = 12
3x1 + 4x2 = 11
and x1 is unrestricted, x2, x3 ≥ 0
Solution:The standard linear programming problem
Maximize Z = 3(x1 − x1 ) + 2x2 + x3 − MA1 − MA2
subject to 2(x1 − x1 ) + x2 + x3 + A1 = 12
3(x1 − x1 ) + 4x2 + A2 = 11
x1, x1 , x2, x3, A1, A2 ≥ 0
Maximize Z = 3x1 − 3x1 + 2x2 + x3 − MA1 − MA2
subject to 2x1 − 2x1 + x2 + x3 + A1 = 12
3x1 − 3x1 + 4x2 + A2 = 11
x1, x1 , x2, x3, A1, A2 ≥ 0

Cj 3 -3 2 1 -M -M
CB XB x1 x1 x2 x3 A1 A2 b(RHS)
-M A1 2 -2 1 1 1 0 12
-M A2 3 -3 4 0 0 1 11
Zj − Cj ↑ −5M − 3 5M + 3 -5M -2 -M- 1 0 0 Z = -23M
Cj 3 -3 2 1 -M -M
CB XB x1 x1 x2 x3 A1 A2 b(RHS) m
-M A1 0 0 -5/3 1 1 × 14/3 14
3 x1 1 -1 4/3 0 0 × 11/3 -
Zj − Cj 0 0 5/3M+1 ↑ −M − 1 0 × Z = −14M
3 + 11

Cj → 3 -3 2 1 -M -M
CB ↓ XB ↓ x1 x1 x2 x3 A1 A2 b(RHS) min ratio
1 x3 0 0 -5/3 1 × × 14/3
3 x1 1 -1 4/3 0 × × 11/3 -
Zj − Cj → 0 0 1/3 0 × × Z = 47/3
Since all Zj − Cj ≥ 0, optimal basic feasible solution is obtained
x1 = 11/3, x1 = 0
x1 = x1 − x1 = 11/3 − 0 = 11/3
Therefore the solution is Max Z = 47/3, x1 = 11/3, x2 = 0, x3 = 14/3.

Some Complications and Their Resolution
Unrestricted Variables
Let variable xr be unrestricted in sign. We deﬁne two new variables say xr
and xr such that
xr = xr − xr ; &xr , xr ≥ 0

Tie for Entering Basic Variable (Key Column)
While solving an LP problem using simplex method two or more
columns of simplex table may have same zj − cj value (positive or
negative depending upon the type of LP problem).
In order to break this tie, the selection for key column (entering
variable) can be made arbitrary.
However, the number of iterations required to arrive at the optimal
solution can be minimized by adopting the following rules.
(i) If there is a tie between two decision variables, then the selection can
be made arbitrarily.
(ii) If there is a tie between a decision variable and a slack (or surplus)
variable, then select the decision variable to enter into basis.
(iii) If there is a tie between two slack (or surplus) variables, then the
selection can be made arbitrarily

Tie for Leaving Basic Variable (Key Row) Degeneracy
While solving an LP problem a situation may arise where either the
minimum ratio to identify the basic variable to leave the basis is not
unique or value of one or more basic variables in the xB becomes
zero.This causes the problem of degeneracy.
In order to break the tie in the minimum ratios, the selection can be
made arbitrarily.
However, the number of iterations required to arrive at the optimal
solution can be minimized by adopting the following rules.
(i) Divide the coefficients of slack variables in the simplex table where
degeneracy is seen by the corresponding positive numbers of the key
column in the row, starting from left to right.
(ii) Compare the ratios in step (i) from left to right column wise, select the
row that contains the smallest ratio.
Remark When there is a tie between a slack and artificial variable to
leave the basis, preference should be given to the artificial variable for
leaving the basis.

Example
Solve the following LP problem
x1 + 4x2 ≤ 8,
x1 + 2x2 ≤ 4
and x1, x2 ≥ 0
Solution Adding slack variables s1 and s2 to the constraints, the problem
can be expressed as
Maximize Z = 3x1 + 9x2 + 0s1 + 0s2
x1 + 4x2 + s1 = 8,
x1 + 2x2 + s2 = 4
and x1, x2, s1, a2 ≥ 0

The initial basic feasible solution is given in Table . Both variables s1 and
s2 are eligible to leave the basis as the minimum ratio is same, i.e. 2, so
there is a tie among the ratio in rows s1 and s2.This causes the problem of
degeneracy. To obtain the key row for resolving degeneracy, apply the
following procedure:
cj 3 9 0 0
CB XB x1 x2 s1 s2 b(RHS) min. ratio
0 s1 1 4 1 0 8 8/4=2
0 s2 1 2 0 1 4 4/2=2
Zj − Cj -3 ↑-9 0 0 0
Dividing the coeﬃcients of slack variables s1 and s2 by the corresponding
elements in the key column as shown below in the table.
x2 column
Row (Key Column) s1 s2
s1 4 1/4=1/4 0/4=0
s2 2 0/2=0 1/2=1/2

Comparing the ratios of Step (ii) from left to right column wise, the
minimum ratio (i.e., 0/2 = 0) occurs in the s2-row. Thus, variable s2 is
selected to leave the basis. The new solution is shown in Table
cj 3 9 0 0
CB XB x1 x2 s1 s2 b(RHS)
0 s1 -1 0 1 -2 0
9 x2 1/2 1 0 1/2 2
Zj − Cj 3/2 0 0 9/2 18
In Table above , all Zj − Cj ≤ 0. Hence, an optimal solution is arrived at.
The optimal basic feasible solution is: x1 = 0, x2 = 2 and Max Z = 18.

Types of Linear Programming Solutions
While solving any LP problem using simplex method, at the stage of
optimal solution, the following three types of solutions may exist:
Unbounded Solution
Infeasible Solutions

The zj − cj values in the simplex table indicates the contribution in
the objective function value by each unit of a variable chosen to enter
into the basis.
Also, Also, an optimal solution to a maximization(minimization) LP
problem is reached when all zj − cj ≥ 0(zj − cj ≤ 0).
But, if zj − cj = 0 for a non-basic variable column in the optimal
simplex table and such nonbasic variable is chosen to enter into the
basis, then another optimal solution so obtained will show no
improvement in the value of objective function.

Example
Solve the following LP problem.
2x1 + 3x2 ≤ 30,
3x1 + 2x2 ≤ 24
,
x1 + x2 ≥ 3
and
x1, x2 ≥ 0

Solution Adding slack variables s1, s2, surplus variable s3 and artiﬁcial
variable A1 in the constraint set,the LP problem becomes
Maximize Z = 6x1 + 4x2 + 0s1 + 0s2 + 0s3 − MA1
2x1 + 3x2 + s1 = 30,
3x1 + 2x2 + s2 = 24
,
x1 + x2 − s3 + A1 = 3
and
x1, x2, s1, s2, A1 ≥ 0
The optimal solution: x1 = 8, x2 = 0 and Max Z = 48 for this LP problem
is shown in Table below.

Cj 6 4 0 0 0
CB XB x1 x2 s1 s2 s3 b(RHS) Ratio
0 s1 0 5/3 1 -2/3 0 14 42/5 ←
0 s3 0 -1/3 0 1/3 1 5
6 x1 1 2/3 0 1/3 0 8 12
zj − cj 0 ↑ 0 0 2 0 48
In Table above , z2 − c2 = 0 corresponds to a non-basic variable, x2. Thus, an
alternative optimal solution can also be obtained by entering variable x2 into the
basis and removing basic variable, s1 from the basis. The new solution is shown in
Table below.
Cj 6 4 0 0 0
4 x2 0 1 3/5 -2/5 0 42/5
0 s3 0 0 1/5 1/5 1 39/5
6 x1 1 0 -2/5 3/5 0 12/5
zj − cj 0 0 0 2 0 48
The optimal solution shown in Table above is: x1 = 12/5, x2 = 42/5 and Max Z
= 48. Since this optimal solution shows no change in the value of objective
function, it is an alternative solution. Once again, z3 − c3 = 0 corresponds to
nonbasic variable, s1. This indicates that an alternative optimal solution exists.

Unbounded Solution
In a maximization LP problem, if zj − zj < 0(zj − cj > 0 for a
minimization case) corresponds to a non-basic variable column in
simplex table, and all aij values in this column are negative, then
minimum ratio to decide basic variable to leave the basis can not be
calculated.
It is because negative value in denominator would indicate the entry
of a non-basic variable in the basis with a negative value (an
infeasible solution).
Also,a zero value in the denominator would result in a ratio having an
inﬁnite value and would indicate that the value of non-basic variable
could be increased inﬁnitely with any of the current basic variables
being removed from the basis.

Example
x1 − 2x2 ≤ 6,
3x1 ≤ 10
,
x2 ≥ 1
and
x1, x2 ≥ 0
Solution:Adding slack variables s1, s2 , surplus variable s3 and artiﬁcial
variable A1 in the constraint set.

Then the standard form of LP problem becomes
Maximize Z = 3x1 + 5x2 + 0s1 + 0s2 + 0s3 − MA1
x1 − 2x2 + s1 = 6,
3x1 + s2 = 10
,
x2 − s3 + A1 = 1
and
x1, x2, s1, s2, s3, A1 ≥ 0
The initial solution to this LP problem is shown in Table below.
Cj 3 5 0 0 0 -M
CB xB x1 x2 s1 s2 s3 A1 b(RHS) min. ratio
0 s1 1 -2 1 0 0 0 6 -
0 s2 1 0 0 1 0 0 10 -
-M A1 0 1 0 0 -1 1 1 1 ←
zj − cj -3 ↑-5-M 0 0 M 0 M

Iteration 1: Since z2 − c2 ≥ 0, non-basic variable x2 is chosen to enter into
the basis in place of basic variable A1. The new solution is shown in Table
below.
Cj 3 5 0 0 0 -M
CB xB x1 x2 s1 s2 s3 A1 b(RHS)
0 s1 1 0 1 0 -2 2 8
0 s2 1 0 0 1 0 0 10
5 x2 0 1 0 0 -1 1 1
zj − cj -3 0 0 0 -5 M+5 5
In Table above, z5 − c5 = −5 is largest negative, so variable s3 should enter
into the basis. But, coefficients in the s3 column are all negative or zero.
This indicates that s3 cannot be entered into the basis. However,the value
of s3 can be increased infinitely without removing any one of the basic
variables. Further, since s3 is associated with x2 in the third constraint, x2
will also be increased infinitely because it can be expressed as
x2 = 1 + s3 − A1. Hence, the solution to the given problem is unbounded.

Infeasible Solution
Lp models with inconsistent constraints have no feasible solution.
This situation can never occur if all the constraints are of the type ≤
with nonnegative right hand sides because the slacks provide a
feasible solution.
For other types of constraints we use artificial variables.
Although the artificial variables are penalized in the objective function
to force them to zero at the optimum, this can occur only if the
model has a feasible space.
Otherwise, at least one artificial variables will be positive in the
optimum iteration.
If LP problem solution does not satisfy all of the constraints, then
such a solution is called infeasible solution.
Also, infeasible solution occurs when all zj − cj values satisfy optimal
solution condition but at least one of artificial variables appears in the
basis with a positive value.
This situation may occur when an LP model is either improperly
formulated or more than two of the constraints are incompatible.

Example
x1 + x2 ≤ 5,
x2 ≥ 8
and
x1, x2 ≥ 0
Solution:Adding slack, surplus and artiﬁcial variables, the standard form
of LP problem becomes
Maximize Z = 6x1 + 4x2 + 0s1 + 0s2 − MA1
x1 + x2 + s1 = 5,
x2 − s2 + A1 = 8
and x1, x2, s1, s2, A1 ≥ 0

The initial solution to this LP problem is shown in Table below.
Cj 6 4 0 0 -M
CB XB x1 x2 s1 s2 A1 b(RHS) min. ratio
6 x1 1 1 1 0 0 5 5/1←
-M A1 0 1 0 -1 1 8 8/1
zj − cj 0 ↑2-M 6 M 0 30-8M
Iteration 1: Since z2 − c2 = 2 − M(≤ 0), non-basic variable x2 is chosen to enter
into the basis to replace basic variable x1. The new solution is shown in Table
below.
Cj 6 4 0 0 -M
CB XB x1 x2 s1 s2 A1 b(RHS)
4 x2 1 1 1 0 0 5
-M A1 -1 0 -1 -1 1 3
zj − cj M-2 0 4+M M 0 20-3M
In table above, since all zj − cj ≥ 0, the current solution is optimal. But this
solution is not feasible for the given LP problem because values of decision
variables are: x1 = 0 and x2 = 5 violates second constraint,x2 ≥ 8. The presence
of artiﬁcial variable A1 = 3 in the solution also indicates that the optimal solution
violates the second constraint (x2 ≥ 8) by 3 units.

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