Simplex Method - Haldia Institute of Technology

Simplex Method
Introduction:
In the previous chapter, we discussed about the graphical method for solving
problems. Although the graphical method is an invaluable aid to understand the
programming models, it provides very little help in handling practical problems.
concentrate on the simplex method for solving linear programming problems with
variables.
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linear programming
properties of linear
In this chapter, we
a larger number of
Many different methods have been proposed to solve linear programming problems, but simplex method
has proved to be the most effective. This method is applicable to any problem that can be formulated in
terms of linear objective function, subject to a set of linear constraints.
Basic Terminology
Slack variable
It is a variable that is added to the left-hand side of a less than or equal to type constraint to convert the
constraint into an equality. In economic terms, slack variables represent left-over or unused capacity.
Specifically:
ai1x1 + ai2x2 + ai3x3 + .........+ ainxn bi can be written as
ai1x1 + ai2x2 + ai3x3 + .........+ ainxn + si = bi
Where i = 1, 2, ..., m
Surplus variable
It is a variable subtracted from the left-hand side of a greater than or equal to type constraint to convert
the constraint into equality. It is also known as negative slack variable. In economic terms, surplus
variables represent over fulfillment of the requirement.
Specifically:
ai1x1 + ai2x2 + ai3x3 + .........+ ainxn bi can be written as
ai1x1 + ai2x2 + ai3x3 + .........+ ainxn - si = bi
Where i = 1, 2, ..., m
Artificial variable
It is a non negative variable introduced to facilitate the computation of an initial basic feasible solution. In
other words, a variable added to the left-hand side of a greater than or equal to type constraint to convert
the constraint into an equality is called an artificial variable.
Basic Variables and Non-Basic Variables: The variables attached to the independent column vectors of
the basis matrix are known as basic variables and the remaining (n-m) variables whose values are
assumed to be zero are known as non-basic variables.
Basic Solution: Given a system of m simultaneous linear equations containing n variables (n>m) and the
set of equations be AX=b, R(A)=m. If any m*n, non singular matrix be arbitrating selected from A and if
we assume all (n-m) variables zero which are not associated with column matrix, the solution so obtained
is basic solution.
Non-degenerate basic solution: If all components of a solution set corresponding to a basic variable
are zero quantities then the basic solution is known as non-degenerate basic solution.
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Simplex Method
Degenerate basic solution:
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If some components of the solution set corresponding to the basic variables are zero the basic solution is
known as degenerate solution.
A Basic feasible solution is said to be degenerate solution if one or more than one basic variable are zero.
Non- Degenerate Solution:
A Non- Degenerate Solution feasible solution is the basic feasible solution which has exactly m positive xi
(i=1,2,…,m) i.e. None of the basic variables are zero.
At what condition Simplex method has an unbounded solution?
a)
If corresponding to any negative Z j
C j , all elements of Xj column are negative or zero (≤0). Then the
solution is unbounded.
b) If all artificial vectors are driven out from the basis but some non basis victors are less than zero.
At what condition Simplex method has an no feasible solution?
If all Z j
Cj
0 but some artificial variables are present at the positive level in the optimal solution.
At what condition Simplex method has an alternative solution?
If all Z j
Cj
0 , the alternative solution exist if any non-basic Z j
C j is also zero.
Simplex Method
Consider the general linear programming problem
Maximize z = c1x1 + c2x2 + c3x3 + .........+ cnxn
Subject to
a11x1 + a12x2 + a13x3 + .........+ a1nxn b1
a21x1 + a22x2 + a23x3 + .........+ a2nxn b2
.........................................................
am1x1 + am2x2 + am3x3 + .........+ amnxn bm
x1, x2,....., xn 0
Where:
cj (j = 1, 2, ...., n) in the objective function are called the cost or profit coefficients.
bi (i = 1, 2, ...., m) are called resources.
aij (i = 1, 2, ...., m; j = 1, 2, ...., n) are called technological coefficients or input-output coefficients.
Converting inequalities to equalities
Introducing slack variables to convert inequalities to equalities
a11x1 + a12x2 + a13x3 + .........+ a1nxn + s1 = b1
a21x1 + a22x2 + a23x3 + .........+ a2nxn + s2 = b2
................................................................
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Simplex Method
am1x1 + am2x2 + am3x3 + .....+ amnxn + sm = bm
x1, x2,....., xn 0
s1, s2,....., sm 0
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An initial basic feasible solution is obtained by setting x1 = x2 =........ = xn = 0
s1 = b1
s2 = b2
..............
sm = bm
The initial simplex table is formed by writing out the coefficients and constraints of a LPP in a systematic
tabular form. The following table shows the structure of a simplex table.
Example-01: Using simplex method to solve the following LPP:
Max Z=3x1 + 2x2
Subject to
x1 - x2
x1, x2
x1 + x2
4
2
0.
Solution: Introducing slack variables x3 and x4 in 1st and 2nd constraints respectively.
The reformulated LPP becomes
Max Z=3x1 + 2x2 +0x3+0x4
Subject to
x1 + x2 + x3 =4
x1 -x2 + x4 =2,
x1, x2, x3 ,x4 0.
Using simplex method, we have
Table-
Cj
3
2
0
y3
CB
XB
b
y1
y2
0
0
x3
x4
4
1
1
Zj
0
Cj
x3*
2
1
-1
0
-3*
-2
2
0
2
0
y4
1
0
0
1
Min Ratio
0
4/1=4
1
2/1=2*
0
-1
2
2
1*
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Simplex Method
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Zj
x1
2
Cj
0
-5*
0
1
0
1
0
0
2
x2
1
x1
3
11
Cj
Since all Z j
-1
6
3
Zj
1
0
----
1
0
3
1/2
1/2
5/2
4
-1/2
1/2
1/2
0 .Hence the solution is optimal. The optimal solution is Max Z=11 at x1 =3, x2 =1.
Cj
Example-02: Using simplex method to solve the following LPP:
Max Z=60x1 + 50x2
Subject to
x1 + 2x2
40
3x1 +2x2
x1, x2
60
0.
Solution: Introducing slack variables x3 and x4 in 1st and 2nd constraints respectively.
The reformulated LPP becomes
Max Z=60x1 + 50x2 +0x3+0x4
Subject to
x1 + 2x2 + x3 =40
3x1 +2x2 + x4 =60
x1, x2, x3 ,x4 0.
Using simplex method, we have
Table-
CB
XB
0
0
x3
x4
Zj
0
Cj
60
40
b
y1
y2
40
1
2
60
3
Cj
0
x3*
20
-60↑
0
2
0
y3
1
0
y4
Min Ratio
0
40/1=40
20→
1
-50
4/3
0
0
1
0
-1/3
1/3
30
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Simplex Method
60
x1
Zj
20
x2
15
60
x1
10
Since all
Zj
Cj
1
0
0
-10↑
0
130
Cj
2/3
0
C j 120
50
Zj
1
1
0
0
0
4/3
-1/2
30/4
1/3
20
5
-1/4
1/2
70/4
0 .Hence the solution is optimal and the optimal solution is Max Z=1350 at x1 =10,
x2 =15.
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