Simplex Method

Special Cases

4. Multiple Optimal Solutions

The optimal solution may not be unique, if the non basic variables have a zero coefficient in the index row (zj-cj). This implies that bringing the non basic variable into the basis will neither increase nor decrease the value of the objective function. Thus, the problem has multiple optimal solutions (Alternative optimal solutions).

Example

Maximize 2000x1 + 3000x2

subject to

6x1 + 9x2 £ 100
2x1 + x2 £ 20

x1, x2 ³ 0

Solution.

After introducing slack variables, the corresponding equations are

6x1 + 9x2 + x3 = 100
2x1 + x2 + x4 = 20
x1, x2, x3, x4 ³ 0

Where x3 and x4 are slack variables.

Table 1

  cj 2000 3000 0 0  
cB Basic variables
B
x1 x2 x3 x4 Solution values
b (=XB)
0 x3 6 9 1 0 100
0 x4 2 1 0 1 20
zj-cj   -2000 -3000 0 0  

Table 2

  cj 2000 3000 0 0  
cB Basic variables
B
x1 x2 x3 x4 Solution values
b (=XB)
3000 x2 2/3 1 1/9 0 100/9
0 x4 4/3 0 -1/9 1 80/9
zj-cj   0 0 3000/9 0  

Since zj-cj ³ 0 for all variables, x1 = 0, x2 = 100/9 is an optimum solution of the problem. The maximum value of the objective function is 100000/3. However, the zj-cj value corresponding to the non basic variable x1 is zero. This indicates that there is more than one optimal solution of the problem. Thus, by entering x1 into the basis we may obtain another alternative optimal solution.

  cj 2000 3000 0 0  
cB Basic variables
B
x1 x2 x3 x4 Solution values
b (=XB)
3000 x2 0 1 1/6 -1/2 20/3
2000 x1 1 0 -1/12 3/4 20/3
zj-cj   0 0 1000/3 0  

However, the value of the objective function will not be improved, although the present solution x1 = 0, x2 = 100/9 is shifted to x1 = 20/3 and x2 = 20/3.

5. Degeneracy

In some cases, there may be ambiguity in selecting the variable that should be introduced into the basis, i.e., there is a tie between the replacement ratio of two variables. To resolve degeneracy, we select one of them arbitrarily.

Consider the following example.

Example

Maximize 3x1 + 9x2

subject to

x1 + 4x2 £ 8
x1 + 2x2 £ 4

x1, x2 ³ 0

Solution.

After introducing slack variables, the corresponding equations are:

x1 + 4x2 + x3 = 8
x1 + 2x2 + x4 = 4
x1, x2, x3, x4 ³ 0

Where x3 and x4 are slack variables.

Table 1

  cj 3 9 0 0  
cB Basic variables
B
x1 x2 x3 x4 Solution values
b (=XB)
0 x3 1 4 1 0 8
0 x4 1 2 0 1 4
zj-cj   -3 -9 0 0  

Minimum positive value (8/4, 4/2) = (2, 2)

There is a tie between the two values. So you are at liberty to break the tie arbitrarily.

In the following material, we will consider both the cases.

Case I

x4 departs & x2 enters.

Table 2

  cj 3 9 0 0  
cB Basic variables
B
x1 x2 x3 x4 Solution values
b (=XB)
0 x3 -1 0 1 -2 0
9 x2 1/2 1 0 1/2 2
zj-cj   3/2 0 0 9/2  

x1 = 0, x2 = 2, z = 18

Case II

x3 departs & x2 enters.

Table 2

  cj 3 9 0 0  
cB Basic variables
B
x1 x2 x3 x4 Solution values
b (=XB)
9 x2 1/4 1 1/4 0 2
0 x4 1/2 0 -1/2 1 0
zj-cj   -3/4 0 9/4 0  

x4 departs & x1 enters.

Table 3

  cj 3 9 0 0  
cB Basic variables
B
x1 x2 x3 x4 Solution values
b (=XB)
9 x2 0 1 1/2 -1/2 2
3 x1 1 0 -1 2 0
zj-cj   0 0 3/2 3/2  

x1 = 0, x2 = 2, z = 18


In this chapter, you learned the mechanics of obtaining an optimal solution to a linear programming problem by the simplex method. The simplex method is an appropriate method for solving a £ type linear programming problem with more than two decision variables. Two phase and M-method are used to solve problems of ³ or £ type constraints. Further, the simplex method can also identify multiple, unbounded and infeasible problems.


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