# Simplex Method: Special Cases

In this section, we will discuss some special cases of simplex method in linear programming (LP).

## 1. Unrestricted (unconstrained) Variables

Sometimes decision variables are unrestricted in sign (positive, negative or zero). In all such cases, the decision variables can be expressed as the difference between two non-negative variables. For example, if x1 is unrestricted in sign, then

Put x1 = x1' - x1''

### Unrestricted Variables: Simplex Method Examples

Maximize z = 2x1 + 3x2

subject to

-x1 + 2x2 ≤ 4
x1 + x2 ≤ 6
x1 + 3x2 ≤ 9

x1, x2 are unrestricted in sign

Solution.

Since x1 and x2 are unrestricted in sign, we can replace them by non-negative variables x1' , x1'', x2' , x2'' .

Put x1 = x1' - x1''
x2 = x2' - x2''

The given problem can be written as

Max. z = 2(x1' - x1'') + 3(x2' - x2'')

subject to

-(x1' - x1'') + 2(x2' - x2'') ≤ 4
(x1' - x1'') + (x2' - x2'') ≤ 6
(x1' - x1'') + 3(x2' - x2'') ≤ 9

Introducing slack variables

Max. z = 2x1' - 2x1'' + 3x2' - 3x2''

subject to

-x1' + x1'' + 2x2' - 2x2'' + x3 = 4
x1' - x1'' + x2' - x2'' + x4 = 6
x1' - x1'' + 3x2' - 3x2'' + x5 = 9

Where x3, x4 and x5 are slack variables

#### Simplex Method: Table 1

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cj 2 -2 3 -3 0 0 0
cB Basic variables
B
x1' x1'' x2' x2'' x3 x4 x5 Solution values
b (=XB)
0 x3 -1 1 2 -2 1 0 0 4
0 x4 1 -1 1 -1 0 1 0 6
0 x5 1 -1 3 -3 0 0 1 9
zj-cj   -2 2 -3 3 0 0 0

Key column = x2' column.
Minimum (4/2, 6/1, 9/3) = 2
Key row = x3 row.
Pivot element = 2
x3 departs and x2' enters.

Table 2

cj 2 -2 3 -3 0 0 0
cB Basic variables
B
x1' x1'' x2' x2'' x3 x4 x5 Solution values
b (=XB)
3 x2' -1/2 1/2 1 -1 1/2 0 0 2
0 x4 3/2 -3/2 0 0 -1/2 1 0 4
0 x5 5/2 -5/2 0 0 -3/2 0 1 3
zj-cj   -7/2 7/2 0 0 3/2 0 0

Table 3

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cj 2 -2 3 -3 0 0 0
cB Basic variables
B
x1' x1'' x2' x2'' x3 x4 x5 Solution values
b (=XB)
3 x2' 0 0 1 -1 1/5 0 1/5 13/5
0 x4 0 0 0 0 2/5 1 -3/5 11/5
2 x1' 1 -1 0 0 -3/5 0 2/5 6/5
zj-cj   0 0 0 0 -3/5 0 7/5

Table 4

cj 2 -2 3 -3 0 0 0
cB Basic variables
B
x1' x1'' x2' x2'' x3 x4 x5 Solution values
b (=XB)
3 x2' 0 0 1 -1 0 -1/2 1/2 3/2
0 x3 0 0 0 0 1 5/2 -3/2 11/2
2 x1' 1 -1 0 0 0 3/2 -1/2 9/2
zj-cj   0 0 0 0 0 3/2 1/2

The optimal solution is:
x1' = 9/2, x1'' = 0, x2' = 3/2, x2'' = 0.

Solution of the original problem is:
x1 = x1' - x1'' = 9/2 - 0 = 9/2
x2 = x2' - x2'' = 3/2 - 0 = 3/2
z = 2 X 9/2 + 3 X 3/2 = 27/2.