# Degeneracy: Transportation Problem

If the basic feasible solution of a transportation problem with m origins and n destinations has fewer than m + n – 1 positive xij (occupied cells), the problem is said to be a degenerate transportation problem. Degeneracy can occur at two stages:

1. At the initial solution
2. During the testing of the optimal solution

To resolve degeneracy, we make use of an artificial quantity (d). The quantity d is assigned to that unoccupied cell, which has the minimum transportation cost.

For calculation purposes, the value of d is assumed to be zero.

The use of d is illustrated in the following example.

## Example: Degeneracy in Transportation Problem

Factory Dealer Supply
1 2 3 4
A 2 2 2 4 1000
B 4 6 4 3 700
C 3 2 1 0 900
Requirement 900 800 500 400

Solution.

An initial basic feasible solution is obtained by Matrix Minimum Method.

Table 1

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Factory Dealer Supply
1 2 3 4
A 2 4 1000
B 4 4 3 700
C 3 2 900
Requirement 900 800 500 400

Number of basic variables = m + n – 1 = 3 + 4 – 1 = 6
Since number of basic variables is less than 6, therefore, it is a degenerate transportation problem.

To resolve degeneracy, we make use of an artificial quantity(d). The quantity d is assigned to that unoccupied cell, which has the minimum transportation cost.

The quantity d is so small that it does not affect the supply and demand constraints.

In the above table, there is a tie in selecting the smallest unoccupied cell. In this situation, you can choose any cell arbitrarily. We select the cell C2 as shown in the following table.

Table 2

Factory Dealer Supply
1 2 3 4
A 2 4 1000
B 4 4 3 700
C 3 900 + d
Requirement 900 800 + d 500 400 2600 + d

Now, we use the stepping stone method to find an optimal solution.

Calculating opportunity cost

Unoccupied cells Increase in cost per unit of reallocation Remarks
A3 +2 – 2 + 2 – 1 = 1 Cost Increases
A4 +4 – 2 + 2 – 0 = 4 Cost Increases
B1 +4 – 6 + 2 – 2 = –2 Cost Decreases
B3 +4 – 6 + 2 – 1 = –1 Cost Decreases
B4 +3 – 6 + 2 – 0 = –1 Cost Decreases
C1 +3 – 2 + 2 – 2 = 1 Cost Increases

The cell B1 is having the maximum improvement potential, which is equal to -2. The maximum amount that can be allocated to B1 is 700 and this will make the current basic variable corresponding to cell B2 non basic. The improved solution is shown in the following table.

Table 3

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Factory Dealer Supply
1 2 3 4
A 2 4 1000
B 6 4 3 700
C 3 900
Requirement 900 800 500 400 2600

The optimal solution is

2 X 200 + 2 X 800 + 4 X 700 + 2 X d + 1 X 500 + 0 X 400 = 5300 + 2d.

Notice that d is a very small quantity so it can be neglected in the optimal solution. Thus, the net transportation cost is Rs. 5300