The **Vogel approximation method
(Unit cost penalty method)** is an iterative procedure for computing a basic feasible solution of
a transportation problem.

This method is preferred over the two methods
discussed in the previous sections, because the initial basic feasible
solution obtained by this method is either optimal or very close to
the **optimal solution**.

The standard instructions are paraphrased below:

- Identify the boxes having minimum and next to minimum transportation cost in each row and write the difference (penalty) along the side of the table against the corresponding row.
- Identify the boxes having minimum and next to minimum transportation cost in each column and write the difference (penalty) against the corresponding column
- Identify the maximum penalty. If it is along the side of the table, make maximum allotment to the box having minimum cost of transportation in that row. If it is below the table, make maximum allotment to the box having minimum cost of transportation in that column.
- If the penalties corresponding to two or more rows or columns are equal, you are at liberty to break the tie arbitrarily.
- Repeat the above steps until all restrictions are satisfied.

Consider the **transportation problem** presented in the following table:

Destination | |||||
---|---|---|---|---|---|

Origin | 1 | 2 | 3 | 4 | Supply |

1 | 20 | 22 | 17 | 4 | 120 |

2 | 24 | 37 | 9 | 7 | 70 |

3 | 32 | 37 | 20 | 15 | 50 |

Demand | 60 | 40 | 30 | 110 | 240 |

Solution.

Calculating penalty for table 1

17 - 4 = 13, 9 - 7 = 2, 20 - 15 = 5

24 - 20 = 4, 37 - 22 = 15, 17 - 9 = 8, 7 - 4 = 3

Table 1

Destination | ||||||
---|---|---|---|---|---|---|

Origin | 1 | 2 | 3 | 4 | Supply | Penalty |

1 | 20 | 17 | 4 | 13 | ||

2 | 24 | 37 | 9 | 7 | 70 | 2 |

3 | 32 | 37 | 20 | 15 | 50 | 5 |

Demand | 60 | 30 | 110 | 240 | ||

Penalty | 4 | 15 | 8 | 3 |

The highest penalty occurs in the second column. The minimum cij in
this column is c_{12} (i.e., 22). So x_{12} = 40 and
the second column is eliminated. The new reduced matrix is shown
below:

Now again calculate the penalty.

Table 2

Origin | 1 | 3 | 4 | Supply | Penalty |
---|---|---|---|---|---|

1 | 20 | 17 | 13 | ||

2 | 24 | 9 | 7 | 70 | 2 |

3 | 32 | 20 | 15 | 50 | 5 |

Demand | 60 | 30 | 110 | ||

Penalty | 4 | 8 | 3 |

The highest penalty occurs in the first row. The minimum c_{ij} in this row is c_{14} (i.e., 4). So x_{14} = 80 and
the first row is eliminated. The new reduced matrix is shown below:

Table 3

Origin | 1 | 3 | 4 | Supply | Penalty |
---|---|---|---|---|---|

2 | 24 | 7 | 70 | 2 | |

3 | 32 | 20 | 15 | 50 | 5 |

Demand | 60 | 30 | |||

Penalty | 8 | 11 | 8 |

The highest penalty occurs in the second column. The minimum cij in
this column is c_{23} (i.e., 9). So x_{23} = 30 and
the second column is eliminated. The reduced matrix is given in the
following table.

Table 4

Origin | 1 | 4 | Supply | Penalty |
---|---|---|---|---|

2 | 17 | |||

3 | 15 | 17 | ||

Demand | ||||

Penalty | 8 | 8 |

The following table shows the computation of penalty for various rows and columns.

Destination | |||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|

Origin | 1 | 2 | 3 | 4 | Supply | Penalty | |||||

1 | 20 | 17 | 13 | 13 | - | - | - | - | |||

2 | 37 | 2 | 2 | 2 | 17 | 24 | 24 | ||||

3 | 37 | 20 | 15 | 5 | 5 | 5 | 17 | 32 | - | ||

Demand | 240 | ||||||||||

Penalty | 4 | 15 | 8 | 3 | |||||||

4 | - | 8 | 3 | ||||||||

8 | - | 11 | 8 | ||||||||

8 | - | - | 8 | ||||||||

8 | - | - | - | ||||||||

24 | - | - | - |

22 X 40 + 4 X 80 + 24 X 10 + 9 X 30 + 7 X 30 + 32 X 50 = 3520.