In the previous section, we provided the steps in **MODI method (modified distribution method)** to solve a transportation problem.

In this section, we provide an example. Let's solve the following example:

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

Distribution centre | ||||||
---|---|---|---|---|---|---|

D1 | D2 | D3 | D4 | Supply | ||

Plant | P1 | 19 | 30 | 50 | 12 | 7 |

P2 | 70 | 30 | 40 | 60 | 10 | |

P3 | 40 | 10 | 60 | 20 | 18 | |

Requirement | 5 | 8 | 7 | 15 |

Determine the optimal solution of the above problem.

Solution.

An initial basic feasible solution is obtained by Matrix Minimum Method and is shown in table 1.

Table 1

Distribution centre | ||||||
---|---|---|---|---|---|---|

D1 | D2 | D3 | D4 | Supply | ||

Plant | P1 | 19 | 30 | 50 | 7 | |

P2 | 30 | 60 | 10 | |||

P3 | 60 | 18 | ||||

Requirement | 5 | 8 | 7 | 15 |

12 X 7 + 70 X 3 + 40 X 7 + 40 X 2 + 10 X 8 + 20 X 8 = Rs. 894.

Calculating
u_{i} and v_{j} using u_{i }+ v_{j} = c_{ij}

Substituting u_{1} = 0, we get

u_{1 }+ v_{4} = c_{14} ⇒
0 + v_{4} = 12 or v_{4} = 12

u_{3 }+ v_{4} = c_{34} ⇒
u_{3} + 12 = 20 or u_{3} = 8

u_{3 }+ v_{2} = c_{32} ⇒
8 + v_{2} = 10 or v_{2} = 2

u_{3 }+ v_{1} = c_{31} ⇒
8 + v_{1} = 40 or v_{1} = 32

u_{2 }+ v_{1} = c_{21 }⇒
u_{2} + 32 = 70 or u_{2} = 38

u_{2 }+ v_{3} = c_{23} ⇒
38 + v_{3} = 40 or v_{3} = 2

Table 2

Distribution centre | |||||||
---|---|---|---|---|---|---|---|

D1 | D2 | D3 | D4 | Supply | u_{i} |
||

Plant | P1 | 19 | 30 | 50 | 7 | 0 | |

P2 | 30 | 60 | 10 | 38 | |||

P3 | 60 | 18 | 8 | ||||

Requirement | 5 | 8 | 7 | 15 | |||

v_{j} |
32 | 2 | 2 | 12 |

Calculating **opportunity cost** using c_{ij} – ( u_{i }+ v_{j })

Unoccupied cells | Opportunity cost |
---|---|

(P_{1}, D_{1}) |
c_{11} – ( u_{1 }+ v_{1 })
= 19 – (0 + 32) = –13 |

(P_{1}, D_{2}) |
c_{12} – ( u_{1 }+ v_{2 })
= 30 – (0 + 2) = 28 |

(P_{1}, D_{3}) |
c_{13} – ( u_{1 }+ v_{3 })
= 50 – (0 + 2) = 48 |

(P_{2}, D_{2}) |
c_{22} – ( u_{2 }+ v_{2 })
= 30 – (38 + 2) = –10 |

(P_{2}, D_{4}) |
c_{14} – ( u_{2 }+ v_{4 })
= 60 – (38 + 12) = 10 |

(P_{3}, D_{3}) |
c_{33} – ( u_{3 }+ v_{3 })
= 60 – (8 + 2) = 50 |

Table 3

Distribution centre | |||||||
---|---|---|---|---|---|---|---|

D1 | D2 | D3 | D4 | Supply | u_{i} |
||

Plant | P1 | 7 | 0 | ||||

P2 | 10 | 38 | |||||

P3 | 18 | 8 | |||||

Requirement | 5 | 8 | 7 | 15 | |||

v_{j} |
32 | 2 | 2 | 12 |

Now choose the smallest (most) negative value from opportunity cost
(i.e., –13) and draw a closed path from** **P1D1. The following
table shows the closed path.

Table 4

Choose the smallest value with a negative position on the closed path(i.e., 2), it indicates the number of units that can be shipped to the entering cell. Now add this quantity to all the cells on the corner points of the closed path marked with plus signs and subtract it from those cells marked with minus signs. In this way, an unoccupied cell becomes an occupied cell.

Now again calculate the values for u_{i} & v_{j} and opportunity cost. The resulting matrix is shown below.

Table 5

Distribution centre | |||||||
---|---|---|---|---|---|---|---|

D1 | D2 | D3 | D4 | Supply | u_{i} |
||

Plant | P1 | 7 | 0 | ||||

P2 | 10 | 51 | |||||

P3 | 18 | 8 | |||||

Requirement | 5 | 8 | 7 | 15 | |||

v_{j} |
19 | 2 | –11 | 12 |

Choose the smallest (most) negative value from opportunity cost (i.e.,
–23). Now draw a closed path from** **P2D2** ** .

Table 6

Now again calculate the values for u_{i} & v_{j} and opportunity cost

Distribution centre | |||||||
---|---|---|---|---|---|---|---|

D1 | D2 | D3 | D4 | Supply | u_{i} |
||

Plant | P1 | |
7 | 0 | |||

P2 | 10 | 28 | |||||

P3 | 18 | 8 | |||||

Requirement | 5 | 8 | 7 | 15 | |||

v_{j} |
19 | 2 | 12 | 12 |

Since all the current opportunity costs are non–negative, this is the optimal solution. The minimum transportation cost is: 19 X 5 + 12 X 2 + 30 X 3 + 40 X 7 + 10 X 5 + 20 X 13 = Rs. 799