Consider the following transportation problem

Destination | Availability | ||||

D_{1} |
D_{2} |
D_{3} |
D_{4} |
||

O_{1} |
5 | 8 | 3 | 6 | 30 |

O_{2} |
4 | 5 | 7 | 4 | 50 |

O_{3} |
6 | 2 | 4 | 6 | 20 |

Requirement | 30 | 40 | 20 | 10 |

Determine an initial basic feasible solution using Vogel’s approximation method

#### Solution

Let ‘a_{i}‘ denote the availability and ‘b_{j}‘ denote the requirement.

Then, `sum"a"_"i"` = 30 + 50 + 20 = 100 and `sum"b"_"j"` = 30 + 40 + 20 + 10 = 100

Since `sum"a"_"i" = sum"b"_"j"`.

The given problem is a balanced transportation problem and we can get an initial basic feasible solution.

Vogel’s approximation method:

**First allocation:**

D_{1} |
D_{2} |
D_{3} |
D_{4} |
(a_{i}) |
penalty | |

O_{1} |
5 | 8 | 3 | 6 | 30 | (2) |

O_{2} |
4 | 5 | 7 | 4 | 50 | (1) |

O_{3} |
6 | ^{(20)}2 |
4 | 6 | 20/0 | (2) |

(b_{j}) |
30 | 40/20 | 20 | 10 | ||

penalty | (1) | (3) | (1) | (2) |

The largest penalty = 3.

So allocate 20 units to the cell (O_{3}, D_{2}) which has the least cost in column D_{2}

**Second allocation:**

D_{1} |
D_{2} |
D_{3} |
D_{4} |
(a_{i}) |
penalty | |

O_{1} |
5 | 8 | ^{(20)}3 |
6 | 30/10 | (2) |

O_{2} |
4 | 5 | 7 | 4 | 50 | (1) |

(b_{j}) |
30 | 20 | 20/0 | 10 | ||

penalty | (1) | (3) | (4) | (2) |

Largest penalty = 4

So allocate min (20, 30) units to the cell (O_{1}, D_{3}) which has the least cost in column D_{3}

**Third allocation:**

D_{1} |
D_{2} |
D_{4} |
(a_{i}) |
penalty | |

O_{1} |
5 | 8 | 6 | 10 | (1) |

O_{2} |
4 | ^{(20)}5 |
4 | 50/30 | (1) |

(b_{j}) |
30 | 20/10 | 10 | ||

penalty | (1) | (3) | (2) |

The largest penalty is 3.

So allocate min (20, 50) to the cell (O_{2}, D_{2}) which has the least cost in column D_{2}.

**Fourth allocation:**

D_{1} |
D_{4} |
(a_{i}) |
penalty | |

O_{1} |
5 | 6 | 10 | (1) |

O_{2} |
4 | ^{(10)}4 |
30/20 | (0) |

(b_{j}) |
30 | 10/0 | ||

penalty | (1) | (2) |

Largest penalty = 2.

So allocate min (10, 30) to cell (O_{2}, D_{4}) which has the least cost in column D_{4}

**Fifth allocation:**

D_{1} |
(a_{i}) |
penalty | |

O_{1} |
^{(10)}5 |
10/0 | – |

O_{2} |
^{(20)}4 |
20/0 | – |

(b_{j}) |
30/10/0 | ||

penalty | (1) |

Largest penalty = 1.

Allocate min (30, 20) to cell (O_{2}, D_{1}) which has the least cost in column D_{1}.

Finally allot the balance 10 units to cell (O_{1}, D_{1})

We get the final allocation table as follows.

D_{1} |
D_{2} |
D_{3} |
D_{4} |
Availability | |

O_{1} |
^{(10)}5 |
8 | ^{(20)}3 |
6 | 30 |

O_{2} |
^{(20)}4 |
^{(20)}5 |
7 | ^{(10)}4 |
50 |

O_{3} |
6 | ^{(20)}2 |
4 | 6 | 20 |

Requirement | 30 | 40 | 20 | 10 |

**Transportation schedule:**

O_{1} → D_{1}

O_{1} → D_{3}

O_{2} → D_{1}

O_{2} → D_{2}

O_{2} → D_{4}

O_{3} → D_{2}

i.e x_{11} = 10

x_{13} = 20

x_{21} = 20

x_{22} = 20

x_{24} = 10

x_{32} = 20

Total cost is given by = (10 × 5) + (20 × 3) + (20 × 4) + (20 × 5) + (10 × 4) + (20 × 2)

= 50 + 60 + 80+ 100 + 40 + 40

= 370

Hence the minimum cost by YAM is ₹ 370.