The **North West corner rule** is a method for computing
a basic feasible solution of a **transportation problem**, where the basic
variables are selected from the North – West corner ( i.e., top left
corner ).

The standard instructions for a transportation model are paraphrased below. Please read them carefully.

- Select the upper left-hand corner cell of the transportation table and allocate as many units as possible equal to the minimum between available supply and demand, i.e., min(s1, d1).
- Adjust the supply and demand numbers in the respective rows and columns.
- If the demand for the first cell is satisfied, then move horizontally to the next cell in the second column.
- If the supply for the first row is exhausted, then move down to the first cell in the second row.
- If for any cell, supply equals demand, then the next allocation can be made in cell either in the next row or column.
- Continue the process until all supply and demand values are exhausted.

The Amulya Milk Company has three plants located throughout a state
with production capacity 50, 75 and 25 gallons. Each day the firm must
furnish its four retail shops R_{1}, R_{2}, R_{3},
& R_{4} with at least 20, 20 , 50, and 60 gallons respectively.
The transportation costs (in Rs.) are given below.

Plant | Retail Shop | Supply | |||
---|---|---|---|---|---|

R_{1} |
R_{2} |
R_{3} |
R_{4} |
||

P_{1} |
3 | 5 | 7 | 6 | 50 |

P_{2} |
2 | 5 | 8 | 2 | 75 |

P_{3} |
3 | 6 | 9 | 2 | 25 |

Demand | 20 | 20 | 50 | 60 |

The economic problem is to distribute the available product to different retail shops in such a way so that the total transportation cost is minimum

Solution.

Starting from the North west corner, we allocate min (50, 20) to P_{1}R_{1},
i.e., 20 units to cell P_{1}R_{1}. The demand for the
first column is satisfied. The allocation is shown in the following
table.

Table 1

Plant | Retail Shop | Supply | |||
---|---|---|---|---|---|

R_{1} |
R_{2} |
R_{3} |
R_{4} |
||

P_{1} |
5 | 7 | 6 | ||

P_{2} |
2 | 5 | 8 | 2 | 75 |

P_{3} |
3 | 6 | 9 | 2 | 25 |

Demand | 20 | 50 | 60 |

Now we move horizontally to the second column in the first row and
allocate 20 units to cell P_{1}R_{2}. The demand for
the second column is also satisfied.

Table 2

Plant | Retail Shop | Supply | |||
---|---|---|---|---|---|

R_{1} |
R_{2} |
R_{3} |
R_{4} |
||

P_{1} |
7 | 6 | |||

P_{2} |
2 | 5 | 8 | 2 | 75 |

P_{3} |
3 | 6 | 9 | 2 | 25 |

Demand | 50 | 60 |

Proceeding in this way, we observe that P_{1}R_{3} = 10, P_{2}R_{3} = 40, P_{2}R_{4} =
35, P_{3}R_{4} = 25. The resulting feasible solution
is shown in the following table.

Final Table

Plant | Retail Shop | Supply | |||
---|---|---|---|---|---|

R_{1} |
R_{2} |
R_{3} |
R_{4} |
||

P_{1} |
6 | ||||

P_{2} |
2 | 5 | |||

P_{3} |
3 | 6 | 9 | ||

Demand |

Here, number of retail shops(n) = 4, and

Number of plants (m) = 3

Number of basic variables = m + n – 1 = 3 + 4 – 1 = 6.

The total **transportation cost** is calculated by multiplying each x_{ij} in an occupied cell with the corresponding c_{ij} and adding
as follows:

20 X 3 + 20 X 5 + 10 X 7 + 40 X 8 + 35 X 2 + 25 X 2 = 670