In a transportation problem, consignments are always transported from an origin to a destination.

But, there could be several situations where
it might be economical to transport items via one or more intermediate
centres (or stages). In a **transshipment problem**, the available commodity
is not sent directly from sources to destinations, i.e., it passes through
one or more intermediate points before reaching the actual destination.

For instance, a company may have regional warehouses that distribute
the products to smaller district warehouses, which in turn ship to the
retail stores. Succinctly, the transshipment model is an extension of
the **classical transportation model** where an item available at point
i is shipped to demand point j through one or more intermediate points.
## Example of Transshipment Model

#### Example

A company has nine large stores located in several states. The sales department is interested in reducing the price of a certain product in order to dispose all the stock now in hand. But, before that the management wants to reposition its stock among the nine stores according to its sales expectations at each location.

The above figure shows the numbered nodes (9 stores). A positive value next to a store represents the amount of inventory to be redistributed to the rest of the system. A negative value represents the shortage of stock. Thus, stores 1 and 4 have excess stock of 10 & 2 items respectively. Stores 3, 6 & 8 need 3, 1, and 8 more items respectively. The inventory positions of stores 2, 5 & 7 are to remain unchanged.

An item may be shipped through stores 2, 4, 5, 6, 7 & 8. These locations are known as transshipment points. Each remaining store is a source if it has excess stock, and a sink if it needs stock. In the above figure, store 1 is a source and store 3 is a sink.

The value c_{ij} is the cost of transporting items. To transport
an item from store 1 to store 3, the total shipping cost is

c_{12} + c_{23}

In the following example, you will learn how to convert a transshipment problem to a standard transportation problem.

Consider a transportation problem where the origins are plants and destinations are depots. The unit transportation costs, capacity at the plants, and the requirements at the depots are indicated below:

Table 1

Plant | Depot | |||
---|---|---|---|---|

X | Y | Z | ||

A | 1 | 3 | 15 | 150 |

B | 3 | 5 | 25 | 300 |

150 | 150 | 150 | 450 |

When each plant is also considered a destination and each depot is also considered an origin, there are altogether five origins and five destinations. Some additional cost data are also necessary. These are presented in the following Tables.

Table 2

Unit Transportation Cost from Plant to Plant | ||
---|---|---|

From Plant | To | |

Plant A | Plant B | |

A | 0 | 65 |

B | 1 | 0 |

Table 3

Unit Transportation Cost from Depot to Depot | |||
---|---|---|---|

From Depot | To | ||

Depot X | Depot Y | Depot Z | |

X | 0 | 23 | 1 |

Y | 1 | 0 | 3 |

Z | 65 | 3 | 0 |

Table 4

Unit Transportation Cost from Depot to Plant | ||
---|---|---|

Depot | Plant | |

A | B | |

X | 3 | 15 |

Y | 25 | 3 |

Z | 45 | 55 |

Solution.

From Table 1, Table 2, Table 3 and Table 4 we obtain the transportation
formulation of the *transshipment problem*.

Table 5

Transshipment Table | ||||||
---|---|---|---|---|---|---|

A | B | X | Y | Z | Capacity | |

A | 0 | 65 | 1 | 3 | 15 | 150 + 450 = 600 |

B | 1 | 0 | 3 | 5 | 25 | 300 + 450 = 750 |

X | 3 | 15 | 0 | 23 | 1 | 450 |

Y | 25 | 3 | 1 | 0 | 3 | 450 |

Z | 45 | 55 | 65 | 3 | 0 | 450 |

Requirement | 450 | 450 | 150 + 450 =600 | 150 + 450 =600 | 150 + 450 =600 | 2700 |

The transportation model is extended and now it includes five supply
points & demand points. To have a supply and demand from all the
points, a fictitious supply and demand quantity (**buffer stock**)
of 450 is added to both supply and demand of all the points. An initial
basic feasible solution is obtained by the **Vogel's Approximation
method** and is shown in the final table.

Final Table

Transshipment Table | ||||||
---|---|---|---|---|---|---|

A | B | X | Y | Z | Capacity | |

A | 65 | 15 | 600 | |||

B | 3 | 5 | 25 | 750 | ||

X | 3 | 15 | 23 | 450 | ||

Y | 25 | 3 | 1 | 3 | 450 | |

Z | 45 | 55 | 65 | 3 | 450 | |

Requirement | 450 | 450 | 600 | 600 | 600 | 2700 |

The **total transhipment cost** is:

0 X 150 + 1 X 300 + 3 X 150 + 1 X 300 + 0 X 450 + 0 X 300 + 1 X 150
+ 0 X 450 + 0 X 450 = 1200