Model of Transportation with Transshipment solved with Excel Solver

One of the classic uses of Operations Research and, in particular of Linear Programming is to propose optimum alternatives for the logistics or transport of inputs and products from a group of suppliers to a group receivers or petitioners. When we consider that, in this process of transportation, intermediaries can participate, this is an extension of the basic model of transport which is commonly known as a model of transportation with transshipment. We will then present an applied case of this model.

Example Transportation with Transshipment

20 million barrels of oil should be shipped from Dhahran in Saudi Arabia to the cities of Rotterdam, Marseille and Naples in Europe. The orders of these three cities are 4, 12 and 4 million barrels, respectively. Below is a diagram with the possible routes:


Note that for each city there is the possibility of direct shipment, i.e., that the barrels are transported directly from Dhahran. However, because of certain trade agreements, they cannot ship more than 3 million barrels through the route between Dhahran and Marseille.

On the other hand, there is a possibility of a delay either in the port of Alexandria or Suez, where storage capacity is 8 and 10 million .

Finally, note that it is possible to send barrels of oil from Marseilles to Naples. However, in Naples it is forbidden to receive more oil from Marseille than directly from Dhahran. Formulate and solve a model of linear programming that allows you to find the optimal transportation policy to comply with the requirements of demand from all the harbors.

Decision Variables:

  • X1: Barrels shipped from Dhahran to Rotterdam
  • X2: Barrels shipped from Dhahran to Marseille
  • X3: Barrels shipped from Dhahran to Naples
  • X4: Barrels shipped from Dhahran to Alexandria
  • X5: Barrels shipped from Dhahran to Suez
  • X6: Barrels shipped from Alexandria to Rotterdam
  • X7: Barrels shipped from Alexandria to Marseille
  • X8: Barrels shipped from Suez to Marseille
  • X9: Barrels shipped from Suez to Naples
  • X10: Barrels shipped from Marseille to Naples

Objective Function:

Minimize the total costs of transport given by the following expression: 7X1 + 8X2 + 15X3 + 6X4 + 5X5 + 8X6 + 7X7 + 2X8 + 6X9 + 1X10

Satisfy the Demand of the Harbors:

  • X1 + X6 = 4.000.000   (Rotterdam)
  • X2 + X7 + X8 – X10 = 12.000.000   (Marseille)
  • X3 + X9 + X10 = 4.000.000   (Naples)

Note that Marseille could eventually receive more than 12 million barrels of oil (demand) since this port has the ability to cater to Naples.

Balance in Transshipment:

  • X4 = X6 + X7   (Alexandria)
  • X5 = X8 + X9   (Suez)

The number of barrels that Alexandria and Suez receive, must be equal to the number each of them ship to the ports, i.e., intermediaries do not accumulate inventory at the end of the planning period. At this point it is important to note that, if it is considered an extended model which seeks to satisfy the requirements of demand for various periods, it could be admissible to store inventory in Alexandria and Suez, by changing the shape of the optimization model in this case.

Processing capacity in the Transshipment:

  • X4 <= 8.000.000   (Alexandria)
  • X5 <= 10.000.000   (Suez)

Both Alexandria and Suez cannot receive a bigger amount of barrels than the one that they can process.

Capacity of the Route between Dhahran and Marseille:

  • X2 <= 3.000.000

Through the route between Dhahran and Marseille, more than 3 million barrels cannot be shipped because of trade agreements.

Amount received in Naples:

  • X3 >= X10

In Naples it is forbidden to receive more oil from Marseille than, directly from Dhahran.

Non Negativity:

  • Xi >= 0 For All i

When implementing the previous model with Solver of Excel, the following results are obtained:


The solution obtained has the following structure (the value of the optimal solution is outlined above the arches):


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