¡Descarga The transportation problem y más Ejercicios en PDF de Administración de Empresas solo en Docsity!
THE TRANSPORTATION PROBLEM
The transportation problem is very easy to formulate and has a very elegant (for those who appreciate the beauty of mathematical models) dual. There is a specialised algorithm, which is based on SIMPLEX that is very fast. The transportation problem also offers a link with graph theory.
We have seen the transportation problem in the case study of Brisbane airport. There are many situations in which it can be applied, and we will look at some of them. Here we will look at a simple case that I have taken from Taha (Problem 9, chapter 5 in my edition).
“Three refineries with daily capacities of 6, 5, and 8 million gallons, respectively, supply three distribution areas with daily demands of 4, 8, and 7 million gallons, respectively. Gasoline is transported to the three distribution areas through a network of pipelines. The transportation cost is 10 cents per 1000 gallons per pipeline mile. Refinery 1 is not connected to the distribution area 3. The mileage between refineries and distribution areas is given in the table below:”
Area 1 Area 2 Area 3 Available
Refinery 1 120 180 NA 6
Refinery 2 300 100 80 5
Refinery 3 200 250 120 8
Required 4 8 7
This is known as a balanced problem: the sum of available is equal to the sum of required. Balanced problems do not exist in practice. If the amount available is greater than the amount required, we create an extra destination that we can call “nowhere”, and we enter a cost of zero associated with not sending a given amount to nowhere. If the amount required is higher than the amount available we create an extra source, which we can call “out of the blue sky”. There will be a cost associated with not supplying a destination with one unit (supplying it out of the blue sky), and this is the amount we will enter in the table.
In this exercise we have a missing value, because it is not possible to send from source 1 to destination 3. Never mind, we just enter a very large number. The table then becomes:
Area 1 Area 2 Area 3 Available
Refinery 1 120 180 999999 6
Refinery 2 300 100 80 5
Refinery 3 200 250 120 8
Required 4 8 7
We can now formulate the problem. It is best visualised as a graph.
Decision variables are the amount to be sent from source (i) to destination (j), x (^) ij.
There are two sets of constraints:
Everything available at a given source has to go to a destination
Everything that arrives at a destination must have a source
The objective function is long but straightforward. Since cost is proportional to distance travelled multiplied by amount sent, we will just minimise the product of amount sent by distance travelled.
This is completed with the non-negativity constraints, as we cannot send negative amounts of oil through the pipeline.
One of the constraints is clearly redundant. This is easy to see. We have a balanced problem, which means that the sum of available at all sources is equal to the sum of required by all the destinations. Once we have decided how to allocate what is available at sources 1 and 2, what is available at source 3 is automatically allocated to the destinations in order to make up the shortfall. Having redundant constraints does not affect the solution, but it creates problems with the dual variables. Dual variables are not only intellectually challenging, they are also notoriously uncooperative, as they get upset quite easily if the primal problem presents interesting characteristics. For the moment we will just turn a blind eye to these aspects (as most people, including some very respectable authorities in the matter, do).
We will now think about the dual. Notice that the constraints are of the equality type, which means that their associated dual variables will be unconstrained in sign. Remember also that we treat equality constraints as if they were of the relevant type. As we are minimising, the constraints are assumed to be of the more or equal type.
The matrix form of the problem, in standard form, is:
Where
It is precisely the interesting form of matrix A that allows us to develop a specialised algorithm for this type of problem. We will denote u (^) i as the dual variables associated with the sources (the extra cost that will be incurred if we decide to send one extra unit from source i), and with vj the dual variable associated with destination j (the extra cost that will be incurred if destination j insists in getting one extra unit). The dual, after doing all the algebraic manipulations, will be:
