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THE TRANS-SHIPMENT PROBLEM
This is an extension of the transportation problem, but we now have some warehouses as well as factories and customers. In the easiest form of the problem, we send from factories to warehouses and from warehouses to customers. Here I will complicate the problem a little bit more. First, I will start with the example given in Taha.
In this example we have two factories (P1 and P2), two warehouses (T1 and T2), and three customers (D1, D2, and D3). We also know the routes that can be used and the distances. The problem is balanced; the amount required exactly matches the amount available. This is not a normal situation, but we know how to deal with excess supply by creating dummy demand points, or with excess demand, by creating dummy supply points.
We want to send from factories to customers in the cheapest possible way.
This is a problem that we can formulate as a transportation problem. To do this, we need to realise that intermediate points, T1 and T2 act both as sources and as destinations of goods. I can, therefore, produce the following table which mirrors what we did in the transportation situation.
T1 T2 D1 D2 D3 Available
P1 3 4 M M M 100
P2 2 5 M M M 120
T1 0 7 8 6 M
T2 M 0 M 4 9
D1 M M 0 5 M
D2 M M M 0 3
Required 80 90 50
Things have been fairly simple up to now. When it was possible to send material, we entered the location as a source. P1 and P2 are sources for T1 and T2. T1 and T2 are sources for D1, D2, and D3. D1 is a source for D2. Finally, D2 is a source for D3. The same reasoning is applied to destinations. I have also entered routes such as D1 to D with cost zero, why did I do this? I did it because sending from D1 to D2 means that we do not send it, and not sending does not cost anything.
Distances in the table are as given when it is possible to use the route, but we set them to a very large number (M) when we cannot use that route.
The available at sources P1 and P2 is given in the problem. The required at destinations D1, D2, and D3 is also given in the problem. What about the numbers we have left blank?
We need to think. We concentrate on a given location, say T1 and ignore the rest of the graph.
I have represented only T1 and its connections. T1 receives material from P1 and from P2 and sends it to T2, D1, and D2. But I have also represented a route with cost zero from T1 to T1 (without the end point because I did not manage to find out how to draw it in Powerpoint).
We now think in scientific terms. T1 does not create items and does not consume them either. So, all that arrives into T1 has to leave from it. How much arrives into T1? Taha says, do not worry about it and call it Buffer, B. But I do not agree with that view, because it will limit our modelling capacity. It is easier to thing as follows: if everything that is produced by P1 and P2 is sent to T1, T1 can receive 100 + 120 = 220 items. All these items are then shipped to D1, D2, or T2. Hence, as a destination, we can set the requirement of T1 as 220, and as a source we can set the availability from T at 220. And what about route T1 to T1? This will represent unused capacity. Imagine that the final solution involves sending 120 from P2 to T1 and nothing from P1 to T (as in the solution given by Taha). Then, T1 can demand 220 but receives only 120. To make up the shortfall, T1 ships to itself 100 units. Then the amount received at T1 is the 120 that it receives from P2 plus the 100 that it receives from itself. In the same way, the amount shipped by T1 is the 120 that arrived from P2 (it goes to D1 and D2) plus the 100 it ships to itself. Everything is balanced: T1 receives 220 and ships 220, even if we have to do a small amount of cheating to balance the books. Up to now, there is no difference between Taha’s approach and our approach, but I prefer this one because it can be extended. The table, then becomes:
T1 T2 D1 D2 D3 Available
P1 3 4 M M M 100
P2 2 5 M M M 120
T1 0 7 8 6 M 220
T2 M 0 M 4 9
D1 M M 0 5 M
D2 M M M 0 3
Required 220 80 90 50
We can think in similar terms about T2. Everything available in the system can go to T2, and T2 can ship it to D2 or D3. Actually, not everything will be shipped because T can have up to 220 but D2 and D3 only require 140, but the difference will be shipped from T2 to T2 at zero cost.
T1 T2 D1 D2 D3 Available
P1 3 4 M M M 100
P2 2 5 M M M 120
T1 0 7 8 6 M 220
T2 M 0 M 4 9 220
D1 M M 0 5 M
D2 M M M 0 3
Required 220 220 220 90 50
We move on and think about D1. D1 is both a destination (gets items from T1) and as a source (it ships items to D2). As a destination it can receive all that T1 has to offer,
!Every source must send all that it has available; @FOR (SOURCES (I) :@SUM (DESTINATIONS (J): SENT (I,J) )= AVAILABLE (I) ); !Every destination must receive all that it desires; @FOR (DESTINATIONS (J) : @SUM (SOURCES (I) : SENT (I,J))= DESIRED (J) );
END
The solution is:
With a cost of 2070. In a nice graph it looks:
This is the same solution that you can find in Taha’s book. This approach requires you to do a little bit more thinking, but it can deal with the case where there are some items available at the warehouses, or consumed inside the warehouses. I am surprised that Taha does not provide you with the tools to model more complex situations than the most basic one.
Cecilio Mar Molinero October 2012.
