The Transshipment Problem. This section is a slight modification of the transportation problem.
This section is a slight modification of the transportation problem.
You might recall in the transportation problem that we had a source of product and a destination for product. An analogy might be you had production locations and sales (retail) locations. We looked for the cheapest way to get products from the production places to the sales places.
Here we have the possibility of an intermediate step. Maybe between the production and the sales places we have warehouses.
Warehouses would then get stuff from production places and move the stuff on to the sales places. Here is a visual to have you “see” the situation.
Production warehouse sales
Call xij the amount shipped from place i to place j. Note in my visual the flow is from left to right, but in the real world we would have to keep track of the “type” of entity of each location.
Subscripts 1 and 2 here refer to production places, 3 and 4 to warehouses and 5 through 8 for sales places. On page 439 of the book we have cost data and from this we have the following objective function in linear programming.
Min 2x13 + 3x14 + 3x23 + 1x24 + 2x35 + 6x36 + 3x37 + 6x38 + 4x45 + 4x46 + 6x47 + 5x48
You might note here that we do not have a smooth flow in our subscripts like x11, x12, x13, x21, and so on because, for example, x12 does not make sense since locations 1 and 2 are both production places and we would not ship from production place to production place (this is in context of outputs).
Now, the production places have a limit to what can be produced and we will have a constraint for each in the form
xi3 + xi4 ≤ production limit,
Specifically, here we have
x13 + x14 ≤ 600 and x23 + x24 ≤ 400.
Next, what comes out of a warehouse has to first come in. This means each warehouse will have a constraint that has an equality of what comes in and what goes out. For warehouse 3 we have
x13 + x23 = x35 + x36 + x37 + x38 or to be in the form for the computer we have
x13 + x23 – x35 – x36 – x37 – x38 = 0. Similarly for warehouse 4 we have
x14 + x24 – x45 – x46 – x47 – x48 = 0.
As in the transportation problems we have the destinations get what they need, so those constraints would be
x35 + x45 = 200,
x36 + x46 = 150,
x37 + x47 = 350, and
x38 + x48 = 300.
On the next slide I have the results of the linear program from the software that comes with our book.
Note the cost of shipping on x36 is 6 but the reduced cost is only 1. If you forced a unit to be shipped on x36 you would think costs would go up 6, but is only 1 here. The reason is when you force 1 shipped on x36 destination 6 now has one more unit than needed, so the program reduces shipments from other places and the net is only 1. Neat, huh?