A Man with a Van Goes to Market: Optimise his Profit

He has only two products to sell. OK, you do not often come across companies which make only two products, but there are many situations in larger companies where optimisation for two products can be applied. Note: They need to be totally isolated from everything else you make and not compete for any resources with the rest of production but two in isolation, with care is a valuable and profitable tool.

Firstly, let us state the problem in Layman's Terms:

The van is an odd shape, but this is an example, not quite reality! It's inside size is 3 meters x 3 meters by 5 meters, which gives it a total usable volume of 45 cubic meters.

It has a load limit of only 1 metric ton or 1000 Kg.

There are two products The first is in a box which is a cube with 1 meter sides so it has a volume of 1 cubic meter. This is product x

The other product has a volume of 0.1 cubic meters so it has the seemingly weird dimensions of 0.464159m*0.464159m*0.464159m . 0.464159m is the cube root of 0.1 so the volume of the smaller cube is 0.1 cubic meters. This is product y

Note that the sides of the smaller box does not go quite into the sides of the larger box as a whole number. This is not important in this problem, but in future we will look at integer optimisations which are a bit harder to do but they are useful.

You cannot after all do very much with 0.243 of a vacuum-cleaner or 0.65 of a computer!

Now, as we have established the volume of the two cubic boxes as 1 cubic meter and 0.1 cubic meters we need to establish their weights, so as not to overload the van. We also need to establish the profit from selling each type of box.

The small box y is heavy and valuable, weighing 30 Kg and has a profit of ï¿¡500 on sale.

The large box x is less heavy at 5 Kg and has a lower profit of ï¿¡100 on sale.

There is a further problem though. The factory cannot make more than 44 units of the big box, lets call it x so that we can use equations. It also cannot make more then 30 units of the small box which we will call y.

In fact the factory uses the same machine to make both x and y, a fact that will become important when we ask our model to calculate what further investments they can make to increase profits beyond just packing the van properly or optimally.

The above facts can now be set out as a linear equation. (Do not worry, we do this bit for you) It is however stated as follows.

Maximize Z=100x + 500y

Or in English, The profit Z, equals the profit (ï¿¡100) from selling x units of product x, plus the profit (ï¿¡500) on selling y units of product y.

This is subject to the constraints imposed by the volume and load of the van and the ability of the factory to make units x and y. This is formally stated below:

Maximize Z=100x + 500y Total profit, Z is the profit on x (ï¿¡100) times the number of x sold plus the profit on y (ï¿¡500) times the number of y sold.

Subject to: (These are called the constraints)

x+0.1y <= 45. The combined volume of x and y must be less than or equal the volume of the van.

5x+30y <= 1000. The combined weight of x and y must be less than or equal to the load limit of the van.

y <= 30 The factory cannot make more than 30 of product y.

x <= 44 The factory cannot make more then 44 of product x.

Now, by re-arranging the equations and drawing a graph from them, we can solve the problem by walking around the feasible area of the graph.

y<=(45-x)/0.1 = 450-10x

y<=(1000-5x)/30 = 1000/30-5/30x= 33.3333-1/6x

x<=30 is a straight line perpendicular to the x axis

y<=44 is a straight line perpendicular to the Y axis.

Here, then is the graph of all the above

Let us go for a wander towards optimal profit, starting at the point 0,0 in the bottom left hand corner when there is nothing in the van. That is a profit of ï¿¡0!!

So, seeking a greater profit than ï¿¡0 we take a wander up the y axis of the graph, where we go as far as is allowed by the constraints and we reach 30 units of y, the production limit. Just imagine if the factory could produce 60 of product y, could we go there? Well, no as we hit the load weight limit of 1000 Kg at 33.33333 units, leaving no load for product x and a 2/3 empty van which is actually fully loaded! So we have a profit of ï¿¡15,000 at coordinate 0,30 or the black point where we are not making any of product x - the large and light one.

Let us now travel in the most profitable direction. Going back to 0,0 is obviously not an option, so we go in the direction of x, adding the large lighter product x to the van until we meet a limit - the line made by the equation 5x+30y = 1000 and y=30. We cannot exceed the weight limit. However a quick calculation of profit at this point 20,30 or the green point the profit is ï¿¡17,000 which is much better than ï¿¡15,000

We travel now down the sloping line (green line on the graph) until we reach the intersection of the green line and the red one.The point of intersection is the blue point which represents 42.37 of product x and 26.27 of product y. This assumes that both products can, if necessary be carved up. The profit here is ï¿¡17,372.88, so it is more profitable, still.

Let us now test the points on either side of the blue point. Going backwards is going back to a profit of ï¿¡17,000 so that is not on. Going down the red line however to the red point of 44,10 represents a profit of ï¿¡9,400, so it is worse than the blue point. Again, we need to test points on either side of the red point. Going backwards we get back to the blue point and a profit of ï¿¡17,372.88 which is starting to look like an optimum profit.

However we need to test the model to be sure, so now we travel down the purple line to the yellow point 44,0 which represents a van filled with 1 meter cubes and nothing else. The profit here is only ï¿¡4,400. One large box too many and your profit dives from ï¿¡17,372.88 to ï¿¡4,400 so optimisation is worth a fortune for this particular company!

The blued point is therefore the optimum profit that can be achieved with this van and these products. ï¿¡17,372.88 is as good as it gets! WE have optimised the van for market.

OK Steve, I have been told in a meeting. Now that you have shown us your fancy maths, we can do it for ourselves. "OK," I say. "Now try it for three products!" "and as your company makes 32 products, try it for all 32"

There is a very important lesson here. All of the points tested lie on a vertex of a polygon bounded by limits. This is called the feasible area and you can test anywhere in this area for profit, but it will never exceed ï¿¡17,372.88 Usually, clients are packing their van by guesswork and in the past they will have used some point withing the polygon, probably a very unprofitable one.

That is usually when I get the order to optimise a company for profit using linear programming.

Even with this simple graph, we can look at some useful decision making such as "Is it more profitable to buy a bigger van" (no) and is it profitable to buy a better machine to make more of x and y (Only if you also buy a bigger van or a second van!!)