This is a mini-series on how revenue sharing and operations research are linked. You might want to read part 1 on the specific business problem I was solving. In this part, I’ll be telling you about the maths behind the business solution.
But first, I have to tell you about a couple of maths concepts.
Converging from opposite directions
Let’s say someone asked you how tall you are. Instead of giving a straight answer, you decide to give a simple maths puzzle. You say you are at least 6 feet tall. Then you also say you are at most 6 feet tall. So what’s your height? Let’s put those 2 conditions into mathematical form:
To satisfy those 2 conditions, there’s only 1 answer: You must be 6 feet tall.
It looks elementary, but it will come into play later on. Just keep in mind that an equality can be split into 2 inequalities. This actually reminds me of the squeeze theorem, but I digress.
Reversing the direction of an inequality
The next maths concept is how to reverse the direction of an inequality. For example, we have
x >= 5
This can also be expressed as
-x <= -5
This will also come in handy later on.
System of linear inequalities
My experience in operations research had been confined to course work in an academic semester. Operations research mainly is about maximising or minimising some objective. You are given a series of conditions to fulfill. Then you’re given an objective to maximise or minimise. And you’re to translate those conditions and objective into a mathematical model formulation. Let’s look at an example.
Jake’s mom gave him $15 to buy some sweets (because he did awesomely in a maths test. Parental note: I don’t think this is a good reward, but hey, it’s your kid). But mom told Jake that he cannot buy more than 3 lollipops (although Jake already decided he’s not buying more than 4). There’s this contest where if you submit wrappers from peppermints and lollipops, you win an iPad (I’m totally making this up!). Jake needed just 4 more wrappers, and he didn’t want more than 4 because he wanted to buy more chocolates. Also, Jake had secretly decided to buy at least $10 worth of sweets.
Now, chocolates cost $2 each. Peppermints cost $1 each. And lollipops cost $3 each.
Let C be the number of chocolates bought.
Let P be the number of peppermints bought.
Let L be the number of lollipops bought.
Let the objective be to maximise the number of sweets bought. So in maths form, we typically write it as:
max C + P + L
Let’s formulate the conditions:
2C + P + 3L <= 15 (mom only gave Jake $15)
L <= 3 (mom said Jake can't buy more than 3 lollipops)
L <= 4 (Jake independently decided he's not buying more than 4)
P + L = 4 (Jake just wanted 4 more wrappers)
2C + P + 3L >= 10 (Jake wanted to waste at least $10)
Now, logically speaking the condition for the number of wrappers doesn’t make sense. We could say the number of wrappers for peppermints and lollipops be at least 4, which makes better sense. But this is a hypothetical example. More to the point, it’s my hypothetical example, and I needed an equality condition. So there.
You might also note that there are redundant conditions.
L <= 3
L <= 4
can be represented by just
L <= 3
because if it's satisfied, then L is definitely <= 4. This is to show you that in your mathematical formulation, you can get redundant conditions. It's up to you to eliminate them so you work with less conditions. Real life problems are messy, don't you know?
Let's clean up the formulation.
Objective: max C + P + L
2C + P + 3L <= 15
-2C - P - 3L <= -10
P + L <= 4
-P - L <= -4
L <= 3
I've rearranged the conditions a little for clarity. Note the inequality reversal for the "more than $10" condition. Note the split of the equality to 2 inequalities. Why are we doing this? So we get the general form Ax <= B where
A is the coefficient matrix
x is the variable vector
B is the value vector
We need to do the inequality reversal so the "direction" for all the inequalities is the same. Otherwise we cannot have Ax <= B.
In this case, A is
| 2 1 3 |
|-2 -1 -3 |
| 0 1 1 |
| 0 -1 -1 |
| 0 0 1 |
| C |
| P |
| L |
| 15 |
| 4 |
| -4 |
| 3 |
With this Ax <= B form, we're dealing with matrices and vectors. That means we can use all the mathematical techniques for solving such formulations, such as Gaussian elimination or even Gauss-Jordan elimination. The idea is to solve all of our unknowns at one go.
You might notice that we have more inequalities than unknowns. This means we have multiple solutions.
For example, a possible solution for (C, P, L) is (2, 3, 1). That solution satisfies all the conditions. It doesn’t mean it’s optimal, but it’s a solution. This means solutions exist for the problem (sometimes just knowing this is reassuring…).
I want you to know that the objective can always be stated in the opposite manner. If the objective is to “maximise X”, it is equivalent to “minimising -X”. This is known as the dual of the problem. For example, maximising profit is equivalent to minimising the negative value of profit. NOTE: Maximising profit is NOT equivalent to minimising cost. Profit and cost are generally not the opposite of each other (they’re different business terms), even if they sound like they are (and sometimes give the same solutions). The mathematical formulation and solution is different for both of them. Know what you’re solving.
This duality property is useful if your program is optimised to solve only minimisation problems. So you just convert a maximisation problem to a minimisation problem, and your program works just fine!
And you might also note that the objective and the conditions are all linear, meaning unknowns are up to power of 1 (as opposed to quadratic or cubic). Sometimes you get a non-linear condition, and depending on the situation, you might be able to form linear conditions that are equivalent to that non-linear condition.
For example, CP = 2 (number of chocolates multiply by number of peppermints equal to 2).
This set of conditions might be equivalent:
- C >= 1
- P >= 1
- C <= 2
- P <= 2
- C + P = 3
The first 2 conditions are inferred from the fact that C and P cannot be zero. If they are, then CP = 2 is never satisfied. If they’re never zero, then they must be at least 1.
The 3rd and 4th conditions are inferred from the fact that C and P cannot be more than 2. If they are, then the other unknown is fractional. For example, if C is 4, then P must be 1/2. But these are number of items, so they must be integer. Therefore, C and P must be at most 2.
Based on the first 4 conditions, C and P can only take on either 1 or 2 as values. The only combination of permutations that still satisfies the “CP = 2” condition is one unknown must be 1 and the other unknown be 2. Hence the sum of the two unknowns must be equal to 3.
Not all non-linear conditions can be translated into linear conditions. Go ask an operations research professor for more.
And we come to the worst part. All the unknowns must be integer (you can’t buy 2.37 chocolates. Well, at least not without the candy store owner going crazy). This integral condition makes the problem much harder to solve. If you solve the Ax <= B form, you generally get fractional values for the unknowns. This is the optimum solution, but doesn't satisfy the integral condition.
If I remember correctly, you solve Ax <= B as usual. Then you take one of the unknowns and work with the 2 integers that are the floor and ceiling values of that unknown. Then you solve iteratively again, with reduced and reformulated conditions for Py <= Q.
For example, if you get C = 2.37, then you split off 2 scenarios with C = 2 and C = 3 as the optimum solutions. Scenario 1 is
Objective: max P + L
P + 3L <= 11
-P - 3L <= -6
P + L <= 4
-P - L <= -4
L <= 3
Scenario 2 is
Objective: max P + L
P + 3L <= 9
-P - 3L <= -4
P + L <= 4
-P - L <= -4
L <= 3
Basically, you just substitute C = 2 and C = 3. For each scenario, you continue splitting with the other unknowns. There are 3 unknowns, so there are eventually 8 scenarios (2^3 = 8). It's a binary tree. Well, it's at most 8 scenarios, because you might get lucky and hit an original optimised solution with one or more of the unknowns having an original value that's already an integer (now that's a long sentence...).
For each of the scenarios, you either get a solution or you get a contradiction somewhere (meaning it's unsolvable). The contradiction can come from the splitting, because when you substitute a possible value into one of the unknowns, the resulting set of conditions contain a contradiction. For example, out of the set of conditions, you get 2 conditions as follows:
x <= 4
-x <= -8 ( meaning x >= 8 )
Since an unknown variable cannot be both (less than or equal to 4) AND be (more than or equal to 8), the scenario becomes unsolvable.
Then you examine all those scenarios with solutions, substitute the solutions to the unknowns, and compare the result. If it’s a maximising problem, you check which scenario gave a solution that’s the largest. If it’s a minimisation problem, you check which scenario gave a solution that’s the smallest. And that particular solution becomes the optimum integral solution.
Let me tell you, you do not want to do this by hand. This is why software is written to handle the Gaussian eliminations and checking all the possible scenarios due to the integral condition. With just 3 unknowns, you’re expected to do 9 Gaussian eliminations (1 for the original, 8 for the branch scenarios), and check through all 8 scenarios. Imagine having hundreds of variables and conditions…
Mathematics and programming in one. Having skills in one but not the other makes writing the software quite difficult. Imagine a mathematician who cannot express his ideas in programming code, or a programmer who cannot understand the maths involved.
Back to the business problem
After that extremely long mathematical discussion, we can now turn back to the original business problem. If you haven’t read part 1, you should do so now, otherwise you’d be lost.
Let R be the total revenue for a content provider for that particular month
Let R1 be the total revenue for ProductA
Let R2 be the total revenue for ProductB
Let R3 be the total revenue for ProductC
(so R1 + R2 + R3 = R)
Let X1 be the revenue share for the telecommunications company for ProductA
Let X2 be the revenue share for the telecommunications company for ProductB
Let X3 be the revenue share for the telecommunications company for ProductC
Let Y1 be the revenue share for the content provider for ProductA
Let Y2 be the revenue share for the content provider for ProductB
Let Y3 be the revenue share for the content provider for ProductC
Let the revenue sharing split for the telecommunications company be 30%
Let the revenue sharing split for the content provider be 70%
The objective is to minimise rounding errors. Actually, there should be no errors, but we can live with a possible solution first. If it’s non-zero, then we panic…
The unknowns in this case are X1, X2, X3, Y1, Y2 and Y3.
Objective: minimise ABS(R – X1 – X2 – X3 – Y1 – Y2 – Y3)
X1 + Y1 = R1
X2 + Y2 = R2
X3 + Y3 = R3
X1 = 0.30 * R1
X2 = 0.30 * R2
X3 = 0.30 * R3
Y1 = 0.70 * R1
Y2 = 0.70 * R2
Y3 = 0.70 * R3
And Xi’s and Yi’s must be integers, for i = 1, 2, 3
And ABS stands for the absolute function.
As you can see, there’s a lot of equal conditions. And it’s a problem with integer conditions. Frankly speaking, I don’t want to program a solution either. Just looking at what I’ve written is already giving me a headache…
I’m glad my ex-colleague said no. *whew* This was too complicated to implement. And the content provider might have more products, so the number of unknowns is unknown (no pun intended).
There’s actually a little bit more to the maths formulation, but I’ll tell you about it in the 3rd part. I had to come up with a solution that made mathematical sense and could be understood by other people. Mostly, the users (that is, “normal” people) must be able to understand the logic behind it.
Leave a comment if you have any questions about the maths involved (or even tell me I’m wrong), and I’ll do my best to answer them.