Optimal Product Portfolio

1. Prerequisites

Assume one owns n>0 different products, n\in\mathbb{N}, usually n\approx 100. Each product has a cost c_i>0, c_i\in\mathbb{R}, associated with this ownership, i=1,\ldots, n. Let N=\left\{1,\ldots,n\right\}, and \mathbb{P}(N) be the powerset of N, i.e., the set of all subsets. The powerset \mathbb{P}(N) has 2^n elements.

There are two real-valued functions f and g operating on \mathbb{P}(N), i.e., f,g\colon\mathbb{P}(N)\to\mathbb{R^+}. Function g(I) denotes the sum of the cost of the product set given by I\in\mathbb{P}(N), g is therefore easy to evaluate. I.e.,

\displaystyle{      g(I) = \sum_{i\in I} c_i  }

Function f(I) denotes the buying cost if one can dispose the set of products given by I\in\mathbb{P}(N). This means, it costs money to get rid of the products given by set I. As the products have interrelationships, f is more costly to evaluate than g and is given by some table lookup. Function f does not depend on c_i.

Some notes on f: Assume a non-negative matrix A=(a_{ij}), where a_{ij} denotes the cost to decouple the i-th product from the j-th. Then

\displaystyle{      f(I) = \sum_{i\in I} \sum_{j\in N} a_{ij}  }

Usually a_{ij}=0, if i<j, i.e., A is upper triangular, because decoupling product i from product j does not need to be done again for decoupling the other direction from j to i. More zeros in above sum are necessary if decoupling product i is sufficient if done once and once only. In practical application cases f depends on a real-valued parameter \gamma\in[0,1] giving different solution scenarios.

Example: For three products let c_1=8, c_2=4, c_3=5, thus cost of ownership is g(\emptyset)=0, g(\left\{1\right\})=8, g(\left\{1,2\right\})=8+4=12, g(\left\{1,3\right\})=8+5=13, g(\left\{1,2,3\right\})=8+4+5=17, and so on for the rest of 2^3-5=3 sets. Function f gives positive values in a similar fashion, although, as mentioned before, these values are not related to c_i.

2. Problem statement

Find the set I of products which gives the least cost, i.e., find I for

\displaystyle{      \min_{I\in\mathbb{P}(N)}\left(f(I)-g(I)\right)  }

Rationale: One invests f(I) to get rid of the products in set I but saves ownership costs given by g(I).

I does not need to be unique. If f depends on \gamma then I will likely also depend on \gamma.

3. Challenges

As n can be “large”, evaluating all possible combinations becomes prohibitive on todays computers. Maybe one of our gentle readers knows a method to find the optimum, or maybe just a nearby solution to above problem, or a probabilistic approach. Any comment is welcome.

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This blog was viewed more than 60.000 times since its inception and had more than 45.000 visitors. I wrote about the development of this blog as follows:

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