30th November 2017

# 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.,

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

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

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

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

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

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.

Categories: mathematics
Author: Elmar Klausmeier