# When matrix theory explains consolidation …

# When matrix theory explains the process of group consolidation

*by Allen White (1), Sigma Conso co-founder & administrator
*

### Abstract

One of the tasks of the consolidation process of a group of companies requires the calculation of percentages. If the structure of the group, which is not a planar graph, presents a significant number of companies with, between them, a significant number of equity links, the use of a matrix algorithm becomes necessary. This article explains the mathematical approach.

## I. Description of the context

The diagram below shows a group of four companies consisting of S_{0} named parent company, consolidating company or holding company and companies S_{i} with i ∈ [1, 3] named owned companies or subsidiaries.

The percentages appearing on this diagram represent the level of detention of shares of a company into another company, that is the ratio between the number of shares owned and the number of shares issued. It is called the direct percentage.

But the consolidation process needs to calculate indirect percentages owned by parent company S_{0} into each company S_{i} , i ≠ 0.

For example, the indirect percentage owned in company S_{3} is calculated as follows

- 10% owned directly by company S
_{0} - 32% = 80% ∗ 40% calculated along the path S
_{0}→ S_{1}→ S_{3} - 2.4% = 60% ∗ 10% ∗ 40% calculated along the path S
_{0}→ S_{2}→ S_{1}→ S_{3} - 12% = 60% ∗ 20% calculated along the path S
_{0}→ S_{2}→ S_{3}

The addition of these four percentages, which is 56.4%, gives the expected indirect percentage.

That algorithm seems simple to apply because, for each company S_{i} , we just need to inventory all the paths starting at S_{0} and arriving at S_{i} and multiply the direct percentages all along these paths. Summation of these calculated percentages will give the answer.

However, the size of the group (number of companies) and the complexity of its structure (number of participations, thus paths) can make the algorithm difficult to apply.

Finally, the existence of crossed participations will make the algorithm unapplicable definitely as in the following situation where we would have a participation from S_{2} in S_{3} and from S_{3} in S_{2} .

And it is here that matrices bring a very elegant help.

## II. Writing conventions

- S
_{0}represents the parent company of a group which consists of n + 1 companies, including that parent company - S
_{i}with i ∈ [1, n] represents any company of the group, excepted the parent company - D
_{ij}represents the direct percentage of company S_{i}in company S_{j}with i ∈ [0, n], j ∈ [1, n] and i ≠ j - D
_{ii}= 0 ∀i ∈ [0, n] means that company S_{i}does not own a number of its shares. If so, these shares would be ignored in the calculations. - D
_{i0}= 0 ∀i ∈ [1, n] means that no company S_{i}owns shares of the parent company. If so, these shares would be ignored in the calculations. - 0 ≤ D
_{ij}≤ 1 ∀i, j ∈ [0, n] because D_{ij}represents a participation percentage and we assume that 100% = 1. - N
_{i}∀i ∈ [0, n] represents the indirect percentage in company S_{i}.

We assume that N_{0}= 1 which means that the group owns 100% of the parent company.

In this article, we agree to write .

Considering these conventions, with the direct percentages we build a square matrix D of dimensions (n +1) × (n + 1) containing the elements D_{ij} as follows

Obviously, this matrix can never be equal to the unit matrix I .

## III. Let us translate the algorithm into an equation

In the most general case, we can suppose that a company is held by all the other companies of the group. So we have

In such a situation, the indirect percentage in S_{i} can be written

*N _{i} = N_{0} D_{0i} + N_{1} D_{1i} + N_{2} D_{2i} + … + N_{n} D_{ni}*

This relation being true for each company, we can build a system of n + 1 equations whose n + 1 unknowns are the indirect percentages N_{i} . So

If we define

*N = (1 N _{1} N_{2} … N_{n} )*

as a vector 1 × (n + 1), then we can write the following matrix relation

*N = (1 0 0 0 … 0) + N D*

and by setting *U = (1 0 0 0 … 0)*, we get

*N = U + N D*

## IV. Let us calculate the vector solution N

The vector N can be explicitly calculated using an iterative method by building a sequence N_{t} = U +

N_{t−1} D with an initial value *N _{0} = U* , and for which we have to check the convergence.

Iteration 1 : *N _{1} = U + N_{0} D = U + U D*

Iteration 2 : *N _{2} = U + N_{1}D = U + (U + UD)D = U + UD + UD^{2}*

…

Iteration t : *
*

By increasing the number of iterations infinitely and knowing that (refer to section VI for a complete proof), we can also write

and then

This last relation shows that

Let us reconsider the relation (*) which gives

and which is the answer we were looking for.

## V. Let us apply this result to our example

The group’s structure of our example which consists in four companies gives the following D matrix

and the following results

this last vector contains the indirect percentages.

We shall notice that, for example, the matrix element (3,4) of (I − D)^{−1} equal to 24% is the indirect percentage of company S_{2} in company S_{3} .

## VI. Some properties of the matrix D

In this section, we show some properties of matrix D based on the fact that it is connected to a group’s structure of companies. In particular, these properties allow to deduct that .

### A. Norm of a matrix D

We define as being the summation of its elements. It is indeed a norm by considering the following properties.

- ||D|| ≥ 0 because D
_{ij}≥ 0 ∀i, j ∈ [0, n] - If ||D|| = 0, then and as D
_{ij}≥ 0 we must have D_{ij}= 0 ∀i, j ∈ [o, n], so D = 0. - Consider . Then .
- Considering two square matrix A and B of same dimension, we have = . It would have been sufficient to prove that ||A + B|| ≤ ||A|| + ||B||.

### B. Summation of all elements of a column of matrix D

Let’s call the sum of all elements of column j of matrix D. As D is connected to a group’s

structure, we can deduct the following properties:

- σj represents the total of percentages owned by group shareholders of company S
_{j}∀j ∈ [1, n]. - σ
_{0}= 0 because the parent company is supposed to have no shareholders in the group’s structure. - 0 < σj ≤ 1 ∀j ∈ [1, n] because each company, excepted the parent company, has at least one shareholder in the group’s structure. Moreover, the maximum of percentages owned by all shareholders cannot of course exceed 100 % = 1.

### C. What is the meaning of the term of the matrix D^{t}?

We define as the element of line i and column j of the matrix D^{t}. We then can write for example for t = 2

and each element D_{ik} D_{kj} of that summation for a certain value of k represents the product of direct percentages owned by S_{i} in S_{k} and S_{k} in S_{j} . In other words, it is the product of direct percentages along the path starting from S_{i} and arriving at S_{j} by passing through S_{k}. In such a situation, we speak about a path of length two.

By making a summation on k, we simply consider the sum of the products of the percentages along all the paths of length two from S_{i} to S_{j}. This is the meaning of .

The extension appears quite naturally when we consider the term and its development

This expression represents the product of the direct percentages along all the possible paths of length t

starting from the company S_{i} and arriving at the company S_{j}.

If = 0, we can assert that there is no path of length t between S_{i} and S_{j} .

This situation can occur for example in the following group of four companies.

Indeed, a path of length two starting at company S_{0} and arriving at company S_{3} does not exist and thus

We shall finally notice that in the group of three companies below

that represents crossed participations between S_{1} and S_{2}, the element contains in particular the term

which corresponds to the path of length five S_{0} → S_{1} → S_{2} → S_{1} → S_{2} → S_{1}. This means that in a group, generally, the paths from a company to another company can present an infinite length when there are crossed or circular participations.

### D.

We have

Considering property B paragraph 3 above, we can write

and thus .

### E. If, for t = T , ||D^{T} || = 0, then ||D^{u} || = 0 ∀u ≥ T

This is obvious because if ||D^{T} || = 0, then D^{T} = 0 and we have D^{u} = D^{u−T} D^{T} = 0, so ||D^{u} || = 0.

### F. If, for t = T , ||D^{T +1} || = ||D^{T} ||, then = 0 ∀j ∈ [0, n]

On one hand we have

On the other hand

and we suppose that ||D^{T +1} || = ||D^{T} ||, thus

and as σ_{0} = 0

In this last relation, the left-hand side member is lower or equal to zero because σ_{k} − 1 ≤ 0 and but the right-hand side member is greater than or equal to zero because ≥ 0. This implies that the equality is realized only if or

We conclude that, under the condition of the statement, the group does not contain a path of length T

connecting S_{0} to S_{j} . And this conclusion is then verified for u ≥ T because

### G.

We showed that the sequence ||D^{t} || is decreasing and limited by 0. But nothing proves that the lower bound of this sequence is zero, value to which it would converge.

It is what we show below by considering successively two complementary hypothesis.

*Let us suppose that the group does not contain cycles*

In this case the longest paths between two companies consist of n segments for a group of n + 1 companies. Thus, from T > n, we have

summation in which there should be in every product of factors at least one term equal to zero, otherwise there would exist a path exceeding the length n. Thus ||D^{T} || = 0 and D^{t} = 0 for all the values of t ≥ T .

*Let us suppose that the group contains at least one cycle*

Moreover, let us suppose that . This implies

As the terms D_{αβ} ≥ 0, it must exist at least one product in these summations such as

*D _{ik1} D_{k1} k_{2} …D_{kt1j} > 0 (**)*

but as t can increase without limit and considering the finite number of companies in a group, there must necessarily exist one cycle which repeats indefinitely.

Practically, we must meet a situation of this type

where, at a certain step, a path starting at S_{0} via some companies S_{a} meets S_{b} which is an entry point into a cycle leading back to S_{b}.

It is important to note that at the entry point S_{b}, there exists at least two shareholders, which implies that D_{ab} < 1 and D_{db} < 1.

In the term considered above, it is clear that one factor will appear with an exponent increasing towards infinity. As that factor is strictly less than 1, it converges to zero. This contradicts our initial hypothesis that the term (**) is strictly positive.

And so we have .

See also: Financial valuation versus valuation of the consolidated numbers. Two identical approaches?

(1) Author of:

- “Direct Consolidation – Consolidate your expertise” – i6doc.com
- and: “Il était une fois … Le Dernier Théorème de Fermat” – i6doc.com