When matrix theory explains the process of group consolidation
by Allen White (1), Sigma Conso co-founder & administrator
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 S0 named parent company, consolidating company or holding company and companies Si 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 S0 into each company Si , i ≠ 0.
For example, the indirect percentage owned in company S3 is calculated as follows
- 10% owned directly by company S0
- 32% = 80% ∗ 40% calculated along the path S0 → S1 → S3
- 2.4% = 60% ∗ 10% ∗ 40% calculated along the path S0 → S2 → S1 → S3
- 12% = 60% ∗ 20% calculated along the path S0 → S2 → S3
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 Si , we just need to inventory all the paths starting at S0 and arriving at Si 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 S2 in S3 and from S3 in S2 .
And it is here that matrices bring a very elegant help.
II. Writing conventions
- S0 represents the parent company of a group which consists of n + 1 companies, including that parent company
- Si with i ∈ [1, n] represents any company of the group, excepted the parent company
- Dij represents the direct percentage of company Si in company Sj with i ∈ [0, n], j ∈ [1, n] and i ≠ j
- Dii = 0 ∀i ∈ [0, n] means that company Si does not own a number of its shares. If so, these shares would be ignored in the calculations.
- Di0 = 0 ∀i ∈ [1, n] means that no company Si owns shares of the parent company. If so, these shares would be ignored in the calculations.
- 0 ≤ Dij ≤ 1 ∀i, j ∈ [0, n] because Dij represents a participation percentage and we assume that 100% = 1.
- Ni ∀i ∈ [0, n] represents the indirect percentage in company Si .
We assume that N0 = 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 Dij 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 Si can be written
Ni = N0 D0i + N1 D1i + N2 D2i + … + Nn Dni
This relation being true for each company, we can build a system of n + 1 equations whose n + 1 unknowns are the indirect percentages Ni . So
If we define
N = (1 N1 N2 … Nn )
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 Nt = U +
Nt−1 D with an initial value N0 = U , and for which we have to check the convergence.
Iteration 1 : N1 = U + N0 D = U + U D
Iteration 2 : N2 = U + N1D = U + (U + UD)D = U + UD + UD2
Iteration t :
By increasing the number of iterations infinitely and knowing that (refer to section VI for a complete proof), we can also write
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 S2 in company S3 .
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 Dij ≥ 0 ∀i, j ∈ [0, n]
- If ||D|| = 0, then and as Dij ≥ 0 we must have Dij = 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 Sj ∀j ∈ [1, n].
- σ0 = 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 Dt?
We define as the element of line i and column j of the matrix Dt. We then can write for example for t = 2
and each element Dik Dkj of that summation for a certain value of k represents the product of direct percentages owned by Si in Sk and Sk in Sj . In other words, it is the product of direct percentages along the path starting from Si and arriving at Sj by passing through Sk. 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 Si to Sj. 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 Si and arriving at the company Sj.
If = 0, we can assert that there is no path of length t between Si and Sj .
This situation can occur for example in the following group of four companies.
Indeed, a path of length two starting at company S0 and arriving at company S3 does not exist and thus
We shall finally notice that in the group of three companies below
that represents crossed participations between S1 and S2, the element contains in particular the term
which corresponds to the path of length five S0 → S1 → S2 → S1 → S2 → S1. 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.
Considering property B paragraph 3 above, we can write
and thus .
E. If, for t = T , ||DT || = 0, then ||Du || = 0 ∀u ≥ T
This is obvious because if ||DT || = 0, then DT = 0 and we have Du = Du−T DT = 0, so ||Du || = 0.
F. If, for t = T , ||DT +1 || = ||DT ||, then = 0 ∀j ∈ [0, n]
On one hand we have
On the other hand
and we suppose that ||DT +1 || = ||DT ||, thus
and as σ0 = 0
In this last relation, the left-hand side member is lower or equal to zero because σk − 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 S0 to Sj . And this conclusion is then verified for u ≥ T because
We showed that the sequence ||Dt || 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 ||DT || = 0 and Dt = 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 Dαβ ≥ 0, it must exist at least one product in these summations such as
Dik1 Dk1 k2 …Dkt1j > 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 S0 via some companies Sa meets Sb which is an entry point into a cycle leading back to Sb.
It is important to note that at the entry point Sb, there exists at least two shareholders, which implies that Dab < 1 and Ddb < 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 .
(1) Author of:
- “Direct Consolidation – Consolidate your expertise” – i6doc.com
- and: “Il était une fois … Le Dernier Théorème de Fermat” – i6doc.com