When matrix theory explains consolidation ...

By Allen White (1), Sigma Conso co-founder & administrator
Adobe Stock 268784201


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.

Schema I Description

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 companySi with i ∈ [1, n] represents any company of the group, excepted the parent companyDij represents the direct percentage of company Si in company Sj with i ∈ [0, n], j ∈ [1, n] and i ≠ jDii = 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

II Writing Convention

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

II Convention d écriture MATRICE

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:

IV Calculons le vecteur

By increasing the number of iterations infinitely and knowing that

IV Calculons le vecteur 2

(refer to section VI for a complete proof), we can also write

IV Calculons le vecteur 3

And then

IV Calculons le vecteur 4

This last relation shows that

IV Calculons le vecteur 5

Let us reconsider the relation (*) which gives

IV Calculons le vecteur 6

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

VI Matrice 1

And the following results

VI Matrice 2
VI Matrice 3
VI Matrice 4

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