Two Proofs and One Algorithm Related to the Analytic Hierarchy Process

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Introduction
Literature regarding the Analytic Hierarchy Process is rather extensive.The well-known database Current Contents Connect provides a growing number of records in all document types (articles, books, reports, reviews, etc.) regarding the acronym AHP in a publication title.For the last 15 years, the number of records can be seen in Figure 1.These publications are mainly focused on different AHP applications, for example, a brown coal deposit [1], a comparative analysis of group aggregation techniques [2], a lab fire prevention management system [3], a green vendor evaluation and selection in production outsourcing in mining industry [4], an approximation of risk assessment [5], an evaluation of healthcare equipment [6], or many others.
Recently, the mathematical principles of AHP were published by Saaty [7].This monograph involves a lot of mathematical findings that were collected in the area of decision-making processes, particularly, in the area of the AHP, starting from first publication [8] in 1980.
In this paper, we add two proofs regarding the main statement in AHP theory as well as a corresponding algorithm for practical use.Neither proofs nor algorithm has been involved in [7].
Let us consider a consistent matrix ,  ×  ( ≥ 2) in the form where  1 ,  2 , . . .,  −1 are given positive constants, known in AHP theory as priorities.Let the column vector  = ( 1 ,  2 , . . .  ) T reflect the probabilities   (  ≥ 0) of disjoint events   = 1, 2, . . .,  so that the space Ω of elementary events   is given as a union Ω =  1 ∪ 2 ∪. . .  of such events   , which are disjoint (  ∩   = ⊘,  ̸ = ).Elementary events represent different criteria or alternatives in the AHP scheme and probabilities   = (  ) are known as the weights.As usual, it must be required for probability components   .Product  is a column vector whose components are sums of many addends if  is sufficiently large natural number.Therefore, it is better to find a real number  such that the simpler product  is equal to the complex product .Hence, we have the equation  = .Certainly, such equation has the trivial solution  = (0 0 . . .0) T for arbitrary real number  but this is not interesting for us because this zero solution does not satisfy condition (2).If somebody is looking for nontrivial solutions  ̸ = (0 0 . . .0) T , then the system  = , rewritten into the form ( − ) =  → 0 must have a singular matrix  − , i.e., determinant      −      = 0. Here,  marks the unit matrix.

Derivation
Now, we will demonstrate a way for how to derive the determinant      −      for arbitrary consistent matrix  of type  × .By the calculation of determinant | − | [9][10][11], resp.by means of Laplace expansion the validity of following formula can be evaluated The cases  = 2 and  = 3 are considered as special using fundamental rules like Sarrus.Next, we present two different ways for how to prove statement (3), which are involved neither in [7,12] nor in other works known to the authors of this paper.The first proof is done by the direct calculation of the determinant on the left hand side of statement (3).The second proof is based on the mathematical induction method.

Proof by Direct Calculation of Determinant
At the beginning, we arrange the determinant of (3).We choose the factors 1/ 1 , 1/ 2 , . . ., 1/ −1 from the second, third, fourth, up till -th row.Thus, we get By this way, we have arranged the determinant of (3) to the simple determinant of type  × , where  = 1-.
Next, we will express such determinant.By elimination in the first column and by selection of the common factor ( − 1)/ from corresponding rows, we will get The similar elimination in the second, third, fourth up till th column provides sequentially If we realize that the last determinant is equal to product of all diagonal elements and using elementary operations, the final result can be expressed which is the same result as the right hand side of statement (3), provided that  = 1 − .

Proof by Mathematical Induction Method
The equation ( 3) is valid for  = 2. Next, we suppose that statement (3) holds for arbitrary natural  ( ≥ 2) and it will be proved that the statement is valid also for the next natural, i.e., n + 1.So, we will prove that Now, we arrange all determinants of type  ×  so that the row of units is the first row.This can be achieved by switching adjacent rows around, which operation leads to a sign change of the determinant.Thus, we have The multiple common determinant (with units in the first row) can be chosen after bracket and due to this, we obtain We arrange the sum near the first determinant in the form Then, the second determinant is substituted according to the induction assumption (3) or provided  = 1−.It is proven that the determinant on the left hand side of ( 10) is equal to the expression on its right hand side.Thus, statement (3) holds for arbitrary natural  ( ≥ 2) as it results from mathematical induction methodology.

Main Statement
Both the proof by direct calculation of determinant in Section 3 and the proof by mathematical induction method in Section 4 lead to the same result; namely, the determinant      −      is equal to (−) −1 ( − ) for arbitrary consistent matrix  with arbitrary natural  ( ≥ 2).
Hence, the maximal eigenvalue is  max =   = .The reason why we are not interested in the other eigenvalues is simple: the components of eigenvectors  = ( 1  2 . . .  ) T for the other eigenvalues do not fulfil the conditions   ≥ 0,  = 1, 2, . . ., .We demonstrate this by the next simple example.
Example 1.Let us consider the following consistent matrix of type 3 × 3.According to Theorem 1, the matrix has the following eigenvalues:  1 =  2 = 0 and  3 = 3. Determine the eigenvectors of nonnegative components for eigenvalues  2 = 0 and  3 = 3.
On the other hand,  3 = 3.Then, the system ( − 3) =  → 0 by the same elimination provides and including condition (2) one can get the unique solution w = (2/9 2/3 1/9) T , where all components are nonnegative.Thus, the result  =  max is evident.

Algorithm for Determination of Components
In this section, we will demonstrate an algorithm which determines all components   ≥ 0,  = 1, 2, . . ., , of the solution  = ( 1  2 . . .  ) T if a consistent matrix A of type  ×  is given by arbitrary natural n (n≥ 2); see (1).As we have seen in Section 4, it must be solved by (−) =  → 0 with  =  max = ; otherwise some components   can be negative and therefore these components do not represent probabilities of any event.In this case, the system ( − ) =  → 0 has the explicit form which is arranged by the following way.First, we multiply the second, third, up to n-th row by  1 ,  2 , . . .,  −1 .The following is obtained Second, we rewrite the last system in the form to get a simpler matrix.Then, we use the elimination method as ( ( ( We can see that the last row is zero which is evident due to determinant      −      with the eigenvalue  =  being of eigenvalues of  =  with a consistent matrix A of type  ×  ( ≥ 2), and it demonstrates how to find the corresponding eigenvector components presented as some probabilities of events for maximal eigenvalue  =  max = .
The derived and suggested algorithm can be easily programmed in different languages (C++, C, FORTRAN, MATHEMATICA, etc.) and can be used in AHP methodology for determination of weights (probabilities) when comparing the criteria to a goal or when comparing the alternatives to an individual criterion.During such comparison, it is important to make sure that the criteria or alternatives present a set of disjoint events (  ∩   = ⊘,  ̸ = ).If some events, say   and   , are such that   ∩   = ⊘,  ̸ = , then condition (2) does not hold and the calculated weights do not reflect any real examined problem.In such a case, we suggest considering the intersection   ∩   as a new event, say   =   ∩   , and incorporating this new event into the previous set of events.Next, we note that the statement and algorithm suggested in the paper hold for arbitrary natural  ( ≥ 2).This enables formulating and solving the relatively complex AHP problems with large number of criteria and huge number of alternatives like those in [1][2][3][4][5][6].
During practical evaluation of AHP, it can happen that the matrix  is not consistent but is close to a consistent matrix   .In this case, certainly,  max (A) differs from  max (  ) and the corresponding difference is measured by means of, socalled, consistency index that was introduced by Saaty in [8].If the consistency index is close to zero, then the suggested algorithm can be applied, but condition (2) holds only approximately.The usual problem of how to obtain  max () for a nonconsistent general matrix A can be solved by means of the software packages which provide eigenvalues for the general full matrices.Such access, however, does not suppose any special matrix structure, like AHP matrix (1), and the eigenvalue software calculations in this general case are computationally difficult, especially when  is large natural number.The special structure nonconsistent matrices which are close to AHP matrix are not considered in this paper and can be analyzed in the future research.
It can be said that there are some methods to repair consistency of matrices.Xu [13] defined a criterion to find unusual and false element and proposed method based on finding such elements in matrix and repair it.The repaired matrix has (after calculating new element instead of unusual one) an acceptable consistency.Saaty [12] proposed method based on additive perturbation using     /  ,  ̸ = , where  is an eigenvector, and focused their attention on element of the matrix  which provokes inconsistency.The consistency is generally more acceptable by substituting this element.

Figure 1 :
Figure 1: Records of AHP term in the database.