## 5 Conclusions
They depend on and can influence higher-level factors, while at the same time exerting some influence on lower-level factors. In other words, factors within the same level are interdependent, since they are subject to the influence of both higher and lower factors. The use of a hierarchical structure within the AHP helps us to organize the factors to be evaluated logically and systematically, taking into account the interdependencies between them. This enables more informed and thoughtful decisions to be made, based on a complete and accurate assessment of the relevant factors [48]. The general scheme of comparative hierarchy is shown in Figure 11.  
Once the criteria have been established, the AHP allows you to assign a relative weight to each criterion based on its importance concerning the objective (superordinate level). To determine the weight of the criteria, it is first necessary to compile a pairwise comparison matrix, through which each criterion is compared with the other criteria of the same hierarchy level using a dominance coefficient or preference index (\(\mathrm{a_{ij}}\)). The comparison takes place through a numerical scale of preference called [PERSON]'s semantic scale, as shown in Table 11.  
Starting from the m selected criteria and performing all pairwise comparisons, the matrix of pairwise comparisons \(\mathbf{A}\) is defined, in which: the element \(\mathrm{a_{ij}}=\mathrm{w_{i}}/\mathrm{w_{j}}\) relating to the i-th row and the j-th column expresses the comparison between the i-th criterion and the j-th criterion; \(\mathrm{w_{i}}\) is the weight of criterion i with respect to the objective; and \(\mathrm{w_{j}}\) is the weight of criterion j with respect to the objective. To make a comparison between any pair of m criteria, \(\mathrm{m(m-1)/2}\) comparisons are needed. This means that each element is compared to all other elements, except itself. In comparison matrix \(\mathbf{A}\), the items on the main diagonal are equal to 1, since an item cannot be compared with itself. The other elements in the comparison matrix are the reciprocals of the previous comparisons, since the comparison between element i and element j is the same as the comparison between element j and element i, but with inverse values. Matrix \(\mathbf{A}\) is shown below [49; 50; 51; 52]:  
\[\mathbf{A}= \begin{bmatrix}\frac{\mathrm{w_{1}}}{\mathrm{w_{1}}}&\frac{\mathrm{ w_{2}}}{\mathrm{w_{2}}}&\cdots&\frac{\mathrm{w_{1}}}{\mathrm{w_{m}}}\\ \frac{\mathrm{w_{2}}}{\mathrm{w_{1}}}&\frac{\mathrm{w_{2}}}{\mathrm{w_{2}}}& \cdots&\frac{\mathrm{w_{2}}}{\mathrm{w_{m}}}\\ \vdots&\vdots&\ddots&\vdots\\ \frac{\mathrm{w_{m}}}{\mathrm{w_{1}}}&\frac{\mathrm{w_{m}}}{\mathrm{w_{2}}}& \cdots&\frac{\mathrm{w_{m}}}{\mathrm{w_{m}}}\\ \end{bmatrix} . \tag{1}\]  
Usually, in realistic situations, there are no adequate instruments to measure the real value of weights \(\mathrm{w_{z}}\) with \(1\leq\mathrm{z}\leq\mathrm{m}\). Therefore, in real cases, it is not possible to say with certainty that \(\mathrm{a_{ij}}=\mathrm{w_{i}}/\mathrm{w_{j}}\). This implies that the matrix of pairwise comparisons is not consistent, i.e., \(\mathrm{a_{ij}}\
eq\mathrm{a_{ik}}\mathrm{a_{ik}}\mathrm{a_{kj}}\). Thus, the problem is to find a value \(\mathrm{a_{ij}}\) such that \(\mathrm{a_{ij}}\approx\mathrm{w_{i}}/\mathrm{w_{j}}\). In a consistent matrix (the ideal case), the following is shown [49; 50; 51; 52]:  
\[\mathbf{A}\cdot\mathbf{w}=\lambda_{\mathrm{max}}\cdot\mathbf{w}=\mathrm{m} \cdot\mathbf{w} \tag{2}\]  
This is because, in the case of a consistent matrix, the maximum eigenvalue \(\lambda_{\mathrm{max}}\) is equal to m, which corresponds to the order of the square matrix. In the case of a not perfectly consistent matrix, to estimate the vector \(\mathbf{w}\), [PERSON] proposes two methods as follows:  
* Exact: raising the matrix consecutively to power and normalizing the sums of the row values;
* Approximate: normalizing values by columns, adding rows and normalizing by the total.