Analytic network process (ANP) was put forward by Saaty [1] in order to solve Multiple Criteria Decision Making (MCDM) problems. From then on, there are a large number of literatures concerning the theory and application of the ANP. In [2], the author developed a classification of methods based on criteria interaction phenomenon and discussed the Decision-Making Trial and Evaluation Laboratory Model (DEMATEL) and ANP hybridization which was one of the most promising approaches to handle criteria interactions in a Multiple-Attribute Decision Making. In [3], the author applied a model by combining the graph-theory based on DEMATEL method with an ANP to solve performance evaluation problems for hot spring hotels. In [4], the author suggested an improved information system (IS) project by using ANP and a zero-one goal programming model to solve information system project selection problems. In [5], the author proposed an application of the ANP for the selection of product mix for efficiency manufacturing in a semiconductor fabricator and generated a performance ranking of product mixes. Results of the study provided guidance for orders considering all aspects of the product. In [6], The author processed parametric study and performance-based multi-criteria optimization of the indirect-expansion solar-assisted heat pump through the integration of ANP decision-making with multi-objective particle swarm optimization (MOPSO) algorithm.

Since a lot of subjective information may skew the final result. In order to get a better result, many studies have made some improvements on the ANP method [7-12]. One of those, the super-matrix improvements of the ANP have been developed for getting more scientific outcome in [7], Wang studied several main structures of ANP super-matrices. In [8], He Chunchun and Wang researched on ANP form the perspective of algorithm improvements and model combinations. But a row or column in a super-matrix is not noticed as a zero vector. Therefore, in this study, the problem that risk indicators cannot be ranked is solved from the perspective of matrix.

As a common method based on multi-objective decision analysis, ANP aims to compare and sort evaluation criteria by establishing pair comparison matrices among criteria, so as to make the optimal choice. In other words, the higher the relative weight, the larger proportion of the evaluation criterion, and the more attention should be paid to it. However, there may be some deviations in the traditional ANP method, which makes it impossible to sort the evaluation criteria. A row or column in a super-matrix is not noticed as a zero vector. Therefore, in this study, the problem that risk indicators can not be ranked is solved from the perspective of matrix. The improvement of ANP super-matrix is helpful to correct this deviation. Therefore, we transform the judgment matrix and get the relative impact ranking of the evaluation criterion.

The purpose of this paper is to address a particular situation in which risk indicators cannot be ranked, we make some improvements to ANP. The rest of the paper is set out as follows. In Section 2, we present the basics of the ANP. In Section 3, we investigate the existence of a row or column of zero in a super-matrix. Constructing the super-matrix is the core of ANP, when the super-matrix satisfies Theorem 2.5.1, the weight of risk indicators can be obtained through calculation. But it cannot be used in a specific instance when a row or column in a super-matrix is a zero vector. In response to this situation, we apply a method to improve the super-matrix on the basis of ANP as shown in Step 4. Section 4 gives an example.

In the ANP method, the decision problem is formulated into a network structure. A general structure of ANP is shown in Figure 1.

ANP divides system elements into two major parts: control layer and network layer. The elements of these two parts relate to each other. As shown in Figure 1, the direction of the arrows in the network structure plays an important role of representing the right relationship between two parts [8].

The control layer includes the goal of the evaluation program and decision criteria. All decision criteria are considered independent of each other and governed only by objective factors. The control layer can have no decision criteria, but at least one goal should be presented. The network layer is composed of factors that is dominated by the control layer. As shown in Figure 1, risk factor groups $ \{{R_{1}}, {R_{2}}, {R_{3}}, \cdots , {R_{n}}\} $, and there are risk indicators under each risk indicator group, for example: factor group $ {R_{i}} $ includes risk factors $ \{e_{i1}, e_{i2}, \cdots, e_{in_{i}}\} $, an arc with an arrow represents an inner dependence relation in a cluster and an arrow represents a dependence among clusters and factors across different clusters. And there is a case where there is no connection between the elements of two sets.

After determining the network structure, pair-wise comparisons between different clusters and facts are given by the risk assessment experts. The comparison of dominance is as follows: give a criterion and compare the importance of two elements under this criterion. A scale of $ 1-9 $ is used to indicate how important the two elements relative to the criterion. The $ 1-9 $ scale is displayed in Table 1.

Experts put forward the degree of influence that each criterion $ e_{if}(f=1,2,\cdots,n_{i}) $ to $ e_{pq}(p,q=1,2,\cdots,n_{i}) $, which was denoted by $ a_{if} $, using the 1-9 scale. Then the normalized vector is derived though

$ \begin{align} w^{(pq)}_{in_{i}}=\frac{a_{if}}{a_{i1}+a_{i2}+\cdots+a_{in_{i}}}. \end{align} $

(2.1)

Therefore pair-wise comparisons are expressed in Table 2.

After passing the consistency test, the normalized eigenvectors of each pair-wise comparison matrix can be combined into a super-matrix. Let $ W_{ij} $ be the influence degree of indicator set $ R_{i} $ on indicator set $ R_{j} $.

$ \begin{equation*} W_{ij}=\left[ \begin{array}{cccc} w^{(j1)}_{i1}&w^{(j2)}_{i1}&\cdots&w^{(jn_{j})}_{i1}\\ w^{(j1)}_{i2}&w^{(j2)}_{i2}&\cdots&w^{(jn_{j})}_{i2}\\ \vdots&\vdots&\vdots&\vdots\\ w^{(j1)}_{in_{i}}&w^{(j2)}_{in_{i}}&\cdots&w^{(jn_{j})}_{in_{i}} \end{array}\right], \end{equation*} $

where $ w^{(jl)}_{ik} $ ($ k=1,2,...,n_i $, $ l=1,2,...,n_j $) is given as in (2.1),

Due to the number of risk fact sets in the network layer are $ n $, so the number of $ W_{ij} $ is $ n\times n $. If the index $ e_{n_{i}} $ has no influence on the index $ e_{n_{j}} $, the corresponding element in $ W_{ij} $ is $ 0 $. Let $ W_{ij} $ be a element of the super-matrix $ W $:

$ \begin{equation*} W=\left[ \begin{array}{cccc} W_{11}&W_{12}&\cdots&W_{1n}\\ W_{21}&W_{22}&\cdots&W_{2n}\\ \vdots&\vdots&\vdots&\vdots\\ W_{n1}&W_{n2}&\cdots&W_{nn}\\ \end{array}\right]. \end{equation*} $

In the super-matrix $ W $, each $ W_{ij} $ is a matrix whose column is normalized or equal to $ 0 $. But the super-matrix $ W $ is not a column normalized matrix.

In order to let the super-matrix be a column normalized matrix, we need to construct the weighted matrix. Rise the pair-wise comparison between factors to the pair-wise comparison between factor groups $ \{R_{i}\}(i=1,2,3,\cdots,n) $. Let control layer elements $ P $ as guidelines to search the influence degree of the factor groups $ \{R_{i}\} $. In a similar way, if the factor group $ R_{i} $ has no influence on factor group $ R_{j}(j=1,2,3,\cdots,n) $, the corresponding element in normalized vector is $ 0 $. Experts are asked to reach the degree of direct influence that each indicator $ R_{j} $ exerts on each indicator $ R_{i} $, which is denoted by $ b_{mi}(m=1,2,3,\cdots,n) $, using the 1-9 scale. A normalized vector is then derived by $ u_{mi}=\frac{b_{mi}}{b_{1i}+b_{2i}+\cdots+b_{ni}} $ and shown in Table 3.

So, all normalized vectors form the weighted matrix $ U $:

$ \begin{equation*} U=\left[ \begin{array}{cccc} u_{11}&u_{12}&\cdots&u_{1n}\\ u_{21}&u_{22}&\cdots&u_{2n}\\ \vdots&\vdots&\vdots&\vdots\\ u_{n1}&u_{n2}&\cdots&u_{nn}\\ \end{array} \right] \end{equation*} $

Finally, the weighted super-matrix $ W^{\ast} $ is obtained by combining the super-matrix $ W $ and the weighted matrix $ U:W^{\ast}=U\odot W $, $ (\odot: w^{\ast}=u_{ij}W_{ij}) $ [7].

After getting the weighted super-matrix, we can not determine the weight. The weighted super-matrix $ W^{\ast} $ needs a stable tackle that $ W^{\ast} $ is constantly multiplied and normalized to form matrix convergence.

Theorem 2.5.1[7] If the weighted super-matrix $ W^{\ast} $ satisfies

(1) the maximum eigenvalue of $ W^{\ast} $ is $ 1 $;

(2) the maximum eigenvalue $ 1 $ is a single;

(3) modulus of other eigenvalues are all less than $ 1 $.

Then,

(i) there is the limited matrix of $ W^{\ast} $, let $ W^{\infty}=\lim_{n\rightarrow \infty}(W^{*})^{n} $;

(ii) each column of $ W^{\infty} $ is identical and is the normalized eigenvector of $ W^{\ast} $ belonging to the maximum eigenvalue $ 1 $;

(iii) the weighted of indexed is obtained, each column of $ W^{\infty} $ is the relative weighted of indexes.

In the process of solving the weighted super-matrix, there might be a situation: there is no influence between factor group $ R_{i} $ and factor group $ R_{j} $, as in this case, the weighted super-matrix $ W^{\ast} $ must be a column (or a row) where all elements of them are zero. It can be seen that the magnitude of all eigenvalue of $ W^{\ast} $ is less than $ 1 $ and $ 1 $ is not eigenvalue of $ W^{\ast} $. In this study, we will solve the problem that risk indicators cannot be ranked from the perspective of matrix.

In this case, if there is a column (or a row) where all of them are zero in the weighted super-matrix $ W^{\ast} $. On the basis of the obtained weighted super-matrix, we improve it for further calculation. Thus, we can get improved calculation steps of the weighted super-matrix of ANP method:

Step 1 According to Section 2.2, the pair-wise comparisons by decision-makers are obtained.

Step 2 According to Section 2.3, based on Step 1, the super-matrix is obtained.

Step 3 According to Section 2.4, we compute the weighted super-matrix $ W^{\ast} $.

Step 4 We improve the weighted super-matrix $ W^{\ast} $.

$ \exists j $, $ \forall i(i=1,\cdots,n) $, $ w^{\ast}_{ij}=0 $ defining the column (or the row) of the weighted super-matrix $ W^{\ast} $ are $ \frac{1}{n} $. $ n $ is the size of the weighted super-matrix $ W^{\ast} $.

Step 5 By calculating the dominance degree of risk indicators, risk indicators are sorted by the dominance degree.

This approach is based directly on conventional operation of ANP. It is shown that the use of known methods may lead to nothing when using the basic ANP method. It is seen that our method provides a feasible solution under the premise of ensuring the subjectivity of decision makers. The main advantage of the proposed method is that it provides a simple way of solving about a complex situation.

Using numerical examples, it is shown that the proposed improved ANP method may provide the final ranking of risk indicators.

The objective of this case is to solve the complex decision problem by using ANP with LPT (live, play, transportation). We attempt to develop an ANP model about the travel problem. Now, we have four destinations: $ Chengdu $, $ Gansu $, $ Sanya $, $ Xinjiang $. The first includes a control hierarchy or network of criteria that control the interactions: live $ P_{1} $, play $ P_{2} $, transportation $ P_{3} $. The second consists of three kinds of subnetworks: benefits $ R_{1} $, costs $ R_{2} $, risks $ R_{3} $. So, ANP with LPT model about travel problem is shown in figure 2.

Two types of connections between nodes contained in groups are represented in figure 2 as one-way influence and two-way influence. If there is one-way influence between the two groups are represented with directed arrows. The two-way influence is represented by bi-directed arrows. Group $ R_{1} $ includes two risk factors: convenience $ R_{11} $ and experience $ R_{12} $. Group $ R_{2} $ includes two risk factors: traffic $ R_{21} $ and residence $ R_{22} $. And group $ R_{3} $ includes two risk factors: epidemic $ R_{31} $ and surroundings $ R_{32} $.

Step $ 1 $  The pair-wise comparisons by the decision-maker are shown in Table 4 and Table 5.

We can obtain the comparison matrix $ w_{11} $ for the factors in $ R_{1} $ to the factors in $ R_{1} $ through Table 4 and Table 5.

$ \begin{equation*} w_{11}=\left[ \begin{array}{cc} \frac{1}{3}&\frac{1}{3}\\ \frac{2}{3}&\frac{2}{3}\\ \end{array} \right]. \end{equation*} $

In the similar way,

$ \begin{equation*} w_{12}= \left[ \begin{array}{cc} \frac{1}{13}&\frac{5}{11}\\ \frac{12}{13}&\frac{6}{11}\\ \end{array} \right] ,\quad w_{13}= \left[ \begin{array}{cc} 0&\frac{4}{7}\\ 0&\frac{3}{7}\\ \end{array} \right] ,\quad w_{21}= \left[ \begin{array}{cc} \frac{2}{5}&\frac{1}{16}\\ \frac{3}{5}&\frac{15}{16}\\ \end{array} \right] ,\quad w_{22}=\left[ \begin{array}{cc} \frac{1}{8}&\frac{1}{8}\\ \frac{7}{8}&\frac{7}{8}\\ \end{array} \right] , \end{equation*} $

$ \begin{equation*} w_{23}= \left[ \begin{array}{cc} 0&\frac{2}{5}\\ 0&\frac{3}{5}\\ \end{array} \right] ,\quad w_{31}= \left[ \begin{array}{cc} 0&0\\ 1&1\\ \end{array} \right] , \quad w_{32}= \left[ \begin{array}{cc} 0&0\\ 1&1\\ \end{array} \right] ,\quad w_{33}= \left[ \begin{array}{cc} 0&\frac{5}{6}\\ 0&\frac{1}{6}\\ \end{array} \right] \\ \end{equation*} $

Step $ 2 $  According to step $ 1 $ result, the super-matrix is

$ \begin{equation*} W= \left[ \begin{array}{cccccc} \frac{1}{3}&\frac{1}{3}&\frac{1}{13}&\frac{5}{11}& 0 &\frac{4}{7}\\ \frac{2}{3}&\frac{2}{3}&\frac{12}{13}&\frac{6}{11}& 0 &\frac{3}{7}\\ \frac{2}{5}&\frac{1}{16}&\frac{1}{8}&\frac{1}{8}& 0 &\frac{2}{5}\\ \frac{3}{5}&\frac{15}{16}&\frac{7}{8}&\frac{7}{8}& 0 &\frac{3}{5}\\ 0 & 0 & 0 & 0 & 0 &\frac{5}{6}\\ 1 & 1 & 1 & 1 & 0 &\frac{1}{6}\\ \end{array} \right]. \end{equation*} $

Step $ 3 $  The decision-maker gives $ R_{j}(j=1,2,3) $ to a $ R_{i}(i=1,2,3) $ level of direct impact on 1-9 scale.

We can get the weighted matrix $ U $ through Table 6, Table 7 and Table 8.

$ \begin{equation*} U=(u_{ij})=\left[ \begin{array}{ccc} \frac{2}{5}&\frac{1}{3}&\frac{3}{7}\\ \frac{1}{5}&\frac{1}{6}&\frac{1}{7}\\ \frac{2}{5}&\frac{1}{2}&\frac{3}{7}\\ \end{array} \right] ( i=1,2,3,j=1,2,3 ). \end{equation*} $

Thus, the weighted super-matrix $ W^{\ast} $:

$ \begin{equation*} W^{\ast}=U\odot W= \left[ \begin{array}{cccccc} \frac{2}{15}&\frac{2}{15}&\frac{1}{39}&\frac{5}{33}& 0 &\frac{12}{49}\\ \frac{4}{15}&\frac{4}{15}&\frac{4}{13}&\frac{2}{11}& 0 &\frac{9}{49}\\ \frac{2}{25}&\frac{1}{80}&\frac{1}{48}&\frac{1}{48}& 0 &\frac{2}{35}\\ \frac{3}{25}&\frac{3}{16}&\frac{7}{48}&\frac{7}{48}& 0 &\frac{3}{35}\\ 0 & 0 & 0 & 0 & 0 &\frac{5}{14}\\ \frac{2}{5}&\frac{2}{5}&\frac{1}{2}&\frac{1}{2}& 0 &\frac{1}{14}\\ \end{array} \right]. \end{equation*} $

By computation to find the eigenvalue of the weighted super-matrix $ W^{\ast} $.

$ (0, 0.8755, -0.2117, 0.0922i, -0.0922i, -0.0257) $

It can be found that the six output eigenvalues are all less than $ 1 $, and $ 1 $ is not the eigenvalue of the weighted super-matrix $ W^{\ast} $.

Step $ 4 $  We improve the weighted super-matrix $ W^{\ast} $, as follows:

$ \begin{equation*} W^{\ast}=\left[ \begin{array}{cccccc} \frac{2}{15}&\frac{2}{15}&\frac{1}{39}&\frac{5}{33}&\frac{1}{6}&\frac{12}{49}\\ \frac{4}{15}&\frac{4}{15}&\frac{4}{13}&\frac{2}{11}&\frac{1}{6}&\frac{9}{49}\\ \frac{2}{25}&\frac{1}{80}&\frac{1}{48}&\frac{1}{48}&\frac{1}{6}&\frac{2}{35}\\ \frac{3}{25}&\frac{3}{16}&\frac{7}{48}&\frac{7}{48}&\frac{1}{6}&\frac{3}{35}\\ 0 & 0 & 0 & 0 &\frac{1}{6}&\frac{5}{14}\\ \frac{2}{5}&\frac{2}{5}&\frac{1}{2}&\frac{1}{2}&\frac{1}{6}&\frac{1}{14}\\ \end{array} \right]. \end{equation*} $

The eigenvalues $ E $ and corresponding eigenvectors $ V $ of the improved super-matrix can be obtained by calculation.

$ E=(1, -0.1366+0.2058i, -0.1366-0.2058i, 0.0780, 0.0922i, -0.0922i) $

$ \begin{equation*} V=\left[ \begin{array}{cccccc} 0.3722&-0.2837-0.0859i&-0.2837+0.0859i&-0.5640&0.5705&0.5705\\ 0.4931&0.1406-0.0589i&0.1406+0.0589i&-0.4362&-0.5705&-0.5705\\ 0.1293&-0.0290+0.3507i&-0.0290-0.3507i&0.4635&-0.4178i&0.4178i\\ 0.3033&0.0524+0.1322i&0.0524-0.1322i&0.1745&0.4178i&-0.4178i\\ 0.2812&-0.4984-0.3381i&-0.4984+0.3381i&0.4817& 0 &0\\ 0.6562&0.6181&0.6181&-0.1195&0&0\\ \end{array} \right] \end{equation*} $

Step $ 5 $  It can be obtained that the maximum eigenvalue of $ W^{\ast} $ is $ 1 $, and the only normalized eigenvector that corresponds to $ 1 $ is

$ (0.3722, 0.4931, 0.1293, 0.3033, 0.2812, 0.6562). $

Now, Let us do a consistency check. According to the mean random consistency index (RI) of order $ 1-15 $ in table 9, we can obtain that the consistency test for the super-matrix

$ CR=\frac{CI}{RI}=\frac{\frac{\lambda_{max}-n}{n-1}}{RI}=-0.64<0.1. $

Therefore, we consider the consistency of the matrix to be acceptable.

Based on the results from the only normalized eigenvector, the priority of the risk factors in this case is

$ R_{32}>R_{12}>R_{11}>R_{22}>R_{31}>R_{21}. $

Thus, we can solve such situations by the improved Analytic Network Process.

In this study, in order to solve the problem that the ANP of risk factors cannot be ranked, the weight of risk factors can be obtained by transforming the weighted super-matrix and then the influence degree of risk factors of policymaker' subjective can be obtained under a certain criterion. A case is used to validate the applicability and efficacy of the proposed approach. The results show that the proposed method can be an effective method for ANP risk ranking.

Although the proposed method is an effective improvement strategy for ANP, it still has some limitations, because calculations against large amounts of data can be relatively cumbersome.