## 4 Interval VIKOR Method for Estimating Steady-State Power Quality  
Since the weight obtained by the method in Section 3 is an interval number, we use interval VIKOR method for comprehensive power quality assessment. The interval VIKOR method not only has the advantages of the traditional VIKOR method in dealing with multi-objective decision-making problems, but it can also take into account the ambiguity and uncertainty of people in making decisions about certain issues [37]. To comprehensively evaluate steady-state power quality with the interval VIKOR method, it is mainly divided into the following six steps [37; 38; 39]:  
Step 1. Establish the original matrix. The original data matrix is the monitored value (the deviation from standard value) of each power quality index. Suppose there are a total of \(m\) datasets, each set containing \(n\) indices. The original matrix can be obtained according to Equation (27):  
\[X= \begin{bmatrix}x_{11}&x_{12}&\cdots&x_{1n}\\ x_{21}&x_{22}&\cdots&x_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ x_{m1}&x_{m2}&\cdots&x_{mm}\end{bmatrix}  \tag{27}\]  
where \(x_{ij}\) represents the monitoring value of the \(j\)th index on the \(i\)th set. In this paper, we evaluate the power quality of different dates of the same monitoring point. For comparison, we take the level upper boundary values given in Table 2 as raw data to constitute the original matrix. Step 2. Standardize the original matrix. Due to the different dimensions of different indices, the data should be standardized before the evaluation in order to eliminate the impact between different indices. The power quality index is a reverse index: the smaller is the value, the better is the power quality. Thus, we choose the extreme value method to standardize the original matrix by Equation (28) [40]:  
\[y_{ij}=\frac{x_{ij}}{x_{j\max}} \tag{28}\]  
where \(x_{j\max}\) represents the maximum value in the \(j\)th index. Then, the standardized matrix can be presented by Equation (29). \[Y= \begin{bmatrix}y_{11}&y_{12}&\cdots&y_{1n}\\ y_{21}&y_{22}&\cdots&y_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ y_{m1}&y_{m2}&\cdots&y_{mm}\end{bmatrix}  \tag{29}\]  
Step 3. Determine the positive ideal solutions (PIS) and the negative ideal solutions (NIS). The PIS and the NIS can be determined using Equations (30) and (31):  
\[P^{+}=\left\{ \begin{array}{cccc}p_{1}^{+}&p_{2}^{+}&\cdots&p_{n}^{+}\end{array}  \right\}=\left\{\underset{i}{\max}y_{ij}\big{|}j=1,2,\cdots n\right\} \tag{30}\]  
\[P^{-}=\left\{ \begin{array}{cccc}p_{1}^{-}&p_{2}^{-}&\cdots&p_{n}^{-}\end{array}  \right\}=\left\{\underset{i}{\min}y_{ij}\big{|}j=1,2,\cdots n\right\} \tag{31}\]  
where \(P^{+}\) and \(P^{-}\) represent PIS and NIS, respectively. It should be pointed out that, since the power quality index is a reverse index, PIS and NIS obtained here are opposite to the actual expressions. Step 4. Determine the value of \(S_{i}^{\pm}\) and the value of \(R_{i}^{\pm}\). The interval values of \(\left[S_{i}^{-},S_{i}^{+}\right]\) and \(\left[R_{i}^{-},R_{i}^{+}\right]\) can be determined using Equations (32)-(35):  
\[S_{i}^{-}=\sum\limits_{j=1}^{n}\frac{\omega_{j}^{-}\left(p_{j}^{+}-y_{ij} \right)}{\left(p_{j}^{+}-p_{j}^{-}\right)}\quad i=1,2,\cdots,m \tag{32}\]  
\[S_{i}^{+}=\sum\limits_{j=1}^{n}\frac{\omega_{j}^{+}\left(p_{j}^{+}-y_{ij} \right)}{\left(p_{j}^{+}-p_{j}^{-}\right)}\quad i=1,2,\cdots,m \tag{33}\]  
\[R_{i}^{-}=\max\limits_{j}\frac{\omega_{j}^{-}\left(p_{j}^{+}-y_{ij}\right)}{ \left(p_{j}^{+}-p_{j}^{-}\right)}\quad i=1,2,\cdots,m\quad j=1,2,\cdots,n \tag{34}\]  
\[R_{i}^{+}=\max\limits_{j}\frac{\omega_{j}^{+}\left(p_{j}^{+}-y_{ij}\right)}{ \left(p_{j}^{+}-p_{j}^{-}\right)}\quad i=1,2,\cdots,m\quad j=1,2,\cdots,n \tag{35}\]  
where \(S_{i}^{\pm}\) represents the value of group utility of \(i\)th power quality and \(R_{i}^{\pm}\) represents the value of individual regret of \(i\)th power quality. The value of group utility and the value of individual regret are both decision values in a multi-index decision problem [41].