## 2 Method
### Determine the Principle of Initial Value in Selection Iteration Based on the LMS Method  
According to the data [39], whether the LS resistance estimates collapsed is directly related to the number of outliers. The IGG-ELM method can accurately identify the abnormal value when the number of outliers is small, but the performance of resistance ability becomes poorer when the number of outliers increases dramatically. Therefore, the Least Median Squares (LMS) method is chosen to substitute for the traditional LS method in this paper to determine the initial value in selection iteration.  
The main principle of LMS is to minimize the median squared residue of the sample sequence. If you select \(n\) from the \(Q\) observation equations, calculate the parameter \(x\), together with group \(C_{Q}^{n}\) solution, the median of the residual square in the \(i\) group is:  
\[med\Big{[}v_{i}^{2}\Big{]}=median\big{[}v_{i1}^{2}&v_{i2}^{2}&\cdots&v_{in}^{2 }\big{]}^{T} \tag{14}\]  
\begin{table}
\begin{tabular}{c c c c} \hline \hline
**Judgment Condition** & **Monitoring Value** & **Measure** & **Weight** \\ \hline \(|e|<k_{1}\sigma\) & normal & reservation weight & 1 \\ \(\text{k}_{1}\sigma\leq|e|<k_{2}\sigma\) & available & reduce the weight & less than 1 \\ \(|e|\geq k_{2}\sigma\) & abnormal & eliminate & 0 \\ \hline \hline \end{tabular}
\end{table}
Table 2: Treatment measure of IGG method weight.  
where \(i=1,2,\cdots,Q\), and \(median\) is the median operator, then:  
\[v_{med}=\min(\sqrt{med(v_{1}^{2})}\quad\sqrt{med(v_{2}^{2})}\quad\cdots\quad \sqrt{med(v_{Q}^{2})}) \tag{15}\]  
The unit weighted variance expresses as:  
\[\sigma=1.4826v_{med} \tag{16}\]  
This value is estimated as the initial value in the Equation (13).  
Determine the Number of Hidden Layer Nodes through the Incremental Extreme Learning Machine Method (I-ELM)  
Suppose the initial number of nodes in the ELM network structure is \(s-1\), then \(\beta_{s-1}=\left(H_{s-1}^{T}H_{s-1}\right)^{-1}H_{s-1}^{T}T^{\prime}\), when a new node is added in it, the connection weight between the hidden layer and the output layer is \(\beta_{s}=\left(H_{s}^{T}H_{s}\right)^{-1}H_{s}^{T}T^{\prime}\), where \(H_{s}=\left[H_{s-1},v_{s}\right]\), \(v_{s}\) is the hidden layer output vector connected to \(s\) hidden layer node.  
Let \(D_{s-1}=\left(H_{s-1}^{T}H_{s-1}\right)^{-1}H_{s-1}^{T}\), \(D_{s}=\left(H_{s}^{T}H_{s}\right)^{-1}H_{s}^{T}\), then \(\left(H_{s}^{T}H_{s}\right)^{-1}=\left(\left[H_{s-1}^{T}H_{s-1}\quad v_{s} \right]\right)^{-1}=\left[H_{s\overline{T}}^{T}H_{s-1}\quad\frac{H_{s\overline {T}}^{T}}{v_{s}^{T}H_{s-1}}^{T}\right]^{-1}\).  
Due to \(\left(H_{s}^{T}H_{s}\right)^{-1}\) is symmetrical, therefore:  
\[\left(H_{s}^{T}H_{s}\right)^{-1}=\begin{bmatrix}A&B\\ B&E\end{bmatrix}^{-1}=\begin{bmatrix}A^{\prime}&B^{\prime}\\ B^{\prime T}&E^{\prime}\end{bmatrix} \tag{17}\]  
In equation:  
\[\left\{\begin{array}{l}A=H_{s-1}^{T}H_{s-1}\\ B=H_{s-1}^{T}v_{s}\\ C=v_{s}^{T}v_{s}\end{array}\right. \tag{18}\]  
There is:  
\[I=\begin{bmatrix}A&B\\ B^{T}&E\end{bmatrix}\begin{bmatrix}A^{\prime}&B^{\prime}\\ B^{\prime T}&E^{\prime}\end{bmatrix}=\begin{bmatrix}AA^{\prime}+BB^{\prime T} &AB^{\prime}+BE^{\prime}\\ B^{T}A^{\prime}+EB^{\prime T}&B^{T}B^{\prime+}EE^{\prime}\end{bmatrix} \tag{19}\]  
The solution is:  
\[\left\{\begin{array}{l}A^{\prime}=\frac{A^{-1}B(A^{-1}B)}{E-B^{\prime}(A^{-1 }B)}+A^{-1}\\ B^{\prime}=\frac{A^{-1}B}{B^{\prime}A^{-1}B-E}\\ C^{\prime}=\frac{A^{-1}B}{E-B^{\prime}A^{-1}B}\end{array}\right. \tag{20}\]  
Therefore, \(D_{s}\) is expressed as:  
\[D_{s}=\begin{bmatrix}A^{\prime}&B^{\prime}\\ B^{\prime T}&E^{\prime}\end{bmatrix}\begin{bmatrix}H_{s-1}^{T}\\ v_{s}^{T}\end{bmatrix}=\begin{bmatrix}A^{\prime}H_{s-1}^{T}+B^{\prime}v_{s} ^{T}\\ B^{\prime T}H_{s-1}^{T}+E^{\prime}v_{s}^{T}\end{bmatrix} \tag{21}\]  
Let \(D_{s}=\begin{bmatrix}M\\ N\end{bmatrix}\), input Equations (18) and (20) into Equation (21), then \(M\), \(N\) and \(\beta_{s}\) are calculated.