## Appendix A  
Assume that the probability of each uncertain event is independent of each other, the full probability formula can be used to calculate the \(P_{k}^{t}(Y)\). \([Y1,Y2,Y3,Y4,Y5]\) is defined as a division of the event space set \(\Omega\), and define \"faults directly related to ice icing\" as event \(X\), as shown in Figure A1. The distribution line disconnection event and the inverted tower event are inseparable from the stress conditions of the overhead line and the tower. Therefore, it is necessary to perform mechanical analysis on the wire and the corresponding tower. In order to simplify the calculation workload, the distribution line can be regarded as the ideal cable. Therefore, the internal stress \(\sigma_{0}\) can be obtained according to the parabolic formula. The specific calculation process is shown in Equations (A1)-(A7): \begin{table}
\begin{tabular}{c c c c} \hline \hline
**Defense Strategies** & **Switching Action Process** & \begin{tabular}{c} **Number of Switching** \\ **Operations** \\ \end{tabular} & \
\begin{tabular}{c} **Loss of Load** \\ **(10,000 kW)** \\ \end{tabular} \\ \hline no strategy & no & 0 & 29.432 \\ \hline \multirow{2}{*}{only the load transfer strategy} & close ZL1, ZL2; disconnect Z9, Z55, Z58. & 6 & 16.514 \\ \hline \multirow{2}{*}{only the island strategy} & access Z1, Z2, Z3; disconnect & \multirow{2}{*}{7} & \multirow{2}{*}{18.875} \\  & Z9, Z61, Z64, Z67. & & \\ \hline \multirow{2}{*}{
\begin{tabular}{c} coordinated active defense \\ strategy (CADS) \\ \end{tabular} } & access Z2, Z3; close ZL1, ZL2; disconnect Z9, Z29, Z64, Z67. & 8 & 1.61 \\ \hline \hline \end{tabular}
\end{table} 
Table 4: Comparison of three defense strategies. \(F_{1}\) and \(F_{2}\) are the wind load changes caused by the horizontal and vertical wind. The corresponding calculation formulas are shown as:  
\[F_{1}\ =\ 10^{-3}\frac{\rho_{w}}{2}V_{1}^{2}\Big{(}2R_{d}^{j}\ +\ D\Big{)}L_{t} \tag{11}\]  
\[F_{2}\ =\ 10^{-3}\frac{\rho_{w}}{2}V_{2}^{2}\Big{(}2R_{d}^{j-1}\ +\ D\Big{)}L_{t} \tag{12}\]  
In the formula, \(\rho_{w}\) is the air density value, the unit is g/cm\({}^{3}\), which can be obtained from the measured atmospheric pressure. \(V_{1}\) and \(V_{2}\) represent the horizontal and vertical components of the wind speed, the unit is m/s. \(R_{d}^{j}\) is the thickness of the ice accretion. \(D\) is the radius of the distribution line. The self-weight load \(G\) of the distribution line is:  
\[G\ =\ \frac{10^{-3}}{4}\pi D^{2}L_{t}\rho_{1}g_{n} \tag{13}\]  
where \(\rho_{1}\) is the density of the line itself, the unit is g/cm\({}^{3}\). Thus, the predicted value of total load \(H\) can be calculated by the Equation (28), and the total specific load \(\gamma\) of the distribution line is:  
\[\gamma\ =\ \frac{H}{\frac{1}{4}\pi D^{2}L} \tag{14}\]  
In the formula, \(L\) is the distribution line length, the unit is \(m\). The line stress \(\sigma_{n}\) can be calculated by using the force state equations, which are shown as:  
\[\left\{ \begin{array}{c}\sigma_{n}^{2}(\sigma_{n}\ +\ A)\ =\ B\\ A\ =\ \frac{Kg_{n}^{2}L}{24\sigma_{m}}\ -\ \sigma_{m}\ +\ \alpha E(\sigma_{n}\ -\ \theta_{m})\\ B\ =\ \frac{E\gamma^{2}L_{t}^{2}}{24}\end{array} \right. \tag{15}\]  
In the formula, set the subscripts \(m\) and \(n\) are the initial state and current state of the line, respectively. Among them, \(\theta_{n}\) and \(\theta_{m}\) respectively represent the current temperature of the line and the initial temperature during normal operation. \(\sigma_{n}\) and \(\sigma_{m}\) respectively represent the stress that the line is subjected at the current temperature and during normal operation. In addition, \(K\) is the coefficient of elasticity of the line, the unit is N/mm\({}^{2}\). \(\alpha\) is the coefficient of line temperature expansion. Therefore, the internal stress \(\sigma_{0}\) can be calculated by using \(\sigma_{n}\) from the Equation (15) in the manuscript.