# A Fault Section Location Method for Distribution Networks Based on Divide-and-Conquer
## 4 Fault Section Location Method for Distribution Networks Based on Divide-and-Conquer
### Fault Location Case
#### 4.4.1 Test Case 1: Single Fault  
In Figure 6, assume that section \(x_{9}\) in the multi-branch distribution network has a fault. The nodes that detect the fault current report the fault information, which is marked with red boxes.  
The algorithm traverses the sections in the network from the starting section \(x_{1}\), calculates the approximation gain of the sections, and determines the section's type. When accessing section \(x_{2}\), \(x_{2}\) is detected as a T-type section, and the network is divided into subnetwork \(G_{3}\) and subnetwork \(G_{15}\). The algorithm is called recursively for \(G_{15}\), and when the traversal end \(k=18\)\(G_{15}\) is determined to be the base case that does not contain a T-type section. The section with the maximum approximation gain in \(x_{15}\)-\(x_{18}\) is \(x_{15}\). According to Theorem 8, \(X_{15}=\{x_{15}\}\) and \(\Delta_{\mathbf{X}_{15}}=\Delta_{15}=1\) are taken as the fault location results of \(G_{15}\). Call the algorithm recursively for \(G_{3}\), traverse the sections in \(G_{3}\) from \(x_{3}\), detect that \(x_{6}\) is a T-type section, and divide the network into \(G_{7}\) and \(G_{12}\). Call the algorithm recursively for \(G_{7}\), and when the traversal end \(k=11\), determine \(G_{7}\) as the base case and return the result of fault location: \(X_{7}=\{x_{9}\}\), \(\Delta_{\mathbf{X}_{7}}=\Delta_{9}=\ 9\). Similarly, \(X_{12}=\{x_{12}\}\) and \(\Delta_{\mathbf{X}_{12}}=\Delta_{12}=\ 5\) are taken as the fault location results of \(G_{12}\). The divide process of the algorithm and the calculation results of each section approximation gain are shown in Figure 7a.  
Figure 6: Single fault in the multi-branch distribution network.  
The combine process of the algorithm is shown in Figure 7b, and the detailed running procedure is as follows. Combine the fault candidate sets \(G_{3}\) and \(G_{12}\) to create the fault location result \(G_{3}\). Find the section \(x_{6}\) with the largest approximation gain from \(x_{3}\) to \(x_{6}\) and construct the set of sections \(\mathbf{X}_{3-6}=\{x_{6}\}\) with the approximation gain \(\Delta_{\mathbf{X}_{3-6}}=\Delta_{6}=6\). Construct the set \(\mathbf{X}_{7612}=\mathbf{X}_{7}\cup\mathbf{X}_{12}=\{x_{9},x_{12}\}\), whose approximation gain is \(\Delta_{\mathbf{X}_{7612}}=8\). The largest approximation gain of \(\Delta_{\mathbf{X}_{3-6}}\), \(\Delta_{\mathbf{X}_{7}}\), \(\Delta_{\mathbf{X}_{12}}\), and \(\Delta_{\mathbf{X}_{7612}}\) is \(\Delta_{\mathbf{X}_{7}}\); thus, \(\mathbf{X}_{7}\) and \(\Delta_{\mathbf{X}_{7}}\) are taken as the fault location results of \(\mathbf{G}_{3}\), i.e., \(\mathbf{X}_{3}=\mathbf{X}_{7}=\{x_{9}\}\),\(\Delta_{\mathbf{X}_{3}}=\Delta_{\mathbf{X}_{7}}=9\). Combine the fault candidate sets of \(\mathbf{G}_{3}\) and \(\mathbf{G}_{15}\) to create the fault location result \(\mathbf{G}_{1}\). Find the section \(x_{2}\) with the largest approximation gain from \(x_{1}\) to \(x_{2}\) and construct the set of sections \(\mathbf{X}_{1-2}=\{x_{2}\}\) with the approximation gain \(\Delta_{\mathbf{X}_{1-2}}=\Delta_{2}=2\). Construct the set \(\mathbf{X}_{3615}=\mathbf{X}_{3}\cup\mathbf{X}_{15}=\{x_{9},x_{15}\}\), whose approximation gain is \(\Delta_{\mathbf{X}_{3615}}=8\). The largest approximation gain of \(\Delta_{\mathbf{X}_{1-2}}\), \(\Delta_{\mathbf{X}_{3}}\), \(\Delta_{\mathbf{X}_{15}}\), and \(\Delta_{\mathbf{X}_{3615}}\) is \(\Delta_{\mathbf{X}_{3}}\); thus, \(\mathbf{X}_{3}\) and \(\Delta_{\mathbf{X}_{3}}\) are taken as the fault location results of \(G_{1}\), i.e., \(\mathbf{X}_{1}=\mathbf{X}_{3}=\{x_{9}\}\),\(\Delta_{\mathbf{X}_{1}}=\Delta_{\mathbf{X}_{3}}=9\). The algorithm is executed, and \(x_{9}\) is determined to be the fault section.