# 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
#### 4.4.3 Test Case 3: Fault Information Distortion  
In actual projects, FTUs are mostly installed outdoors, where the working environments are harsh and the reported fault information may be misreported, missed, or experiencing other distortions. Therefore, it should be possible to output correct positioning results as much as possible in the case of small-scale distortion of information. This case sets up the small-scale distortion of information to test the information fault tolerance of the algorithm.  
In Figure 10, assume that section \(x_{18}\) has a fault. Node \(s_{10}\) misreports the fault information (i.e., \(I_{10}\): \(0\to 1\)), and node \(s_{2}\) misses the fault information (i.e., \(I_{2}\): \(1\to 0\)). The nodes that reported the fault information are marked with red boxes.  
The divide process of the algorithm is similar to Test Case 1. The divide process of the algorithm and the calculation results of each section approximation gain are shown in Figure 11a.  
Figure 9: Execution process of the algorithm under multiple failures. (**a**) Divide process; (**b**) combine process.  
The combine process of the algorithm is shown in Figure 11b, and the detailed running procedure is as follows. Combine the fault candidate sets \(\mathbf{G}_{7}\) and \(\mathbf{G}_{12}\) to create the fault location result \(\mathbf{G}_{3}\). Find the section \(x_{3}\) with the largest approximation gain from \(x_{3}\) to \(x_{6}\) and construct the set of sections \(\boldsymbol{X}_{3-6}=\{x_{3}\}\) with the approximation gain \(\Delta_{\boldsymbol{X}_{3-6}}=\Delta_{3}=-1\). Construct the set \(\boldsymbol{X}_{7612}=\boldsymbol{X}_{7}\cup\boldsymbol{X}_{12}=\{x_{7},x_{1 2}\}\), whose approximation gain is \(\Delta_{\boldsymbol{X}_{7612}}=-6\). The largest approximation gain of \(\Delta_{\boldsymbol{X}_{3-6}}\), \(\Delta_{\boldsymbol{X}_{7}}\), \(\Delta_{\boldsymbol{X}_{12}}\), and \(\Delta_{\boldsymbol{X}_{7612}}\) is \(\Delta_{\boldsymbol{X}_{3-6}}\); thus, \(\boldsymbol{X}_{3-6}\) and \(\Delta_{\boldsymbol{X}_{3-6}}\) are taken as the fault location results of \(\mathbf{G}_{3}\), i.e., \(\boldsymbol{X}_{3}=\boldsymbol{X}_{3-6}=\{x_{3}\}\),\(\Delta_{\boldsymbol{X}_{3}}=\Delta_{\boldsymbol{X}_{3-6}}=-1\). Combine the fault candidate sets \(\mathbf{G}_{3}\) and \(\mathbf{G}_{15}\) to create the fault location result \(\mathbf{G}_{1}\). Find the section \(x_{1}\) with the largest approximation gain from \(x_{1}\) to \(x_{2}\) and construct the set of sections \(\boldsymbol{X}_{1-2}=\{x_{1}\}\) with the approximation gain \(\Delta_{\boldsymbol{X}_{1-2}}=\Delta_{1}=1\). Construct the set \(\boldsymbol{X}_{3615}=\boldsymbol{X}_{3}\cup\boldsymbol{X}_{15}=\{x_{9},x_{1 5}\}\), whose approximation gain is \(\Delta_{\boldsymbol{X}_{3615}}=\ 3\). The largest approximation gain of \(\Delta_{\boldsymbol{X}_{1-2}}\), \(\Delta_{\boldsymbol{X}_{3}}\), \(\Delta_{\boldsymbol{X}_{15}}\), and \(\Delta_{\boldsymbol{X}_{3615}}\) is \(\Delta_{\boldsymbol{X}_{15}}\); thus, \(\boldsymbol{X}_{15}\) and \(\Delta_{\boldsymbol{X}_{15}}\) are taken as the fault location results of \(\mathbf{G}_{1}\), i.e., \(\boldsymbol{X}_{1}=\boldsymbol{X}_{15}=\{x_{18}\}\), \(\Delta_{\boldsymbol{X}_{1}}=\Delta_{\boldsymbol{X}_{15}}=\ 4\). After the algorithm is executed, \(x_{18}\) is determined to be the fault section.  
As can be seen from Test Case 3, the algorithm proposed in this paper uses the redundancy of fault information in the whole network to tolerate the distorted information, which can accurately locate the fault with small-scale distortion of fault information.  
Figure 11: Execution process of the algorithm under fault information distortion. (**a**) Divide process; (**b**) combine process.  
Figure 10: Fault information distortion in the multi-branch distribution network.