## 4 Discussion
\begin{array}{l}\mathrm{x}_{0}=a=10.2\ \mathrm{km};\ \Delta\mathrm{x}=(b-a)/4=(34.964-10.2)/4=6.191 \ \mathrm{km};\\ f_{0}=f(\mathrm{x}_{0})=\Delta U=\big{|}\big{[}2.0\cdot 10.2\cdot 1.278+40 \big{]}-[0.56\cdot 1.278\cdot(40-10.2+40\cdot 0.85+40\cdot 0.7)+28\big{]}\big{|}=27.63 \ \mathrm{kV};\\ \mathrm{x}_{1}=\mathrm{x}_{0}+(b-a)/4=10.2+(34.964-10.2)/4=16.391\ \mathrm{km}; \\ f_{1}=f(\mathrm{x}_{1})=\Delta U=\big{|}\big{[}2.0\cdot 16.391\cdot 1.278+40 \big{]}-[0.56\cdot 1.278\cdot(40-16.391+40\cdot 0.85+40\cdot 0.7)+28\big{]}\big{|}=7. 37\ \mathrm{kV}.\end{array} \]  
In the second step:  
\[ \begin{array}{l}\mathrm{x}_{0}=16.391\ (\mathrm{km});f_{0}=f(\mathrm{x}_{0})= \Delta U=7.37\ \mathrm{kV};\ \mathrm{x}_{1}=\mathrm{x}_{0}+\Delta\mathrm{x}=16.391+6.191=22.582\ \mathrm{km};\\ f_{1}=f(\mathrm{x}_{1})=\Delta U=\big{|}\big{[}2.0\cdot 22.582\cdot 1.278+40 \big{]}-[0.56\cdot 1.278\cdot(40-22.582+40\cdot 0.85+40\cdot 0.7)+28\big{]}\big{|}=12. 88\ \mathrm{kV};\\ \Delta\mathrm{x}=-\Delta\mathrm{x}/4=-1.55.\end{array} \]  
In the third step:  
\[ \begin{array}{l}\mathrm{x}_{0}=16.391\ (\mathrm{km});f_{0}=f(\mathrm{x}_{0})= \Delta U=7.37\ \mathrm{kV};\ \mathrm{x}_{1}=\mathrm{x}_{0}+\Delta\mathrm{x}=16.391-1.55=14.841\ \mathrm{km};\\ f_{1}=f(\mathrm{x}_{1})=\Delta U=\big{|}\big{[}2.0\cdot 14.841\cdot 1.278+40 \big{]}-[0.56\cdot 1.278\cdot(40-14.841+40\cdot 0.85+40\cdot 0.7)+28\big{]}\big{|}=12. 44\ (\mathrm{kV});\\ \Delta\mathrm{x}=-\Delta\mathrm{x}/4=1.55/4=0.39.\end{array} \]  
The variables obtained in a similar way according to the algorithm in Figure 7a are listed in Table 1. \begin{table}
\begin{tabular}{c c c c c c c} \hline \hline
**Step** & \(\mathrm{x}_{0}\)**, **km** & \(f_{0}\) = \(f(\mathrm{x}_{0})\)**, **kV** & \(\mathrm{x}_{1}\)**, **km** & \(f_{1}\) = \(f(\mathrm{x}_{1})\)**, **kV** & \(f_{0}\) > \(f_{1}\)** & \(\Delta\mathrm{x}\)**, **km** \\ \hline
1. & 10.2 & 27.63 & 16.391 & 7.37 & yes & 6.191 \\
2. & 16.391 & 7.37 & 22.582 & 12.88 & no & \(-1.55\) \\
3. & 16.391 & 7.37 & 14.841 & 12.44 & no & 0.39 \\
4. & 14.841 & 12.44 & 15.231 & 11.17 & yes & 0.39 \\
5. & 15.231 & 11.17 & 15.621 & 9.89 & yes & 0.39 \\
6. & 15.621 & 9.89 & 16.011 & 8.62 & yes & 0.39 \\
7. & 16.011 & 8.62 & 16.401 & 7.34 & yes & 0.39 \\
8. & 16.401 & 7.34 & 16.791 & 6.06 & yes & 0.39 \\
9. & 16.791 & 6.06 & 17.181 & 4.79 & yes & 0.39 \\
10. & 17.181 & 4.79 & 17.571 & 3.51 & yes & 0.39 \\ \hline \hline \end{tabular}
\end{table} 
Table 1: An example of the FL implementation on OHPLs based on the bitwise search method. Thus, the distance to the fault place is \(L_{F}\) = 18.641 km, and its calculation requires \(m_{1}\) = 20 steps of the iterative procedure organized by the bitwise search method (Figure 8). We will refine the inspection zone of the OHPL by the line crew in accordance with the requirements of [74], i.e., (\(\pm\)7%) for the OHPL with a length from 100 to 300 km inclusive (\(\pm\)120\(\cdot\)0.07 = 8.4 km). Thus, the revised survey zone will correspond to the section of the OHPL from 10,241 km to 27,041 km. Since, in the implementation of the FL on OHPL, the alleged fault place can be at any point of an inhomogeneous OHPL, under the conditions of the example for a simple method of sequential enumeration, the required number of iteration steps in the case of fault at the end of the overhead transmission line will be \(m_{2}=L/\delta=120/0.3=400\). The gain in speed when implementing the bitwise search method, compared with the sequential search method, will be \(w=m_{2}/m_{1}=20\) times. The gain in the speed of iterative procedures \(w\) largely depends on the pre-specified error \(\delta\) of the FL line (Figure 8). The smaller the error, the more significant the gain. According to the calculation results (Table 1), the error in the implementation of the FL on OHPL due to the non-uniform distribution of resistivity along the overhead line will be  
\[\Delta L_{F}=L_{F}-L^{\prime}_{F}=22.582-18.641=3.941\text{ km},\]  
or 3.28% of the length of the high-voltage transmission line.