## 4 Test and Results
### Static TNP
#### 4.1.2 IEEE 24-Bus System  
The IEEE 24-bus has 24 buses and 41 right-of-ways for the addition of new circuits. The generation capacity is 10,215 MW, and the demand is 8550 MW. This system has a limit of 5 circuits that can be added in each right-of-way. In this case, a population of 40 individuals (\(k_{p}\)) was considered, with \(k_{s}\) equal to 3%, \(k_{dr}\) equal to 5% and \(k_{m}\) equal to 5% (selection, diversification and mutation rates, respectively). A summary of results for this system is presented in Table 3.  
Test 1: The best solution found by the HEGA for the static TNEP problem without SCC and active power losses has a total investment cost of US$152.00 \(\times\) 10\({}^{6}\) and no load shedding. This solution is equal to the one reported in [15]. The TNEP problem is solved by the HEGA after executing 46 LPs with a computational time of 0.046 min. The configuration found is: \(n_{6-10}=1\), \(n_{7-8}=2\), \(n_{10-12}=1\), and \(n_{14-16}=1\).  
Test 2: The solution found by the proposed approach for the static TNEP problem with SCC and without active power losses is equal to the one of test 1 (no lines compensation).  
Test 3: The best solution found by the HEGA for the static TNEP problem considering active power losses and without SCC devices has a total investment cost of US$182.00 \(\times\) 10\({}^{6}\) and no load shedding. This solution is found by the HEGA after executing 20 LPs with a computational time of 0.06 min. The configuration found is: \(n_{6-10}=1\), \(n_{7-8}=2\), \(n_{10-12}=1\), \(n_{14-16}=1\), and \(n_{20-23}=1\).  
Test 4: The best solution found by HEGA for the static TNEP problem with SCC devices and active power losses has a total investment cost of US$154.40 \(\times\) 10\({}^{6}\) and no load shedding. This solution is found by the HEGA after executing 8170 LPs with a computational time of 3.3 min. The total active power loss is equal to 225.525 MW. The added circuits have an investment cost of US$152.00 \(\times\) 10\({}^{6}\), and are added in the following  
\begin{table}
\begin{tabular}{c c c c c c c} \hline \hline \multirow{2}{*}{**Planning Type**} & **Total Cost (10\({}^{6}\) USS)** & **Power Losses** & **LPs** & **Time** & **Configuration** & **Compensated** \\  & **(10\({}^{6}\) USS)** & **(MW)** & **Executed** & **(min)** & **Lines** \\ \hline \multirow{3}{*}{1} & \multirow{3}{*}{152.00} & \multirow{3}{*}{–} & \multirow{3}{*}{46} & \multirow{3}{*}{0.045} & \(n_{6-10}=1\), & \multirow{3}{*}{–} \\  & & & & & \(n_{7-8}=2\), & \\  & & & & & \(n_{10-12}=1\), & \\  & & & & & \(n_{14-16}=1\) & \\ \hline \multirow{3}{*}{2} & \multirow{3}{*}{152.00} & \multirow{3}{*}{–} & \multirow{3}{*}{120} & \multirow{3}{*}{0.605} & \(n_{6-10}=1\), & \multirow{3}{*}{–} \\  & & & & & \(n_{7-8}=2\), & \\  & & & & & \(n_{10-12}=1\), & \\  & & & & & \(n_{14-16}=1\) & \\ \hline \multirow{3}{*}{3} & \multirow{3}{*}{182.00} & \multirow{3}{*}{216.19} & \multirow{3}{*}{20} & \multirow{3}{*}{0.121} & \(n_{6-10}=1\), & \multirow{3}{*}{–} \\  & & & & & \(n_{7-8}=2\), & \\  & & & & & \(n_{10-12}=1\), & \\  & & & & & \(n_{14-16}=1\) & \\ \hline \multirow{3}{*}{4} & \multirow{3}{*}{154.40} & \multirow{3}{*}{225.52} & \multirow{3}{*}{8170} & \multirow{3}{*}{3.3} & \(n_{6-10}=1\), & \multirow{3}{*}{\(n_{15-16}=2\)} \\  & & & & & \(n_{10-12}=1\), & \\  & & & & & \(n_{14-16}=1\) & \\ \hline \hline \end{tabular}
\end{table}
Table 3: Summary of results for the IEEE 24-bus system (static TNEP).  
right-of-ways: \(n_{6-10}=1\), \(n_{7-8}=2\), \(n_{10-12}=1\), and \(n_{14-16}=1\). On the other hand, the SCC devices have an investment cost of US$2.40 \(\times\) 10\({}^{6}\), and the compensated line is: \(n_{15-16}=2\) (type 1).