## 5 Numerical Analysis
### Small Numerical Example  
The small network is used to show the detailed application of this model. As shown in Figure 6, this network includes four nodes, six links, and one OD pair. There are four feasible paths connecting the OD pair: path 1: 1-5, path 2: 1-4-6, path 3: 2-6, and path 4: 2-3-5. The OD demand is set at 100. Additional inputs of this network are shown in Table 2. In this example, a static wayfinding instruction is placed in front of origin O to guide the route along link 1 towards destination D. Thus, link 1 is a guided link, while link 2 is an unguided link. According to Figure 6 and Table 2, the incident matrix and the diagonal static wayfinding instruction matrix can be expressed as follows:  
\[\Delta_{s}=\left( \begin{array}{ccccc}1&1&0&0\\ 0&0&1&1\\ 0&0&0&1\\ 0&1&0&0\\ 1&0&0&1\\ 0&1&1&0\end{array} \right),\ \mathbf{A}_{s}=\left( \begin{array}{ccccc}1&&&&\\ &-1&&&\\ &&1&&\\ &&&1&\\ &&&1&\\ &&&1\end{array} \right)\]  
Based on the method outlined in Section 3.2, \(\mathbf{\Lambda}_{s}\mathbf{\Delta}_{s}\) marks the unguided link. Setting the marked column to 0 column gives the revised incident matrix with elements \(\delta_{nbi}\):  
\[\mathbf{\Delta}_{is}=\left( \begin{array}{ccccc}1&1&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&1&0&0\\ 1&0&0&0\\ 0&1&0&0\end{array} \right).\]  
The length of the four paths are: path 1: 10; path 2: 12; path 3: 11; and path 4: 13, and \(\alpha_{\text{mk}}^{rsi}\) can be obtained by \(\mathbf{\Delta}_{is}\) and Equation (9):  
\(\alpha_{11}^{ODi}=0.600\); \(\alpha_{41}^{ODi}=0.400\); \(\alpha_{12}^{ODi}=0.545\); \(\alpha_{32}^{ODi}=0.182\); \(\alpha_{62}^{ODi}=0.273\), and other inclusion coefficients are all 0. Initially distribute the demand flow evenly across the paths between each OD pair, with each path receiving a traffic flow of 25 units. Then, obtain the flow on each link: link 1: 25 + 25 = 50; link 2: 25 + 25 = 50; link 3: 25; link 4: 25; link 5: 25 + 25 = 50; and link 6: 25 + 25 = 50. Calculate the link travel time cost using Equation (11) and determine the path \begin{table}
\begin{tabular}{c c c c c c} \hline
**Link** & **1** & **2** & **3** & **4** & **5** & **6** \\ \hline capacity & 50 & 50 & 30 & 50 & 40 & 50 \\ length & 6 & 7 & 2 & 2 & 4 & 3 \\ free flow time & 6 & 8 & 4 & 4 & 4 & 4 \\ \hline \end{tabular}
\end{table} 
Table 2: Link attributes input of small example network. Figure 6: Small example network with static wayfinding instructions. travel time cost by summing the cost of links on each path: \(c_{1}=12.4\); \(c_{2}=15.5\); \(c_{3}=13.8\); and \(c_{4}=19.0\). Apply Equation (12) to calculate the path selection probabilities:  
\[p_{1}=0.876;\ p_{2}=0.124;\ p_{3}=0;\ p_{4}=0.\]  
According to Equation (14), obtain decent direction of path flow:  
\[g_{1}^{OD(1)}-f_{1}^{OD(1)}=87.6-25=62.6;\ g_{2}^{OD(1)}-f_{2}^{OD(1)}=12.4-25=- 12.6;\]  
\[g_{3}^{OD(1)}-f_{3}^{OD(1)}=0-25=-25;\ g_{4}^{OD(1)}-f_{4}^{OD(1)}=0-25=-25.\]  
In this example, MSA is used to determine step size:  
\[\eta^{(1)}=1/(n+1)=0.5.\]  
Thereby, we can update path flow to reach a better solution:  
\[f_{1}^{OD(2)}=f_{1}^{OD(1)}+\eta^{(1)}\Big{(}g_{1}^{OD(1)}-f_{1}^{OD(1)}\Big{)} =25+0.5\times(62.6)=56.3;\]  
\[f_{2}^{OD(2)}=f_{2}^{OD(1)}+\eta^{(1)}\Big{(}g_{2}^{OD(1)}-f_{2}^{OD(1)}\Big{)} =25+0.5\times(-12.6)=18.7;\]  
\[f_{3}^{OD(2)}=f_{3}^{OD(1)}+\eta^{(1)}\Big{(}g_{3}^{OD(1)}-f_{3}^{OD(1)}\Big{)} =25+0.5\times(-25)=12.5;\]  
\[f_{4}^{OD(2)}=f_{4}^{OD(1)}+\eta^{(1)}\Big{(}g_{4}^{OD(1)}-f_{4}^{OD(1)}\Big{)} =25+0.5\times(-25)=12.5.\]  
Subsequently, based on the new auxiliary solution, repeat the above procedure until the convergence criterion (r = \(10^{-4}\)) is met, which occurs at the 79 th iteration. Table 3 presents the solution for this example and compares it with the GNL-SUE model under two conditions: condition 1 refers to the network with the consideration of wayfinding instruction and condition 2 means the network without considering wayfinding instruction.