## 6 Example: Results and Discussion
\begin{array}{cccc}0.217&0.749&1.000&1.000\\ 0.250&0.917&1.000&0.833\\ 0.399&0.753&0.333&0.833\\ 0.266&1.000&1.000&0.833\\ 1.000&0.432&1.000&0.833\\ \end{array} \right)\]  
Step 3: Establish the pair-wise comparison matrix by the comparison method, based on the opinions of our sample set of twenty drivers with a range of driving experiences, as described in Section 4. The following shows the response from one of the drivers:  
\[B_{i}=\ \left( \begin{array}{cccc}0.5&0.8&0.9&0.4\\ 0.2&0.5&0.1&0.4\\ 0.1&0.9&0.5&0.9\\ 0.6&0.6&0.1&0.5\\ \end{array} \right)\]  
where, \(i=1,\ldots,10\) for our data set. As shown in \(B_{i}\), all diagonal values are 0.5 as each factor complements itself. The values of \(b_{12}=0.8\) and \(b_{21}=0.2\) indicate that in the view of this driver, the factor \"walking distance\" is more important than \"distance to parking\". \"distance to parking\" is still a factor that the driver would consider when choosing a parking space, but it is just less important. Step 4: Obtain the weight vector for the pair-wise comparison matrix in Step 3 according to Equation (4):  
\[w_{i}=(0.40,\ 0.05,\ 0.35,\ 0.20)\] \begin{table}
\begin{tabular}{c c c c c} \hline \hline
**Parking Space** & **Walking Distance (m)** & **Distance to Parking (m)** & **Status of Parking** & **Status of Available Parking Spaces** \\ \hline
5 & 51.0 & 36.7 & clear & adjacent to the road \\ \hline
9 & 44.3 & 30.0 & clear & both sides are occupied \\ \hline
12 & 27.8 & 36.5 & occupied & both sides are occupied \\ \hline
18 & 41.8 & 27.5 & clear & both sides are occupied \\ \hline
23 & 11.1 & 63.6 & clear & both sides are occupied \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Properties of the available parking spaces in the parking lot of scenario 1. Applying Equation (4), the pair-wise comparison matrix from the twenty drivers is converted into the following weight matrix:  
\[w=\left( \begin{array}{cccc}0.400&0.050&0.350&0.200\\ 0.200&0.325&0.450&0.275\\ 0.500&0.125&0.175&0.175\\ 0.175&-0.050&0.625&0.750\\ 0.150&0.375&0.250&0.225\\ 0.225&0.125&0.250&0.400\\ 0.125&0.550&0.350&0.350\\ 0.375&0.100&0.350&0.175\\ 0.125&0.225&0.375&0.250\\ 0.225&0.200&0.400&0.225\\ 0.250&0.075&0.475&0.200\\ 0.425&0.075&0.475&0.200\\ 0.425&0.075&0.325&0.175\\ 0.125&0.125&0.425&0.325\\ 0.050&0.275&0.225&0.450\\ 0.175&0.250&0.275&0.300\\ 0.300&0.125&0.250&0.325\\ 0.325&0.175&0.100&0.425\\ 0.300&0.000&0.450&0.250\\ 0.150&0.150&0.425&0.325\\ 0.100&0.300&0.325&0.275\end{array} \right)\]  
To make the above matrix representative of the preference of most drivers, we remove the largest and smallest values of each factor in the above matrix and average the remaining values to obtain an average weight vector:  
\[\widetilde{w}=(0.233,\ 0.170,\ 0.336,\ 0.286)\]  
Step 5: With the average weight vector and Equation (5), the priority vector that contains the weights of every available parking space in the given situation can be obtained:  
\[z_{i}(w)=(0.799,\ 0.788,\ 0.571,\ 0.806,\ 0.881)\]  
Step 6: Ranking the priorities of the available parking spaces in this example parking lot according to \(z_{i}(w)\) gives: \(z_{23}(w)>z_{18}(w)>z_{5}(w)>z_{9}(w)>z_{12}(w)\). Based on step 6, parking space 23 has the highest priority for the driver of the orange vehicle. Thus, parking space 18 will be allocated to the blue autonomous vehicle by the central command station. By analyzing the preferences of the human driver, the algorithm is able to provide the optimal parking space for the autonomous vehicle. Since an intelligent parking lot system cannot influence the choice of the driver, a suboptimal parking space, which is space 18, is allocated to autonomous vehicle in this situation. Step 7: Establish the adjacent matrix with all the road intersection nodes and the node of parking space 18 for route selection based on the adjacent matrix assignment rule according to Equation (6) in Section 5:\[D=\left(