# The Efficient Processing of Moving \(k\)-Farthest Neighbor Queries in Road Networks +
## 6 Evaluation of the Example M&FN Query Using MOFA
### Finding the Candidate Facilities for the Query Segment \(\overline{v_{1}v_{2}}\)  
Table 2 summarizes the computation of the candidate facilities at border points \(v_{1}\) and \(v_{2}\) of the query segment \(\overline{v_{1}v_{2}}\) and includes the largest distances between the border points and the facility clusters in \(\overline{F}\), candidate distances, and sets of candidate facilities at the border points. Recall that the \(k\)FN queries are issued only at the border points of the query segment, in which the query point \(q\) stays to find the candidate facilities for the query segment. Thus, for the example M&FN query, MOFA evaluates the \(k\)FN queries at border points \(v_{1}\) and \(v_{2}\) to find the candidate facilities for \(\overline{v_{1}v_{2}}\). First, the \(k\)FN query is evaluated at \(v_{1}\). The largest distances between \(v_{1}\) and the facility clusters in \(\overline{F}\) are calculated. Subsequently, the facility clusters \(\overline{F_{C}}\) are arranged in a descending order based on \(d_{max}(v_{1},\overline{F_{C}})\). Figure 10 depicts that the facility clusters \(\{\overline{f_{1}f_{2}f_{3}}\}\) and \(\{\overline{f_{4}f_{5}},\overline{f_{5}f_{6}}\}\) are sorted using their largest distance to \(v_{1}\) as follows: \(\overline{F}\)=\(\{\{\overline{f_{1}f_{2}f_{3}}\},\,\{\overline{f_{4}f_{5}},\overline{f_{5}f_{6}}\}\}\). \(\{\overline{f_{1}f_{2}f_{3}}\}\) is investigated, followed by \(\{\overline{f_{4}f_{5}},\overline{f_{5}f_{6}}\}\). After exploring \(\{\overline{f_{1}f_{2}f_{3}}\}\), \(v_{1}\) chooses \(f_{1}\), \(f_{2}\), and \(f_{3}\) as the candidate facilities for \(\overline{v_{1}v_{2}}\) because the distance from \(v_{1}\) to the second FN (\(f_{1}\) or \(f_{3}\)) is 21, and the candidate distance for \(v_{1}\) is \(d_{cond}(v_{1})=16\). The distances from \(v_{1}\) to \(f_{1}\), \(f_{2}\), and \(f_{3}\) are \(d(v_{1},f_{1})=21\), \(d(v_{1},f_{2})=22\), and \(d(v_{1},f_{3})=21\), respectively (Figure 6). The other facility cluster \(\{\overline{f_{4}f_{5}},\overline{f_{5}f_{6}}\}\) does not need be explored because \(d_{cond}(v_{1})=16\) is larger than \(d_{max}(v_{1},\{\overline{f_{4}f_{5}},\overline{f_{5}f_{6}}\})=12\) (Figure 10). Finally, a set of candidate facilities at \(v_{1}\) is evaluated as \(\Sigma(v_{1})\)=\(\{f_{1},f_{2},f_{3}\}\). In the same manner, the \(k\)FN query is evaluated at \(v_{2}\) to find the candidate facilities for the query segment \(\overline{v_{1}v_{2}}\). The largest distances between \(v_{2}\) and the facility clusters in \(\overline{F}\) are calculated. The facility clusters \(\overline{F_{C}}\) are then arranged in a descending order based on \(d_{max}(v_{2},\overline{F_{C}})\). Figure 11 shows that the facility clusters \(\{\overline{f_{1}}\overline{f_{2}}\overline{f_{3}}\}\) and \(\{\overline{f_{4}}\overline{f_{5}}\overline{f_{6}}\}\) are sorted using their largest distance to \(v_{2}\) as follows: \(\overline{F}\)=\(\{\{\overline{f_{1}}\overline{f_{2}}\overline{f_{3}}\},\{\overline{f_{4}} \overline{f_{5}}\overline{f_{6}}\}\}\). \(\{\overline{f_{1}}\overline{f_{2}}\overline{f_{3}}\}\) is investigated, followed by \(\{\{\overline{f_{4}}\overline{f_{5}},\overline{f_{5}}\overline{f_{6}}\}\}\). After exploring \(\{\overline{f_{1}}\overline{f_{2}}\overline{f_{3}}\}\), \(v_{2}\) chooses \(f_{1}\), \(f_{2}\), and \(f_{3}\) as the candidate facilities for the query segment \(\overline{v_{1}v_{2}}\) because the distance from \(v_{2}\) to the second FN \(f_{1}\) is \(17\), and the candidate distance for \(v_{2}\) is \(d_{cond}(v_{2})=12\). The distances from \(v_{2}\) to \(f_{1}\), \(f_{2}\), and \(f_{3}\) are \(d(v_{2},f_{1})=17\), \(d(v_{2},f_{2})=18\), and \(d(v_{2},f_{3})=16\), respectively (Figure 8). The other facility cluster \(\{\{\overline{f_{4}}\overline{f_{5}},\overline{f_{5}}\overline{f_{6}}\}\}\) does not have to be explored because \(d_{cond}(v_{2})=12\) is larger than \(d_{max}(v_{2},\{\overline{f_{4}}\overline{f_{5}},\overline{f_{5}}\overline{f_ {6}}\})=7\) (Figure 11).