## 4 Example of Formulating a Vector Criterion
Both criteria are criteria from the economics assessment group control; therefore, the first two coordinates of the reduction vector assume a value of zero. \begin{table}
\begin{tabular}{c c c c c c c c c c} \hline  & \(F_{1}(\alpha)\) & \(F_{2}(x)\) & \(F_{3}(x)\) & \(F_{4}(x)\) & \(F_{5}(x)\) & \(F_{6}(x)\) & \(F_{7}(x)\) & \(F_{8}(x)\) & \(F_{9}(x)\) & \(F_{10}(x)\) \\ \hline \(\alpha_{n,1}\) & 0.98 & 0.98 & 0.85 & 0.78 & 0.76 & 0.93 & 0.92 & 0.95 & 0.95 & 0.93 \\ \hline \(\alpha_{n,2}\) & 0.92 & 0.96 & 0.884 & 0.87 & 0.85 & 0.95 & 0.892 & 0.97 & 0.955 & 0.94 \\ \hline \(\alpha_{n,3}\) & 1.0 & 1.0 & 1.0 & 0.97 & 0.97 & 0.95 & 0.9 & 0.95 & 0.95 \\ \hline \(\alpha_{n}\) & 0.9667 & 0.98 & 0.9113 & 0.8733 & 0.86 & 0.9433 & 0.904 & 0.9567 & 0.9517 & 0.9067 \\ \hline \end{tabular}
\end{table} 
Table 1: Confidence levels for scalar criteria. The criterion assessing road loss and the criterion assessing the deviation from a given trajectory affect both time lost and fuel consumption; therefore, the reduction vector will take the form:  
\[\delta=[0,\ 0,\ 0.012,\ 0.01]\]  
For such a reduction vector, we check the fulfillment of the condition resulting from Theorem 3 for criteria \(F_{8}(x)\) and \(F_{9}(x)\). Theorem 3 shows that there is the relation:  
\[\alpha_{8}=0.9567>Q(F)+\frac{1}{3}\sum_{j=1}^{4}\delta_{j}=0.9485+0.0073=0.9558\]  
\[\alpha_{9}=0.9517<Q(F)+\frac{1}{3}\sum_{j=1}^{4}\delta_{j}=0.9485+0.0073=0.9558\]  
Therefore, only criterion \(F_{8}(x)\) can be entered into the vector criterion, because only it meets the condition of Theorem 3. Therefore, the vector criterion function will take the form:  
\[F(x)=[F_{1}(x),\ F_{2}(x),\ F_{6}(x),\ F_{7}(x),\ F_{8}(x)]\]  
For a modified vector criterion, using (9) we determine the total confidence level, which is \(Q(F)=0.95014\). For the new vector criterion, we check the possibility of removing the scalar criterion that meets the condition resulting from Theorem 2. It follows that \(\alpha_{1}>Q(F)\), \(\alpha_{2}>Q(F)\), \(\alpha_{8}>Q(F)\); however, \(\alpha_{6}<Q(F)\), \(\alpha_{7}<Q(F)\) occur for the criteria \(F_{6}(x)\) and \(F_{7}(x)\). Therefore, according to Theorem 2, they can be removed from the vector criterion. Both scalar criteria that we can remove are criteria from the group assessing the economics of control; therefore, the first two coordinates of the reduction vector take a value equal to zero, and the other three values are different from zero. The reduction vector takes the form \(\delta=[0,\ 0,\ 0.012,\ 0.017,\ 0.015]\). For such a reduction vector, we check the fulfillment of the condition resulting from Theorem 4 for criteria \(F_{6}(x)\) and \(F_{7}(x)\). It follows that:  
\[\alpha_{6}=0.9433>Q(F)-\frac{1}{3}\sum_{j=1}^{4}\delta_{j}=0.95014-0.01467=0.9 3547\]  
\[\alpha_{7}=0.904<Q(F)-\frac{1}{3}\sum_{j=1}^{4}\delta_{j}=0.95014-0.01467=0.9 3547\]  
According to Theorem 4, the criterion \(F_{7}(x)\) is removed from the vector criterion. Then, the vector criterion will take the form  
\[F(x)=[F_{1}(x),\ F_{2}(x),\ F_{6}(x),\ F_{8}(x)]\]  
For this criterion, the value of the total confidence level is \(Q(F)=0.961675\). After checking the condition resulting from Theorem 1, we cannot introduce additional scalar criterion into this vector criterion. Thus, the process of forming the vector criterion for the ship's anticollision motion control system has been completed.