# Mathematical Apparatus of Optimal Decision-Making Based on Vector Optimization
## 5 Statement and Optimal Decision Making with Experimental Data in Problems with One Parameter
### The Solution of the Problem
We thus derive the optimum point with respect to the first criterion \(x_{1}^{\text{max}}=18.810\) and the value of the criterion at this point \(f_{1}^{*}=f_{1}(x_{1}^{\text{max}})=-28,969\). Similarly, we have \(x_{2}^{\text{max}}=2192.8\), \(f_{2}^{*}=f_{2}(x_{2}^{\text{max}})=-2063.7\), \(x_{4}^{\text{max}}=630.0\), \(f_{4}^{*}=f_{4}(x_{4}^{\text{max}})=-239.65\), \(x_{3}^{\text{min}}=8389.0\), \(f_{3}^{*}=f_{3}(x_{3}^{\text{min}})=1021.4\) for other criteria. Step 2. We find the worst part for each criterion. We end up with the optimum point \(x_{1}^{\text{min}}=630\) with respect to the first criterion and the value of the criterion at the optimum \(f_{1}^{o}=f_{1}(x_{1}^{o})=4864.9\) (the worst constant part with respect to the first criterion). We use the operator plot(x1 min, f1 min/10,'k\(\times\)') to represent this and other points in \[\min_{A}f(A,X)\equiv\sum_{i=1}^{M}\;(y_{j}-(a_{0}+a_{1}x_{1i}+a_{2}x_{1i}^{2}))^ {2}. \tag{39}\]2 Similarly, we have for other criteria \(x_{2}^{\text{min}}=17,502\), \(f_{2}^{o}=f_{2}(x_{2}^{\text{min}})=365.22\), \(x_{4}^{\text{min}}=18,810\), \(f_{4}^{o}=f_{4}(x_{4}^{\text{min}})=59.838\), \(x_{3}^{\text{max}}=18,810\), \(f_{3}^{o}=f_{3}(x_{3}^{\text{max}})=-2126.3\). Step 3. We analyze the set of Pareto optimal points. In points at an optimum \(X^{*}=\{X_{1}*,X_{2}*,X_{3}*,\)\(X_{4}*,X_{5}*\}\), sizes of criterion functions of \(F(X^{*})=\|f_{q}(X_{k}^{*})\|_{q}^{k}=\frac{1,K}{1,K}\) are determined. We calculate a vector \(D=(d_{1}\ d_{2}\ d_{3}\ d_{4}\ d_{5})^{\text{T}}\) of deviations by each criterion on an admissible set \(S\): \(d_{k}=f_{k}^{*}-f_{k}^{0}\), \(k=\overline{1,5}\), and a matrix of relative estimates \(\lambda(X^{*})=\|\lambda_{q}(X_{k}^{*})\|_{q}^{k}=\frac{1,K}{1,K}\), where \(\lambda_{k}(X)=(f_{k}^{*}-f_{k}^{0})/d_{k}\). In the problem of Equations (44)-(46), criteria in the normalized form \(\lambda_{k}(X^{o})\), \(k=\overline{1,K}\) can be represented as shown in \[\min_{A}f(A,X)\equiv\sum_{i=1}^{M}\;(y_{j}-(a_{0}+a_{1}x_{1i}+a_{2}x_{1i}^{2}))^ {2}. \tag{39}\]3 We calculate the coefficients of the approximating polynomial for the normalized criteria to obtain:  
\(\lambda_{1}(x)=-1.325\times 10^{-9}\ x^{2}+8.0762\times 10^{-5}\ x\times 0.0504\),  
\(\lambda_{2}(x)=5.5735\times 10^{-13}\ x^{3}-1.6467\times 10^{-8}\ x^{2}+6.4179 \times 10^{-5}\ x+0.9325\),  
\(\lambda_{3}(x)=-9.2080\times 10^{-9}\ x^{2}+1.5451\times 10^{-4}\ x+0.3519\),  
\(\lambda_{4}(x)=9.1530\times 10^{-10}\ x^{2}-7.2811\times 10^{-5}\ x+1.0455\). \[\min_{A}f(A,X)\equiv\sum_{i=1}^{M}\;(y_{j}-(a_{0}+a_{1}x_{1i}+a_{2}x_{1i}^{2}))^ {2}. \tag{39}\]4 Approximations of four normalized criteria. Figure 2 shows the optimum points \(X^{0}\) and normalized criteria. The optimal point \(X^{0}\) in Figure 2 can be chosen manually. We solve the \(\lambda\)-problem to find the exact value of \(X^{0}\). Step 4. We construct the \(\lambda\)-problem. Using the obtained function and relative evaluations \(\lambda_{1}(x)\), \(\lambda_{2}(x)\), \(\lambda_{3}(x)\), \(\lambda_{4}(x)\), we construct the \(\lambda\)-problem:  
\[\min_{A}f(A,X)\equiv\sum_{i=1}^{M}\;(y_{j}-(a_{0}+a_{1}x_{1i}+a_{2}x_{1i}^{2}))^ {2}. \tag{39}\]5  
\[ \begin{array}{c}\lambda-\ (-3.1937\ \times\ 10^{-5}x^{2}+1.9467x+\ 3651.1\ -f_{1}^{ 0})/d_{1}\leq 0,\\ \lambda-\ (9.467\ \times\ 10^{-10}x^{3}-2.7968\times\ 10^{-5}x^{2}+\ 0.1090x+\ 1949.2\ -f_{2}^{0})/d_{2}\leq 0, \\ \lambda-\ (1.0174\ \times\ 10^{-5}x^{2}-\ 0.1707x+\ 1737.4\ +f_{3}^{0})/d_{3}\leq 0, \\ \lambda-\ (1.6458\ \times\ 10^{-7}x^{2}-\ 0.01309x+\ 247.83\ -f_{4}^{0})/d_{4}\leq 0, \\ \end{array}  \tag{48}\]  
\[630\leq x\leq 18810 \tag{49}\]  
Step 5. We solve the \(\lambda\)-problem of Equations (47)-(49). We use standard methods, in particular the MATLAB function fmincon (... ).