# Mathematical Apparatus of Optimal Decision-Making Based on Vector Optimization
## 6 Statement and Optimal Decision Making with Experimental Data in Problems with Two Parameters
### The Solution of a Vector Problem of Mathematical Programming-Decision-Making  
For the solution of a vector problem of mathematical programming using the algorithm based on normalization of criteria, the above is used. Step 1. We solve the problem of Equations (53)-(55) with respect to each criterion separately using the function fmincon(... ), resulting in the optimum points:  
\[ \begin{array}{l}X^{*}_{1}\ =\ \{x_{1}\ =\ 0,x_{2}\ =\ 0\},X^{*}_{2}\ =\ \{x_{1}\ =\ 0,x_{2}\ =\ 10\},X^{*}_{3}\ =\ \{x_{1}\ =\ 10,x_{2}\ =\ 8\},\\ X^{*}_{4}\ =\ \{x_{1}\ =\ 8,x_{2}\ =\ 10\},X^{*}_{5}\ =\ \{x_{1}\ =\ 10,x_{2}\ =\ 0\}. \end{array}  \tag{56}\]  
and values of the criteria at these points:  
\[(1)f^{*}_{1}\ =\ -80,\ (2)f^{*}_{2}\ =\ -60,\ (3)f^{*}_{3}\ =\ 50,\ (4)f^{*}_{4}\ =\ 60,\ (5)f^{*}_{5}\ =\ 200. \tag{57}\]  
Step 2. We find the worst constant part by solving the problem of Equations (53)-(55) with respect to each criterion separately, i.e., we minimize the first two criteria and maximize the other criteria. The result is: (1) \(X^{o}_{1}=(10,10)\), \(f^{o}_{1}=-320\), (2), \(X^{o}_{2}=(10,0)\), \(f^{o}_{2}=-240\), (3) \(X^{o}_{3}=(0,0)\), \(f^{o}_{3}=232\), (4) \(X^{o}_{4}=(0,0)\), \(f^{o}_{4}=278.4\), (5) \(X^{o}_{5}=(0,10)\), \(f^{o}_{5}=800\). We represent the domain of admissible points \(S\) given by the restrictions of Equation (55) and the optimum points \(X^{*}_{1}\),..., \(X^{*}_{5}\) in Figure 4.  
Step 3. We analyze the set of Pareto optimal points using the matrix of the values of all objective functions, the column of deviations, and the matrix of the relative estimates at the optimal points:  
\[ \begin{array}{l}F(X^{*})\ =\ \left[\begin{array}{cccc}320.0&150.0&52&62.4&500 \\ 200.0&240.0&82&242.4&200\\ 289.6&163.2&50&88.8&424\\ 289.6&127.2&74&60.0&544\\ 200.0&240.0&82&242.4&200\\ \end{array}\right],\\ \Lambda(X^{*})\ =\ \left[\begin{array}{cccc}1.0000&0.5000&0.9890&0.9890&0.5000 \\ 0.5000&1.0000&0.8242&0.1648&1.0000\\ 0.8733&0.5733&1.0000&0.8681&0.6267\\ 0.8733&0.3733&0.8681&1.0000&0.4267\\ 0.5000&1.0000&0.8242&0.1648&1.0000\\ \end{array}\right].\end{array} \]  
We also show the relative estimates in Figure 4.  
Figure 4: The solution of the vector problem of Equations (53)–(55). Step 4. We construct the \(\lambda\)-problem for the VPMP of Equations (53)-(55)  
\[\lambda^{0}=\max\lambda, \tag{58}\]  
\[\lambda-(-0.4{x_{1}}^{2}-\ 8x_{1}-\ 0.4{x_{2}}^{2}-\ 8x_{2}-\ 80\ -f_{1}^{ 0})/d_{1}\leq 0,\] \[\lambda-(-0.3x_{1}^{2}-\ 6x_{1}-0.3x_{2}^{2}+12x_{2}-\ 150-f_{2}^{ 0})/d_{2}\leq 0,\] \[\lambda-(0.5x_{1}^{2}-\ 20x_{1}+0.5x_{2}^{2}-8x_{2}+\ 232-f_{3}^{ 0})/d_{3}\leq 0, \tag{59}\] \[\lambda-(0.6x_{1}^{2}-\ 9.6x_{1}+0.6x_{2}^{2}-2x_{2}+\ 278.4-f_{4}^{ 0})/d_{4}\leq 0,\] \[\lambda-(x_{1}^{2}-\ 40x_{1}+\ x_{2}^{2}+\ 20x_{2}+\ 500\ -f_{5}^{ 0})/d_{5}\leq 0,\]  
\[0\leq\lambda\leq 1,\ 0\leq x_{1}\leq 10,0\leq x_{2}\leq 10. \tag{60}\]  
Step 5. We solve the \(\lambda\)-problem of Equations (58)-(60) using the same MATLAB function fmincon(... ). This results in:  
* the optimum point \(X^{0}=\{x_{1}=5.0,\ x_{2}=5.0,\lambda^{0}=0.5833\}\),
* the values of criteria: \(f_{1}(X^{0})=-180,f_{2}(X^{0})=-135,f_{3}(X^{0})=117,f_{4}(X^{0})=140.4,f_{4}( X^{0})=450\),
* the values of the relative estimates \(\lambda_{k}(X^{0})\), \(k=\overline{1,K}\): \(\lambda_{1}(X^{0})=0.5833\), \(\lambda_{2}(X^{0})=0.5833\), \(\lambda_{3}(X^{0})=0.6319\), \(\lambda_{4}(X^{0})=0.6319\), \(\lambda_{5}(X^{0})=0.5833\)),
* the maximal relative estimate \(\lambda^{0}=0.5833\) that is the maximal lower level for all relative estimates \(\lambda^{0}=\min\{\lambda_{1}(X^{0}),\ \lambda_{2}(X^{0}),\ \lambda_{3}(X^{0}),\ \lambda_{4}(X^{0}),\ \lambda_{5}(X^{0})\}=0.5833\). Figure 5 shows all found points and relative estimates. From the result of the solution of the \(\lambda\)-problem of Equations (58)-(60), the optimal point \(\mathbf{X}^{0}\) and the maximum relative assessment \(\lambda^{0}\) represent the results of the decision with equivalent criteria.