# Mathematical Apparatus of Optimal Decision-Making Based on Vector Optimization
## Conclusions
### Formation of Technical Specifications (Source Data) for the Numerical Simulation of the System
#### 8.2.4 Construction of a Numerical Model of the System under Certainty and Uncertainty  
For the creation of a numerical model of the system, we use the functions obtained under conditions of definiteness (Equations (92) and (93)) and uncertainty (Equations (97) and (98)), and parametric restrictions (Equations (94) and (99)). We consider the functions of Equations (92), (93), (97) and (98) as the criteria defining the functioning of the system. A set of criteria \(K=4\) includes two criteria \(f_{1}(X),f_{3}(X)\rightarrow\max\) and two \(f_{2}(X),f_{4}(X)\rightarrow\min\). As a result, the model of the functioning of the system is presented as a vector problem of mathematical programming:  
\[ \begin{array}{l}opt\ F(X)\ =\ \{max\ F_{1}(X)\ =\ \{maxf_{1}(X)\equiv 269.867-1.8746 \times x_{1}\\ -1.7469\times x_{2}+0.8939\times x_{3}+1.0937\times x_{4}+0.0484\times x_{1} \times x_{2}-\\ 0.0052\times x_{1}\times x_{3}-0.0141\times x_{1}\times x_{4}+0.0037\times x_{2 }\times x_{3}-0.0052\times x_{2}\times x_{4}\\ -0.0002\times x_{3}\times x_{4}+0.0119\times x_{1}^{2}+0.0035\times x_{2}^{2}-0.002\times x_{3}^{2}-0.0042\times x_{4}^{2},\\ \\ max\ f_{3}(X)\equiv 43.734+0.659\times x_{1}+0.4493\times x_{2}-0.3094\times x_{3}\\ -1.8334\times x_{4}-0.01\times x_{1}\times x_{2}-0.0062\times x_{1}\times x_{3 }+\ 0.0146\times x_{1}\times x_{4}\\ -0.013\times x_{2}\times x_{3}+0.0121\times x_{2}\times x_{4}-0.0004\times x_{3 }\times x_{4}-0.0003\times x_{1}^{2}\\ -0.0002\times x_{2}^{2}+0.0254\times x_{3}^{2}+0.0939\times x_{4}^{2}\\ \end{array}  \tag{101}\]  
\[ \begin{array}{l}min\ F_{2}(X)\ =\ \{min\ f_{2}(X)\equiv 875.3+23.893\times x_{1}-30.866 \times x_{2}-\\ 25.858\times x_{3}-45\times x_{4}-0.6984\times x_{1}\times x_{2}+\ 0.4276\times x_{1}\times x_{3}+\\ 0.6793\times x_{1}\times x_{4}-\ 0.1167\times x_{2}\times x_{3}+0.2969\times x_{2 }\times x_{4}-\\ 0.0093\times x_{3}\times x_{4}+0.0362\times x_{1}^{2}+0.0331\times x_{2}^{2}+ 2.9158\times x_{3}^{2}+2.4052\times x_{4}^{2}.\\ \\ min\ f_{4}(X)\equiv 19.25-0.008\times x_{1}-0.7005\times x_{2}-0.3605\times x_{3 }+\\ 0.977\times x_{4}+0.0126\times x_{1}\times x_{2}+0.0644\times x_{1}\times x_{3 }-\ 0\times x_{1}\times x_{4}+\\ 0.0396\times x_{2}\times x_{3}+0.0002\times x_{2}\times x_{4}+0.0004\times x_{3 }\times x_{4}-0.0016\times x_{1}^{2}\\ +0.0027\times x_{2}^{2}+0.0045\times x_{3}^{2}-0.0235\times x_{4}^{2}],\\ \text{restrictions: }22\leq x_{1}\leq 88\text{, }0\leq x_{2}\leq 66\text{, }2.2\leq x_{3}\leq 8 \text{, }2.2\leq x_{4}\leq 8\text{. }\end{array}  \tag{102}\]  
The vector problem of mathematical programming of Equations (100)-(104) represents the model of decision making under certainty and uncertainty in the aggregate. The Solution of the Vector Problem of Mathematical Programming (VPMP)--Model of the System with Equivalent Criteria  
To solve the vector problem of mathematical programming of Equations (14)-(18), methods based on the axioms of the normalization of criteria and the principle of guaranteed results are presented, which follow from Axiom 1 and the principle of optimality 1. The solution of the vector problem of Equations (14)-(18) follows a sequence of steps. _Step 1_. Equations (100)-(104) are solved for each criterion separately, using the function _fmincon_ (... ) of the MATLAB system, the use of the function _fmincon_ (... ) is considered in [7; 8; 9; 10; 20; 22]. As a result of the calculation for each criterion we obtain optimum points: \(X_{k}^{*}\) and \(f_{k}^{*}=f_{k}(X_{k}^{*})\), \(k=\overline{1,K}\), the sizes of criteria at this point, i.e., the best decision for each criterion:  
\(X_{1}^{*}=\{x_{1}=88.0,\,x_{2}=66.0,\,x_{3}{=}8.8,\,x_{4}=2.2\},f_{1}^{*}=f_{1 }(X_{1}^{*})=-535.06\),  
\(X_{2}^{*}=\{x_{1}=22.0,\,x_{2}=0.0,\,x_{3}{=}2.83,\,x_{4}=6.25\},f_{2}^{*}=f_{2 }(X_{2}^{*})=1301.2\),  
\(X_{3}^{*}=\{x_{1}=88.0,\,x_{2}=0.0,\,x_{3}{=}2.2,\,x_{4}=8.8\},f_{3}^{*}=f_{3} (X_{3}^{*})=-100.15\),  
\(X_{4}^{*}=\{x_{1}{=}22.0,\,x_{2}=62.17,\,x_{3}=2.2,\,x_{4}=2.2\},f_{4}^{*}=f_{4 }(X_{4}^{*})=12.247\).