# Mathematical Apparatus of Optimal Decision-Making Based on Vector Optimization
## 2 Statement of a Problem: Creation of the Mathematical Model \"Acceptance of an Optimal Solution\"
### Vector Optimization with Equivalent Criteria
#### 3.1.2 Mathematical Algorithm of the Solution of a Vector Problem with Equivalent Criteria
The vector target function in the form of a matrix is formed:  
\(\text{disp ('Solution of a vector problem of the linear programming')}\)  
\(\text{cvec}=[\,-4.0\,-5.0\,-9.0\,-11.0\,\%\) Sales volume  
\(-2.\,-10.\,-6.\,-20. ]\) % profit Volume  
\(\text{a}=[1.0\,1.0\,1.0\,1.0,\)  
\(7.0\,5.0\,3.0\,2.0,\)  
\(3.0\,5.0\,10.0\,15.0]\), % matrix of linear restrictions  
\(\text{b}=[15.\,120.\,100]\) % the vector containing restrictions (\(b_{i}\))  
\(\text{Aeq}=[\,]\), \(\text{beq}=[\,]\) % restriction like equality  
\(\text{X0}=[0.\,0.\,0.\,0. ]\), % a vector of variables  
_The algorithm of the solution of VPLP_ is represented as a sequence of steps. Step 1. A decision on each criterion. The decision on the first criterion of the VPLP: \([x1f1]=\text{linprog}(\text{cvec}(1,:),a,b,Aeq,beq,lb,\text{ub})\)  
Decision on the first criterion: \(X_{1}^{*}=x1=\{x_{1}=7.14,\,x_{2}=x_{4}=0,\,x_{3}=7.85\}\), \(f_{1}^{*}=f1=-99.286\). Decision on the second criterion: \(X_{2}^{*}=\{x_{1}=0,\,x_{2}=12.5,\,x_{3}=0,\,x_{4}=2.5\}\), \(f_{2}^{*}=f_{2}=175\). Step 2. The worst point of an optimum is determined for each criterion (anti-optimum) by multiplication of criterion by a minus unit. For the decisions on the first and second criterion: \begin{table}
\begin{tabular}{c c c c c c} \hline \hline \multirow{2}{*}{**Type**} & \multicolumn{3}{c}{**Costs of Resources of One Product**} & \multicolumn{2}{c}{**Possibilities of Firm**} \\ \cline{2-5}  & **Type 1** & **Type 2** & **Type 3** & **Type 4** & **on Resources** \\ \hline Labor (people/week) & 1 & 1 & 1 & 1 & 15 \\ Material (in kg) & 7 & 5 & 3 & 2 & 120 \\ Capacity (per hour) & 3 & 5 & 10 & 15 & 100 \\ \hline \hline \end{tabular}
\begin{tabular}{c} Income from a unit of \\ production \(c_{i}^{1}\) \\ \end{tabular} & 4.0 & 5.0 & 9.0 & 11.0 & maximize \\ \hline Profit \(c_{j}^{2}\), \(j=1\),...,\(\mathcal{A}\) & 2 & 10 & 6 & 20 & maximize \\ \hline Output & \(x_{1}\) & \(x_{2}\) & \(x_{3}\) & \(x_{4}\) & To define \\ \hline \hline \end{tabular}
\end{table} 
Table 1: The consumption of resources and operational performance. \(X_{1}^{0}=x1 min=\{x_{1}=0,\ldots,x_{4}=0\}\), \(f_{1}^{0}=f1 min=0\). \(X_{2}^{0}=x2 min=\{x_{1}=0,\ldots,x_{4}=0\}\), \(f_{2}^{0}=f2 min=0\). Step 3. The system analysis of the criteria in the VPLP is undertaken (i.e., the system of two criteria at optimum points is analyzed). For this purpose, optimum points of \(X_{1}^{*}\), \(X_{2}^{*}\) are defined sizes of criterion functions and relative estimates of: \(F(X^{*})=\|f_{q}(X_{k}^{*})\|_{k}^{q}=\frac{1.\overline{K}}{k=\overline{1.K}}\), \(\lambda(X^{*})=\|\lambda_{q}(X_{k}^{*})\|_{k}^{q}=\frac{\overline{1.K}}{1.K}\), \(\lambda(X^{*})=\frac{f_{1}(X*)-f_{k}^{0}}{f_{k}^{*}-f_{k}^{0}}\),  
\(k=\overline{1.K}\), \(F(X^{*})=\left| \begin{array}{c}f_{1}\left(X_{1}^{*}\right)f_{1}\left(X_{1}^ {*}\right)\\ f_{1}\left(X_{2}^{*}\right)f_{2}\left(X_{2}^{*}\right)\end{array} \right|= \left| \begin{array}{c}99.29\ 61.43\\ 90.175.0\end{array} \right|\), \(\lambda(X^{*})=\left| \begin{array}{c}\lambda_{1}\left(X_{1}^{*}\right)\ \lambda_{2}\left(X_{1}^{*}\right)\\ \lambda_{1}\left(X_{2}^{*}\right)\ \lambda_{2}\left(X_{2}^{*}\right)\end{array} \right|= \left| \begin{array}{c}1.0\ 0.351\\ 0.907\ 1.0\end{array} \right|\),  
Step 4. The \(\lambda\)-problem is constructed as: \(\lambda^{0}=max\ \lambda\),  
with restrictions \(\lambda-(f_{1}(X)-f_{1}^{0})/(f_{1}^{*}-f_{1}^{0})\leq 0\), \(\lambda-(f_{2}(X)-f_{2}^{0})/(f_{2}^{*}-f_{2}^{0})\leq 0\), \(G(X)\leq B\), \(X\geq 0\). Substituting numerical data, we obtain:  
\(\lambda^{0}=\max\ \lambda\),  
with restrictions: \(\lambda-(4.0x_{1}+5.0x_{2}+9.0x_{3}+11.0x_{4}-f_{1}^{0})/(f_{1}^{*}-f_{1}^{0}) \leq 0\),  
\(\lambda-(2x_{1}+10x_{2}+6x_{3}+20x_{4}-f_{2}^{0})/(f_{2}^{*}-f_{2}^{0})\leq 0\),  
\(x_{1}+x_{2}+x_{3}+x_{4}\leq 15\),  
\(7x_{1}+5x_{2}+3x_{3}+2x_{4}\leq 120\),  
\(3x_{1}+5x_{2}+10x_{3}+15x_{4}\leq 100\),  
\(x_{1}\geq 0\), \(x_{2}\geq 0\), \(x_{3}\geq 0\), \(x_{4}\geq 0\). Step 5. Solution of the \(\lambda\)-problem.