# Optimization of Wiring Harness Logistics Flow
## 2 Materials and Methods
### Description of the Proposed Methodology
Thus, we have \(A^{T}u\geq c\), which is equivalent to \(\left(A^{T}u\right)^{T}\geq c^{T}\), and with \(u^{T}A\geq c^{T}\)  
\[\left\{ \begin{array}{c}u^{T}A\geq c^{T}\\ x\geq 0\end{array} \Longrightarrow u^{T}Ax\geq c^{T}x\right.\] (3) Likewise \[\left\{ \begin{array}{c}Ax\leq b\\ u\geq 0\Longleftrightarrow u^{T}\geq 0\end{array} \Longrightarrow u^{T}Ax\leq u^{T}b \;\text{s}\right.\] (4) From the two systems of Equations (2) and (3), the result is that \(b^{T}u\geq c^{T}x\). In this situation, there are three hypotheses:  
* The problem has a finite optimum. In this situation, \(x_{B}\) exists, a basic solution of the primal in standard form, which provides the problem's optimum, meaning \(c^{T}x_{B}=max\;c^{T}x\). All minors corresponding to this basis will be positive, resulting in \(c_{B}^{T}B^{-1}A\geq c^{T}\Longleftrightarrow A^{T}\big{(}c_{B}^{T}B^{-1} \big{)}^{T}\geq c\). This relation indicates that the vector with m components \(u^{*}=\big{(}c_{B}^{T}B^{-1}\big{)}^{T}\) is an admissible basic solution to the dual problem. Thus, we have \(c^{T}x_{B}\leq u^{T}b\) for any feasible solution to the dual problem. On the other hand, \(c^{T}x_{B}=c^{T}B^{-1}b=(u^{*})^{T}b\), from which it results that for any feasible solution to the dual problem, we must check that \((u^{*})^{T}b\leq u^{T}b\). This implies that \(u^{*}\) is the optimal solution to the dual problem, and that \(f(x_{B})=g(u^{*})\);
* The problem has an infinite optimum. In this case, if the dual had admissible solutions \(u,g(u)\) would be an upper bound for the set \(\left\{\!f(x)/x\!\right\}\) admissible, which contradicts the hypothesis of an infinite optimum;
* The problem has no solutions. In this case, if the dual had a finite optimum, according to the statements in variant 1, the primal would also have a finite optimum, which contradicts the hypothesis. Since the three variants cover all possible situations, all leading to one of the previously stated possible outcomes, and considering that it did not matter which problem was chosen for resolution, the previous results have been demonstrated. According to the complementary slackness theorem, there exists \(x^{*}\)\(=\)\((x_{1}^{*}x_{2}^{*},\ldots x_{n}^{*})\) and \(u^{*}\)\(=\)\((u_{1}^{*}u_{2}^{*},\ldots u_{n}^{*})\), the two admissible solutions of the primal and dual problems, which are optimal solutions for both problems and satisfy the system  
\[ \begin{cases}u_{1}^{*}\Big{(}b_{j}-\sum_{j=1}^{n}a_{ij}x_{j}^{*}\Big{)}=0,\ i= \overline{1,m}\\ \Big{(}\sum_{i=1}^{m}(a_{ij}u_{i}^{*}-c_{j})x_{j}^{*}=0,j=\overline{1,n}\end{cases}  \tag{5}\]  
or the matrix-based system  
\[ \begin{cases}(u^{*})^{T}(b-Ax^{*})=0\\ \Big{[}(u^{*})^{T}A-c\Big{]}x^{*}=0\end{cases}  \tag{6}\]  
It is assumed that \(x^{*}\) and \(u^{*}\) are the optimal solutions to the two problems. Therefore,  
\[\begin{array}{c}Ax^{*}\leq b\\ u^{*}\geq 0\end{cases}\Longrightarrow(u^{*})^{T}(b-Ax^{*})\geq 0\\ \\ A^{T}u^{*}\leq c\\ x^{*}\geq 0\end{cases}\Longrightarrow((u^{*})^{T}A-c)(x^{*})\geq 0 \tag{7}\]  
Given that \((u^{*})^{T}=cx^{*}\)(according to the fundamental theorem), \((u^{*})^{T}(b-Ax^{*})+\Big{(}(u^{*})^{T}A-c\Big{)}(x^{*})\) results in \((u^{*})^{T}(b-Ax^{*})=\Big{(}c-(u^{*})^{T}A\Big{)}(x^{*})=0\). If \((u^{*})^{T}(b-Ax^{*})=\Big{(}c-(u^{*})^{T}A\Big{)}(x^{*})=0\Longrightarrow(u^ {*})^{T}b=cx^{*}\), according to the fundamental theorem, \(x^{*}\) and \(u^{*}\) are the optimal solutions to the two problems. The previously described model can be applied within the scope of this study because the time associated with each workstation process depends on previous values. Thus, this simulation aims to optimize time. Applying the previously described model involves establishing three working hypotheses.