## 4 Numerical Example
### Step 1: Transportation with the Subcontracting Strategy Plan  
We consider a UDC system, which has to satisfy customer demand in an urban area in time. Demands of the allocated urban area are depicted in the following Table 1. The unit deliver cost per product equals 10 mu. At this stage, we suppose that this UDC works with the maximal capacity, called \(C_{\max}=20\). The number of periods \(H\) is equal to 10 days. Concerning the subcontracting strategy, there are three other UDCs available to complement the service if needed. The available capacity of deliveries for each of them is presented in Tables 2-4. Table 5 presents the unit service cost and unit emission cost generated by each product delivered by UDC\({}_{i}\). For no delivery of the product, the penalty cost equals 50 mu. Here, you can see, as well, that the product not delivered in time costs more than the subcontractor cost. We notice that our mathematical model contains binary and integer variables presenting quadratic terms, the so-called quadratic program (QP). These quadratic terms appear on the objective functions and constraints. We present some modifications of our mathematical model to linearize these quadratic terms in order to make our model a particular linear programming (LP). Then, we solve our new model using the solver FICO Xpress 8.0. We do a reformulation of the following quadratic terms by using binary integer linearization techniques. The quadratic terms in our models are the product of binary and integer variables. In our case, the integer variables are bounded. We find in the following equations the quadratic terms, shown in blue:  
\[C_{d}=\min\left\{\sum\limits_{t}C_{t}\times x_{t}+\sum\limits_{k}\sum\limits_{i }C_{tt_{i}}\times y_{it}\times AF_{it}+\sum\limits_{t}C_{p}\times Q_{t}\right\} \tag{16}\]  
\[Q_{t}=\max[0,D_{t}-x_{t}-\sum\limits_{i}y_{it}\times AF_{it}] \tag{17}\]  
\(AF_{it}\) is a binary variable, and \(y_{it}\) is an integer and bounded. \begin{table}
\begin{tabular}{c c c c c c c c c} \hline \hline
**J\({}_{1}\)** & **J\({}_{2}\)** & **J\({}_{3}\)** & **Y\({}_{1,4}\)** & **Y\({}_{1,5}\)** & **Y\({}_{1,6}\)** & **Y\({}_{1,7}\)** & **Y\({}_{1,8}\)** & **Y\({}_{1,9}\)** & **Y\({}_{1,10}\)** \\ \hline
4 & 5 & 1 & 7 & 2 & 5 & 2 & 3 & 0 & 1 \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Mean demand of urban area. \begin{table}
\begin{tabular}{c c c c c c c c c} \hline \hline
**Y\({}_{2,1}\)** & **Y\({}_{2,2}\)** & **Y\({}_{2,3}\)** & **Y\({}_{2,4}\)** & **Y\({}_{2,5}\)** & **Y\({}_{2,6}\)** & **Y\({}_{2,7}\)** & **Y\({}_{2,8}\)** & **Y\({}_{2,9}\)** & **Y\({}_{2,10}\)** \\ \hline
0 & 5 & 3 & 8 & 4 & 6 & 9 & 1 & 2 & 0 \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Available capacity for UDC\({}_{1}\), UDC: urban distribution center. \begin{table}
\begin{tabular}{c c c c c c c c} \hline \hline
**T\({}_{3,1}\)** & **Y\({}_{3,2}\)** & **Y\({}_{3,3}\)** & **Y\({}_{3,4}\)** & **Y\({}_{3,5}\)** & **Y\({}_{3,6}\)** & **Y\({}_{3,7}\)** & **Y\({}_{3,8}\)** & **Y\({}_{3,9}\)** & **Y\({}_{3,10}\)** \\ \hline
7 & 0 & 8 & 0 & 2 & 5 & 3 & 1 & 2 & 2 \\ \hline \hline \end{tabular}
\end{table} 
Table 4: Available capacity for UDC\({}_{3}\). \begin{table}
\begin{tabular}{c c c c c} \hline \hline
**Items** & **UDC\({}_{1}\)** & **UDC\({}_{2}\)** & **UDC\({}_{3}\)** \\ \hline
**Unit service cost** & 15 mu & 20 mu & 10 mu \\
**Unit emission cost** & 2 mu & 1 mu & 4 mu \\ \hline \hline \end{tabular}
\end{table} 
Table 5: The unit service cost and unit emission cost for each UDC. mu, monetary unit. \[0\leq y_{it}\leq Y_{it}\ \ \forall t\in H\ i\in L \tag{18}\]  
Then, we introduce a new variable, \(B_{it}\). \[B_{it}=y_{it}\times AF_{it}\ \ \forall t\in H\ i\in L \tag{19}\]  
Furthermore, we add other valid inequalities to our model as follows:  
\[B_{it}\leq Y_{it}\times AF_{it}\ \ \forall t\in H\ i\in L \tag{20}\]  
\[B_{it}\geq 0\ \ \forall t\in H\ i\in L \tag{21}\]  
\[B_{it}\leq y_{it}\ \ \forall t\in H\ i\in L \tag{22}\]  
The new problem is applied to the model, which makes it as a linear programming that is easy to solve.