## Appendix A
\begin{array}{l}E[[I_{i,1}]]\ \ =\sum\limits_{s\geq 0}1-Pr[[I_{i,1}\leq s]]\\ \ \ =\sum\limits_{s\geq 0}1-Pr[[max(\underset{C_{j,2}\in S_{i,1}}{max}(Td_{i,1}+ D_{i,1}+L_{j,2}),\underset{C_{j,2}\in E_{i,1}}{max}(L_{j,2}+X_{j,2}))+St_{i,1}\leq s]]\\ \\ \ =\sum\limits_{s\geq 0}1-Pr[[max(\underset{C_{j,2}\in S_{i,1}}{max}(Td_{i,1}+ D_{i,1}+L_{j,2}),\underset{C_{j,2}\in E_{i,1}}{max}(L_{j,2}+X_{j,2}))\leq s-St_{i,1}]]\\ \\ \ =\sum\limits_{s\geq 0}1-Pr[[\underset{C_{j,2}\in S_{i,1}}{max}(Td_{i,1}+ D_{i,1}+L_{j,2})\leq s-St_{i,1}]]\times Pr[[\underset{C_{j,2}\in E_{i,1}}{max}(L_{j,2}+X_{j,2}) \ \leq s-St_{i,1}]]\\ \\ \ =\sum\limits_{s\geq 0}1-Pr[[\underset{C_{j,2}\in S_{i,1}}{max}(Td_{i,1}+ D_{i,1}+L_{j,2})\leq s-St_{i,1}]]\times Pr[[\underset{C_{j,2}\in E_{i,1}}{max}(L_{j,2}+X_{j,2}) \ \leq s-St_{i,1}]]\end{array} \]  
Moreover, \(\forall\ i\in 1,\ \ldots,N_{1}\) and \(\forall\ j\in 1,\ \ldots,N_{2}\), random variables \(Td_{i,1}+D_{i,1}+L_{j,2}\) and \(L_{j,2}+X_{j,2}\) are independent, so,  
\[E[[I_{i,1}]]=\sum\limits_{s\geq 0}1-(Pr[[Td_{i,1}+D_{i,1}+L_{j,2}\leq s-St_{i, 1},\ldots,]]\times Pr[[L_{j,2}+X_{j,2}\ \leq s-St_{i,1},\ldots,]])\]  
We note by \(P_{j,2}=Td_{i,1}+D_{i,1},\ \forall\ C_{j,2}\in E_{i,1}\), so:  
\[E\left[[I_{i,1}]\right]=\sum\limits_{s\geq 0}\left(1-(\prod\limits_{C_{j,2} \in S_{i,1}}{F_{j,2}(-X_{j,2}-St_{i,1}+s)\times\prod\limits_{C_{j,2}\in E_{i,1 }}{F_{j,2}(-P_{j,2}-St_{i,1}+s)})}\right) \tag{2}\]  
Finally, we calculate \(E[[(I_{PF}+A_{PF})^{+}]]\) and \(E[[(I_{PF}+A_{PF})^{-}]]\) in order to determine after that \(E[[I_{PF}]]:\)  
\[ \begin{array}{l}E[[(I_{PF}+A_{PF})^{+}]]\ \ =E\left[[max\ (I_{PF}+A_{PF},\ T_{PF})]\right]\\ \\ \ =\sum\limits_{s\geq 0}1-Pr\left[[max\ (I_{PF}+A_{PF},\ T_{PF})\leq s]\right] \end{array} \]  
\(I_{PF}+A_{PF}\) and \(T_{PF}\) are independent random variables, so:  
\[\Rightarrow Pr\left[[max(I_{PF}+A_{PF},T_{PF})\leq s]\right]=Pr\left[[I_{PF}+A _{PF}\leq s]\right]\times Pr\left[[T_{PF}\leq s]\right]\]Thus,  
\[E\left[\left[\left(I_{PF}+A_{PF}\right)^{+}\right]\right]=\sum\limits_{s\geq 0}1-Pr \left[\left[I_{PF}\leq s-A_{PF}\right]\right]\times Pr\left[\left[T_{PF}\leq s \right]\right]\]  
\(T_{PF}\) is known and higher than 0, so:  
\[\left\{ \begin{array}{ll}Pr\left[\left[T_{PF}\leq s\right]\right]=1&\forall s \geq T_{PF}\\ Pr\left[\left[T_{PF}\leq s\right]\right]=0&\forall s\leq T_{PF}\end{array} \right.\]  
\[ \begin{array}{ll}E\left[\left[\left(I_{PF}+A_{PF}\right)^{+}\right]\right]&= \sum\limits_{s\geq 0}1-Pr\left[\left[I_{PF}\leq s-A_{PF}\right]\right] \times Pr\left[\left[T_{PF}\leq s\right]\right]\\ &=\sum\limits_{0\leq s\leq T_{PF}}\left(1-Pr\left[\left[I_{PF}\leq s-A_{PF} \right]\right]\times\overbrace{Pr\left[\left[T_{PF}\leq s\right]\right]}^{0} \right)+\\ &\sum\limits_{s\geq 2T_{PF}}\left(1-Pr\left[\left[I_{PF}\leq s-A_{PF} \right]\right]\times\underbrace{Pr\left[\left[T_{PF}\leq s\right]\right]}_{1} \right)\\ &=T_{PF}+\sum\limits_{s\geq T_{PF}}1-Pr\left[\left[I_{PF}\leq s-A_{PF} \right]\right]\end{array} \]  
Knowing that,  
\[\begin{array}{ll}Pr\left[\left[I_{PF}\leq s-A_{PF}\right]\right]&=Pr\left[ \left[\max\limits_{i=1..N_{1}}\left(I_{i,1}+A_{i,1}+L_{i,1}\right)\leq s-A_{ PF}\right]\right]\\ &=Pr\left[\left[\max\limits_{i=1..N_{1}}\left(I_{i,1}+  
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