## Appendix A  
**Proof of Theorem 1. ** Given the solution (sequence of shooting), the objective and constraints to find the worst scenario among the shooting sequences are as follows. \[\text{maximize }\sum_{k=1}^{n}\left\{\tilde{f}_{k}(t_{k}-\tilde{a}_{k})+ \frac{\tilde{f}_{k}^{2}}{2\tau_{k}}\right\} \tag{11}\]  
_subject to_  
\[t_{1}=a_{1} \tag{12}\]  
\[t_{k}=\max\left[\left(t_{k-1}+\frac{\tilde{f}_{k-1}}{\tilde{r}_{k-1}}\right), \,\tilde{a}_{k}\right]\ \ \ \text{k}\geq 2,\ \ \tilde{f}_{k}\in U_{k},\ \tilde{a}_{k}\in U_{k} \tag{13}\]  
In this case, the combination scenario of \(\tilde{f}_{k}\) and \(\tilde{a}_{k}\), which absolutely maximizes the objective equation for all \(k\), is a worst scenario. The following three cases can be considered depending on the value of \(k\). **Case 1. **_When \(\mathbf{k}=1\), Let \(F_{w}\) be Equation (11). Then,_  
\[F_{w}=\sum_{k=1}^{n}\left\{\tilde{f}_{k}(t_{k}-\tilde{a}_{k})+ \frac{\tilde{f}_{k}^{2}}{2\tau_{k}}\right\}=\tilde{f}_{1}(t_{1}-\tilde{a}_{1} )+\frac{\tilde{f}_{1}^{2}}{2\tau_{1}}+\tilde{f}_{2}(t_{2}-\tilde{a}_{2})+ \frac{\tilde{f}_{2}^{2}}{2\tau_{2}}+\ \sum_{k=3}^{n}\left\{\tilde{f}_{k}(t_{k}-\tilde{a}_{k})+ \frac{\tilde{f}_{k}^{2}}{2\tau_{k}}\right\}\] \[t_{1}=a_{1},\ t_{2}=\max\left[\left(t_{1}+\frac{\tilde{f}_{1}}{ \tau_{1}}\right),\,\tilde{a}_{2}\right].\]  
* _If_ \(t_{1}=\tilde{a}_{1}\) _and_ \(t_{2}=t_{1}+\frac{\tilde{f}_{1}}{\tau_{1}}\)_, then_ \(F_{w}=\frac{\tilde{f}_{1}^{2}}{2\tau_{1}}+\tilde{f}_{2}\left(\tilde{a}_{1}+ \frac{\tilde{f}_{1}}{\tau_{1}}-\tilde{a}_{2}\right)+\frac{\tilde{f}_{2}^{2}}{ 2\tau_{2}}+\sum_{k=3}^{n}\left\{\tilde{f}_{k}(t_{k}-\tilde{a}_{k})+\frac{\tilde {f}_{k}^{2}}{2\tau_{k}}\right\}\)_. That is,_ \(t_{1}+\frac{\tilde{f}_{1}}{\tau_{1}}\geq\tilde{a}_{2},\tilde{a}_{1}+\frac{ \tilde{f}_{1}}{\tau_{1}}-\tilde{a}_{2}\geq 0\)_._ \(\forall k\ \tilde{f}_{k},\,\tilde{a}_{k},\,\tau_{k}\)_, and_ \(t_{k}\) _are nonnegative values, and_ \(\forall k\ \tilde{f}_{k},\,\tilde{a}_{k}\) _are independent values. Therefore, when_ \(\tilde{f}_{1}=\tilde{f}_{1},\,\tilde{a}_{1}=\overline{a_{1}},\,F_{w}\) _is maximized, and this is a worst scenario._
* _If_ \(t_{1}=\tilde{a}_{1}\) _and_ \(t_{2}=\tilde{a}_{2}\)_, then_ \(F_{w}=\frac{\tilde{f}_{1}^{2}}{2\tau_{1}}+\frac{\tilde{f}_{2}^{2}}{2\tau_{2}}+ \sum_{k=3}^{n}\left\{\tilde{f}_{k}(t_{k}-\tilde{a}_{k})+\frac{\tilde{f}_{k}^{2} }{2\tau_{k}}\right\}\)_. Therefore, when_ \(\tilde{f}_{1}=\overline{f_{1}},\,F_{w}\) _is maximized and this is a worst scenario._  
**Case 2. **_When \(2\leq\mathbf{k}\leq\mathbf{n}-\mathbf{1}\),_  
\[F_{w}=\sum_{q=1}^{k-1}\left\{\tilde{f}_{q}\left(t_{q}-\tilde{a}_ {q}\right)+\frac{\tilde{f}_{q}^{2}}{2\tau_{q}}\right\}+\tilde{f}_{k}(t_{k}- \tilde{a}_{k})+\frac{\tilde{f}_{k}^{2}}{2\tau_{k}}+\tilde{f}_{k+1}(t_{k+1}- \tilde{a}_{k+1})+\frac{\tilde{f}_{k+1}^{2}}{2\tau_{k+1}}\] \[+\ \sum_{q=k+2}^{n}\left\{\tilde{f}_{q}\left(t_{q}-\tilde{a}_ {q}\right)+\frac{\tilde{f}_{q}^{2}}{2\tau_{q}}\right\}\] \[t_{k}=\max\left[\left(t_{k-1}+\frac{\tilde{f}_{k-1}}{\tau_{k-1}} \right),\,\tilde{a}_{k}\right].\]  
* _If_ \(t_{k}=t_{k-1}+\frac{\tilde{f}_{k-1}}{\tau_{k-1}}\) _and_ \(t_{k+1}=t_{k}+\frac{\tilde{f}_{k}}{\tau_{k}}\)_, then_  
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