## 4 Scheduling Method
### Systems with Two Distillers
: _Assume that: (1) there are two distillers \(DS_{1-2}\) with feeding rates \(f_{ds1}\) and \(f_{ds2}\), respectively, and \(\alpha\) = \(\Psi\)\(\times f_{pmax}\), \(\alpha_{1}\) = \(\Psi\)\(\times f_{ds1}\), and \(\alpha_{2}\) = \(\Psi\)\(\times f_{ds2}\); (2) there are three charging tanks \(CTK_{1-3}\) with capacities \(\xi_{1}\), \(\xi_{2}\), and \(\xi_{3}\), respectively; (3) \(f_{pmax}\) = \(f_{ds1}+f_{ds2}\); and (4) initially, the volume of oil type 1 in \(CTK_{1}\) is \(\xi_{1}\), and the volumes of oil type 2 in \(CTK_{2}\) and \(CTK_{3}\) are \(\xi_{2}\) and \(\zeta_{3}\), respectively. Then, a schedule can be obtained such that \(\delta\) = (\(\mu\) - \(\varpi\))/2\(\mu\) if Condition (4) holds:_  
\[\left\{ \begin{array}{l}\sigma\leq\zeta_{1}\leq\xi_{1}-\alpha_{2}\\ \sigma-\alpha_{2}+\varpi\times\alpha_{1}\times f_{ds1}/f_{pmax}\leq\zeta_{1} \leq\xi_{1}-\mu\times\alpha_{2}\times f_{ds1}/f_{pmax}\\ \zeta_{2}=\alpha_{2}\\ \xi_{3}=\mu\times\alpha_{2}\end{array} \right., \tag{4}\]  
_where \(\mu\) \(\in\) [max(\(\varpi\times f_{ds1}+f_{pmax}\))/\(f_{ds1}\), 2\(f_{pmax}\)/\(f_{ds1}\)], min[\(\xi_{2}\)/\(\alpha_{2}\), \(\xi_{3}\)/\(\alpha_{2}\)]] and \(\varpi\)\(\in\) [0, min[\(\xi_{2}\)/\(\alpha_{1}\), \(\xi_{3}\)/\(\alpha_{1}\)]]._  
Proof. : See Appendix A.2. This schedule is obtained by modifying that created by Proposition 1. By the scheduling method given in the proof of Proposition 2, \(DS_{2}\) is fed by \(CTK_{2}\) and \(CTK_{3}\) under the normal mode during the whole scheduling horizon, which is the same as that in Proposition 1. However, since \(CTK_{2}\) and \(CTK_{3}\) have idle time for feeding \(DS_{2}\), they can feed \(DS_{1}\) under the normal mode during their idle time such that the usage of the SCF mode is reduced. To do so, when the charging tanks are charged with volume \(V_{1}\) = \(\mu\)\(\times\)\(\alpha_{1}\) of oil type 1, we do not charge all the crude oil into \(CTK_{1}\) at one time as done in Proposition 1; instead, \(V_{1}\) is divided into three parcels and charged in the following order: charge \(CTK_{1}\) with volume \(v_{1}\) = \(\alpha_{1}\)\(\rightarrow\) charge \(CTK_{2}\) or \(CTK_{3}\) with volume \(v_{2}\) = \(\varpi\)\(\times\)\(\alpha_{1}\)\(\rightarrow\) charge \(CTK_{1}\) with volume \(v_{3}\) = (\(\mu\) - \(\varpi\) - 1)\(\alpha_{1}\). By doing so, when the crude oil of type 1 in \(CTK_{2}\) or \(CTK_{3}\) is ready for discharging under the normal mode, it can be used to feed \(DS_{1}\) until it is used up. Note that, by doing so, \(DS_{2}\) is still fed under the normal mode. Figure 6: The PN model for a system with three charging tanks and two distillers. For the obtained schedule using Proposition 2, \(CTK_{2}\) is used to feed \(DS_{2}\) and \(DS_{1}\) during [\(\tau_{0}\), \(\tau_{1}\)] and [\(\tau_{3}\), \(\tau_{4}\)] under the normal mode, respectively. In this way, the utilization ratio of \(CTK_{2}\) reaches \(\rho_{2}=1\). Furthermore, \(CTK_{1}\) and \(CTK_{3}\) are occupied during the whole scheduling horizon too. Hence, \(\delta\) is minimized and cannot be reduced any more. It should be pointed out that such a schedule can be obtained only if \(\mu\geq\max\{(\alpha\times f_{ds1}+f_{p\max})/f_{ds1},\,2f_{p\max}/f_{ds1}\}\), otherwise, \(CTK_{2}\) cannot be used to feed \(DS_{1}\) under the normal mode, and the system has to be scheduled using Proposition 1.