## 4 Scheduling Method
### Systems with Two Distillers  
In [21], it is shown that, under the normal mode, no feasible schedule with maximal production rate can be found for a refinery with one distiller and two charging tanks. Thus, it is clear that if there are only two charging tanks for a two-distiller system, to obtain a schedule, the two distillers must be fed by applying the SCF mode all the time, i.e., \(\delta\) = 1. Then, we discuss the case with three charging tanks. For this case, with the PN model shown in Figure 6, we have the following results. **Proposition 1**. : _Assume that: (1) there are two distillers \(DS_{1-2}\) with feeding rates \(f_{ds1}\) and \(f_{ds2}\), respectively, and \(\alpha\) = \(\Psi\)\(\times\)\(f_{p\max}\), \(\alpha_{1}\) = \(\Psi\)\(\times\)\(f_{ds1}\), 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) \(f_{p\max}\) = \(f_{ds1}\) + \(f_{ds2}\); and (4) initially, the volume of oil type 1 in \(CTK_{1}\) is \(\zeta_{1}\), and the volumes of oil type 2 in \(CTK_{2}\) and \(CTK_{3}\) are \(\zeta_{2}\) and \(\zeta_{3}\), respectively. Then, a schedule can be found such that \(\delta\) = 0.5 if Condition (3) holds:_  
\[\left\{ \begin{array}{l}\sigma\leq\zeta_{1}\leq\xi_{1}-\mu\times\alpha_{2} \times f_{ds1}/f_{p\max}\\ \zeta_{2}=\alpha_{2}\\ \zeta_{3}=\mu\times\alpha_{2}\end{array} \right., \tag{3}\]  
_where \(\mu\in\) If\({}_{p\max}\)/\(f_{ds1}\), \(min\{\zeta_{2}\)/\(\alpha_{2}\), \(\zeta_{3}\)/\(\alpha_{2}\)}]._  
**Proof. ** See Appendix A.1. By the proof of the proposition, a method is given to find a schedule. To schedule the system, Condition (3) presents the necessary amount of oil that is initially in the charging tanks, which is easy to meet. In fact, if all the charging tanks are initially empty, no schedule can be found according to the operational requirements, no matter how many charging tanks there are. If there is more oil at the initial state, a schedule can be obtained similarly. By the obtained schedule, in order to keep \(DS_{1}\) in operation, \(CTK_{1}\) has to apply the SCF mode during the whole scheduling horizon due to the limited charging tanks. For \(DS_{2}\), by carefully scheduling the system, it is alternately fed by two charging tanks, \(CTK_{2}\) and \(CTK_{3}\), under the normal mode. If one of them is empty, the other can be used to feed the distiller. Hence, both \(CTK_{2}\) and \(CTK_{3}\) operate under the normal mode during the whole scheduling horizon. To do so, the key is that every time \(CTK_{1}\) (\(CTK_{2}\), \(CTK_{3}\)) is charged, its volume is \(V_{1}\) = \(\mu\)\(\times\)\(\alpha_{1}\) (\(V_{2}\) = \(\mu\)\(\times\)\(\alpha_{2}\), \(V_{2}\) = \(\mu\)\(\times\)\(\alpha_{2}\)). Thus, a schedule can be found simply by using the proposed method. Note that \(CTK_{2}\) is empty during [\(\tau_{1}\), \(\tau_{2}\)], and the utilization ratio of \(CTK_{2}\) is \(\rho_{2}\) = 1 - (\(\mu\)\(\times\)\(f_{ds1}\) - \(f_{p\max}\))/ (\(\mu\)\(\times\)\(f_{p\max}\)) = 1 - \(f_{ds1}\)/\(f_{p\max}\) + 1/\(\mu\). Hence, \(CTK_{2}\) is not fully utilized. If time duration [\(\tau_{1}\), \(\tau_{2}\)can be made longer than \(\Psi\), \(CTK_{2}\) can be used to feed \(DS_{1}\) for some time under the normal mode such that \(\delta\) can be reduced by the full use of \(CTK_{2}\). Let \(\varpi\) = (\(\mu\times f_{ds1}-2\times f_{pmax}\))/(\(f_{ds1}+f_{pmax}\)). We have the following result. **Proposition 2**.