## 4 Scheduling Method
### Scheduling a System with Three Distillers
#### 4.2.1 Case 1: Four Charging Tanks  
The PN model for the system with three distillers and four charging tanks is shown in Figure 7. In this case, only one charging tank can be assigned to each of \(DS_{1}\) and \(DS_{2}\), and two charging tanks are assigned to \(DS_{3}\). Then, we present methods to schedule the system. **Proposition 3**. : _Assume that: (1) there are three distillers \(DS_{1-3}\) with feeding rates \(f_{ds1},f_{ds2},\) and \(f_{ds3}\), respectively, and \(\alpha=\Psi\times f_{p\max},\,\alpha_{1}=\Psi\times f_{ds1},\,\alpha_{2}=\Psi \times f_{ds2},\,and\,\alpha_{3}=\Psi\times f_{ds3}\); (2) there are four charging tanks \(CTK_{1-4}\) with capacities \(\xi_{1}\), \(\xi_{2}\), \(\xi_{3}\), and \(\xi_{4}\), respectively; (3) \(f_{p\max}=f_{ds1}+f_{ds2}+f_{ds3}\); and (4) initially, the volume of oil type 1 (2; 3; 3) in \(CTK_{1}\) (\(CTK_{2}\); \(CTK_{3}\); \(CTK_{4}\)) is \(\zeta_{1}\) (\(\zeta_{2}\); \(\zeta_{3}\); \(\zeta_{4}\)). Then, if Condition (5) holds, we can find a schedule with \(\delta=2/3\)._  
\[\left\{ \begin{array}{l}\sigma\leq\zeta_{1}\leq\zeta_{1}-\mu\times(\alpha- \alpha_{1})\times f_{ds1}/f_{p\max}\\ \sigma+\mu\times\alpha_{1}\times f_{ds2}/f_{p\max}\leq\zeta_{2}\leq\xi_{2}-\mu \times(\alpha-\alpha_{1}-\alpha_{2})\times f_{ds2}/f_{p\max}\\ \zeta_{3}=\alpha_{3}\\ \zeta_{4}=\mu\times\alpha_{3}\end{array} \right., \tag{5}\]  
_where \(\mu\in[f_{p\max}/(f_{ds1}+f_{ds2}),\,min[\zeta_{3}/\alpha_{3},\,\zeta_{4}/ \alpha_{3}]]\)._  
**Proof. ** See Appendix A.3. Similar to Proposition 1, if two charging tanks are assigned to a distiller, this distiller can be fed under the normal mode during the whole scheduling horizon. For the distiller with only one charging tank being assigned, the SCF mode has to be applied. Just as in Proposition 1, the key to the proposed method is that every time crude oil of type 1 (2, 3) is charged, the volume of oil to be charged is \(V_{1}=\mu\times\alpha_{1}\) (\(V_{2}=\mu\times\alpha_{2}\), \(V_{3}=\mu\times\alpha_{3}\)). With the obtained schedule by using the method given by Proposition 3, charging tank \(CTK_{3}\) is emptied at time \(\tau_{0}+\Psi\) and it is kept to be empty until it starts to be charged at time \(\tau_{1}\). Hence, \(CTK_{3}\) is not fully utilized and its utilization ratio is \(\rho_{3}=1-(\mu\times f_{ds1}+\mu\times f_{ds2}-f_{p\max})/(\mu\times f_{p \max})=1-(f_{ds1}+f_{ds2})/f_{p\max}+1/\mu\). Similar to the situation in Proposition 1, \(CTK_{3}\) can be used to reduce the usage of the SCF mode. Let \(\omega=(\mu\times f_{ds1}+\mu\times f_{ds2}-2\times f_{p\max})/(f_{ds2}+f_{p \max})\). We have the following result. **Proposition 4**. : _Assume that: (1) there are three distillers \(DS_{1-3}\) with feeding rates \(f_{ds1},f_{ds2},\) and \(f_{ds3}\), respectively, and \(\alpha=\Psi\times f_{p\max},\,\alpha_{1}=\Psi\times f_{ds1},\,\alpha_{2}=\Psi \times f_{ds2},\,\alpha_{3}=\Psi\times f_{ds3},\,and\,f_{ds1}\geq f_{ds2}\); (2) there are four charging tanks \(CTK_{1-4}\) with capacities \(\xi_{1}\), \(\xi_{2}\), \(\xi_{3}\), and \(\xi_{4}\), respectively; (3) \(f_{p\max}=f_{ds1}+f_{ds2}+f_{ds3}\); and (4) initially,the volume of oil type 1 (2; 3; 3) in \(CTK_{1}\) (\(CTK_{2}\); \(CTK_{3}\); \(CTK_{4}\)) is \(\zeta_{1}\) (\(\zeta_{2}\); \(\zeta_{3}\); \(\zeta_{4}\)). Then, if Condition (6) holds, we can find a schedule with \(\delta=\) (\(2\mu-\varpi\))/3\(\mu\)._  
\[\left\{ \begin{array}{l}\sigma\leq\zeta_{1}\leq\zeta_{1}-\alpha_{2}-\alpha_{3} \\ \sigma-\alpha_{2}-\alpha_{3}+\varpi\times\alpha_{2}\times f_{ds1}/f_{pmax}\leq \zeta_{1}\leq\xi_{1}-(\mu\times\alpha-\varpi\times\alpha_{2}-\mu\times\alpha_ {1})\times f_{ds1}/f_{pmax}\\ \sigma+(\mu\times\alpha_{1}+\varpi\times\alpha_{2})\times f_{ds2}/f_{pmax}\leq \zeta_{2}\leq\xi_{2}-(\mu\times\alpha-\varpi\times\alpha-\mu\times\alpha_{1}- \mu\times\alpha_{2})\times f_{ds2}/f_{pmax}\\ \xi_{3}=\alpha_{3}\\ \zeta_{4}=\mu\times\alpha_{3}\end{array}