## 4 Scheduling Method
### Scheduling a System with Three Distillers
#### 4.2.2 Case 2: Five Charging Tanks  
Now we discuss the situation where there are five charging tanks and three distillers, and the PN model for it is shown in Figure 8. For this system, \(DS_{1}\) has only one charging tank, and the other two distillers have two charging tanks each. Figure 7: PN model for a system with four charging tanks and three distillers. **Proposition 5**. : _Assume that: (1) there are three distillers \(DS_{1-3}\) with feeding rates \(f_{ds1},f_{ds2},\) and \(f_{ds3}\), respectively, and \(\alpha=\Psi\times f_{pmax},\)\(\alpha_{1}=\Psi\times f_{ds1},\)\(\alpha_{2}=\Psi\times f_{ds2},\) and \(\alpha_{3}=\Psi\times f_{ds3};\) (2) there are five charging tanks \(CTK_{1-5}\) with capacities \(\xi_{1-5}\), respectively; (3) \(f_{pmax}=f_{ds1}+f_{ds2}+f_{ds3};\) and (4) initially, the volume of oil type 1 (2; 2; 3; 3) in \(CTK_{1}\) (\(CTK_{2};\)\(CTK_{3};\)\(CTK_{4};\)\(CTK_{5}\)) is \(\zeta_{1}\) (\(\zeta_{2};\)\(\zeta_{3};\)\(\zeta_{4};\)\(\zeta_{5}\)). Then, if Condition (7) holds, a schedule can be found with \(\delta\) = 1/3._  
\[\left\{ \begin{array}{l}\sigma\leq\zeta_{1}\leq\xi_{1}-\mu\times(\alpha-\alpha _{1})\times f_{ds1}/f_{pmax}\\ \zeta_{2}=\alpha_{2}\\ \zeta_{3}=\mu\times\alpha_{2}\\ \zeta_{4}=\alpha_{3}\\ \zeta_{5}=\mu\times\alpha_{3}\\ \end{array} \right., \tag{7}\]  
_where \(\mu\in[f_{pmax}/f_{ds1},\)\(min\)\(\{\xi_{2}/\alpha_{2},\)\(\xi_{3}/\alpha_{2},\)\(\xi_{4}/\alpha_{3},\)\(\xi_{5}/\alpha_{3}\)\(]\)._  
Proof. : See Appendix A.5. By the obtained schedule for this case, the system is operated similarly to Propositions 1 and 3. \(CTK_{2}\) and \(CTK_{4}\) are empty and idle during \([\tau_{1},\)\(\tau_{2}]\) and \([\tau_{1},\)\(\tau_{2}+(\mu\times\alpha_{2})/f_{pmax}]\), respectively, and their utilization ratios for these two charging tanks are \(\rho_{2}=1-(\mu\times f_{ds1}-f_{pmax})/(\mu\times f_{pmax})\) and \(\rho_{4}=1-(\mu\times f_{ds1}+\mu\times f_{ds2}-f_{pmax})/(\mu\times f_{pmax})\), respectively. Thus, it is possible to use \(CTK_{2}\) and/or \(CTK_{4}\) to feed \(DS_{1}\) under the normal mode so as to reduce \(\delta\). By following this idea, we schedule the system as follows. Let \(\omega=(\mu\times f_{ds1}+\mu\times f_{ds2}-2\times f_{pmax})/(f_{ds1}+f_{pmax})\). We have the following result. **Proposition 6**. : _Assume that: (1) there are three distillers \(DS_{1-3}\) with feeding rates \(f_{ds1},f_{ds2},\) and \(f_{ds3}\), respectively, and \(\alpha=\Psi\times f_{pmax},\)\(\alpha_{1}=\Psi\times f_{ds1},\)\(\alpha_{2}=\Psi\times f_{ds2},\)\(\alpha_{3}=\Psi\times f_{ds3},\) and \(f_{ds2}\geq f_{ds3};\) (2) there are five charging tanks \(CTK_{1-5}\) with capacities \(\xi_{1-5}\), respectively; (3) \(f_{pmax}=f_{ds1}+f_{ds2}+f_{ds3};\) and (4) initially, the volume of oil type 1 (2; 2; 3; 3) in \(CTK_{1}\) (\(CTK_{2};\)\(CTK_{3};\)\(CTK_{4};\)\(CTK_{5}\)) is \(\zeta_{1}\) (\(\zeta_{2};\)\(\zeta_{3};\)\(\zeta_{4};\)\(\zeta_{5}\)). Then, if Condition (8) holds, a schedule can be found with \(\delta=(\mu-\omega)/3\mu\)._  
\[\left\{ \begin{array}{l}\sigma\leq\zeta_{1}\leq\xi_{1}-\alpha_{2}-\alpha_{3} \\ \sigma-\alpha_{2}-\alpha_{3}+\omega\times\alpha_{1}\times f_{ds1}/f_{pmax}\leq \zeta_{1}\leq\xi_{1}-(\mu\times\alpha-\omega\times\alpha-\mu\times\alpha_{1}) \times f_{ds1}/f_{pmax}\\ \zeta_{2}=\alpha_{2}\\ \zeta_{3}=\mu\times\alpha_{2}\\ \zeta_{4}=\alpha_{3}\\ \zeta_{5}=\mu\times\alpha_{3}\\ \end{array} \right., \tag{8}\]  
_where \(\mu\in[max|(\alpha\times f_{ds1}+f_{pmax})/f_{ds1},\)\(2f_{pmax}/(f_{ds1}+f_{ds2})]\), \(min|\)\(\xi_{2}/\alpha_{2},\)\(\xi_{3}/\alpha_{2},\)\(\xi_{4}/\alpha_{3},\)\(\zeta_{5}/\alpha_{3}\)\(]\) and \(\omega\in[0,\)\(min|\)\(\xi_{4}/\alpha_{1},\)\(\zeta_{5}/\alpha_{1}\)\(]\)._  
Figure 8: PN model for a system with five charging tanks and three distillers. **Proof. ** See Appendix A.6. The schedule generated by Proposition 6 is based on that obtained by Proposition 5.