## 5 Simulation Results and Discussion
### Simulation Test for Three Sub-Objectives Based on Single-Level and Two-Level MPC
Taking the single-level MPC as an example, the following explains the reason for the first wind curtailment. As shown in Figure 10, the first wind curtailment occurs at minute 435 (interval 88),  
Figure 9: Simulation result for the first sub-objective. \(t_{s}=88\). From Equation (1), we know that pumped hydro storage is in Case 4, _i.e._, it remained unchanged from 420 to 445 min (six scheduling intervals, half hour). Some basic information is provided as follows. Table 4 indicates that the wind power is greater than the grid plan ( \(p_{i}^{\text{wind}}>p_{i}^{\text{plan}}\)), and the excess wind energy must be stored in the storage systems. In Equation (4) ( \(p_{i}^{\text{wind}}=p_{i}^{\text{wd}}+p_{i}^{\text{m}}+p_{i}^{\text{h}}+p_{i}^ {\text{wm}}\) ), \(p_{i}^{\text{wind}}\) is 20.15 MW; \(p_{i}^{\text{wd}}\) is 12.18 MW, due to the third sub-objective and constraints (Equation (17)); \(p_{i}^{\text{wm}}\) is assigned as 6.39 MW according to assumption 1. Based on Equation (7) and \(E_{t_{s},1}^{\text{fw}}=5.0\text{MW}\cdot\text{h}\), \(E_{t}^{\text{fw}}=4.809\text{MW}\cdot\text{h}\), we can calculate \(p_{i}^{\text{h}}\) as 0.66 MW. Finally, based on Equation (4) and the above results (\(p_{i}^{\text{wind}}\), \(p_{i}^{\text{wd}}\), \(p_{i}^{\text{wm}}\), \(p_{i}^{\text{h}}\)), we can calculate \(p_{i}^{\text{wm}}\) as 0.93 MW. The result of the third sub-objective is shown as follows:  
Figure 11 shows the wind power that is directly sent to the grid \(p_{i}^{\text{wd}}\) (wd). Moreover, Figure 11 shows that the single-level \(p_{i}^{\text{wd}}\) is equal to the two-level \(p_{i}^{\text{wm}}\). The reason for this may be as follows: \(p_{i}^{\text{wd}}\) is the third sub-objective. When \(p_{i}^{\text{wind}}<p_{i}^{\text{plan}}\) (the inequality means the wind power is less than the grid demand; thus, the storage systems must release energy to compensate for the missing energy between the wind power and the grid demand--in other words, in this situation, \(p_{i}^{\text{m}}\), \(p_{i}^{\text{h}}\), \(p_{i}^{\text{wm}}\) are equal to 0), to maximize \(p_{i}^{\text{wd}}\), and meet the constraint \(p_{i}^{\text{wind}}=p_{i}^{\text{wd}}+p_{i}^{\text{pa}}+p_{i}^{\text{wm}}\), \(p_{i}^{\text{wind}}\) should be equal to \(p_{i}^{\text{wd}}\). \begin{table}
\begin{tabular}{c c c c c c} \hline \hline \multirow{2}{*}{**Values**} & \multicolumn{5}{c}{**Time(min)**} \\ \cline{2-6}  & **435** & **440** & **445** & **450** & **455** \\ \hline \(p_{i}^{\text{bin}}\) / MW & 12.18 & 12.19 & 12.20 & 12.21 & 12.26 \\ \(p_{i}^{\text{wind}}\) / MW & 20.15 & 20.16 & 16.26 & 18.45 & 17.39 \\ \(p_{i}^{\text{wd}}\) / MW & 12.18 & 12.19 & 9.86 & 12.21 & 12.26 \\ \(p_{i}^{\text{pw}}\) / MW & 6.39 & 6.39 & 6.39 & 4.51 & 4.51 \\ \(p_{i}^{\text{h}}\) / MW & 0.66 & 2.42 & 0 & 1.73 & 0.62 \\ \(E_{i}^{\text{wm}}\) / MW \(\cdot\text{h}\) & 98.58 & 99.03 & 99.48 & 99.80 & 100.1 \\ \(E_{i}^{\text{fw}}\) / MW \(\cdot\text{h}\) & 4.809 & 5.0 & 4.80 & 4.93 & 4.98 \\ \hline \hline \end{tabular}
\end{table} 
Table 4: Analysis on the cause of the wind curtailment. Figure 10: Simulation result for the second sub-objective. Therefore, when \(p_{i}^{\text{wind}}<p_{i}^{\text{plan}}\), the single-level \(p_{i}^{\text{wd}}\) is equal to \(p_{i}^{\text{wind}}\), and the two-level wd is equal to \(p_{i}^{\text{wind}}\). Therefore, the single-level \(p_{i}^{\text{wd}}\) is equal to the two-level \(p_{i}^{\text{wd}}\). In contrast, when \(p_{i}^{\text{wind}}>p_{i}^{\text{plan}}\), to maximize \(p_{i}^{\text{wd}}\) while meeting the constraint \(p_{i}^{\text{wind}}=p_{i}^{\text{wd}}+p_{i}^{\text{rn}}+p_{i}^{\text{n}}+p_{i} ^{\text{w}}\) and \(p_{i}^{\text{wd}}\leq p_{i}^{\text{plan}}\), the optimal \(p_{i}^{\text{wd}}\) should be equal to \(p_{i}^{\text{plan}}\). However, when the fly wheel is almost full, wind curtailments (\(p_{i}^{\text{w}}\)) will occur.