## 5 Example and Discussion
\begin{table}
\begin{tabular}{c c} \hline \hline
**Orbit Elements** & **Values** \\ \hline Semi-major Axis & 6778.14 km \\ Eccentricity & 0 \\ Inclination & 97.0346 deg \\ Argument of Perigee & 0 deg \\ RAAN & 279.066 deg \\ True Anomaly & 0 deg \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Orbit parameters of Sat1. Figure 6: Illustration of the satellite cluster simulation scenario showing the trajectories of Sat2-6 moving with Sat1 (**a**) and distance varying with time from Sat16 to Sat1 (**b**). Figure 8a shows the network topology changing times varying with the time intervals of the snapshot. It is obvious that all of the three topology control algorithms show decreasing trends, and they finally trend to the same when \(\eta=500\) s. The solid line with squares represents the simulation condition without any topology control algorithm, which serves as a contrast to the others. For most situations, it can be seen that when the network works without control, its topology changes as many as 280 times, much higher than the other three. When \(\eta\in(0,100)\), the results from \(\text{FIYG}_{2\pi/3}\) are similar to those of the other three. Figure 7: Network topologies of the satellite cluster with different topology control settings. **(a)** After applying Reliability-Assured Topology (RAT) algorithm Sub-step (1); **(b)** After applying RAT algorithm Sub-step (2); **(c)** After applying RAT algorithm Sub-step (3); **(d)** Without topology control; **(e)** After applying topology control algorithm \(\text{FIYG}_{2\pi/3}\); **(f)** After applying topology control algorithm minimum spanning tree (MST). drops quickly from 335 to 49, also changing much more than RAT at around 40 times. The RAT results vary from 54 times to 20 times, and after the interval \(\eta\) increases to greater than 200 s, it changes little, about 15 times. Thus, compared to the without control and FIYG\({}_{2\pi/3}\) scenarios, the network with RAT or MST make it much simpler for the upper communication protocol layers, like routing protocols, to keep the data transmitting paths stable, because of the fewer topology changes. Figure 8b presents the average topology lifetime from the three different topology control algorithms (MST, FIYG\({}_{2\pi/3}\), and RAT), and also without control. It is evident that the results from the without control condition have nothing to do with the time interval \(\eta\) and holds a constant value, 17.9 s. The other three algorithms show an increasing trend, with MST having the longest average topology lifetime in all situations. In addition, after the interval \(\eta\) increases to greater than 200 s, the results from RAT and FIYG\({}_{2\pi/3}\) trend to be the same. Thus, it can be summarized that the topology control algorithm could greatly increase the average topology lifetime as the time interval grows, which would help the network to transmit more data due to the longer average topology lifetime. In the meantime, this paper analyzes the total received packets and average packet loss rate which can be calculated through the following equation:  
\[\textit{Average packet loss rate}=\frac{P_{\textit{send}}(\textit{Sat1})+P_{ \textit{send}}(\textit{Sat2})-P_{\textit{received}}(\textit{Sat15})-P_{ \textit{received}}(\textit{Sat16})}{(P_{\textit{send}}(\textit{Sat1})+P_{ \textit{send}}(\textit{Sat2}))t} \tag{18}\]  
where \(P_{\textit{send}}(\textit{Sat1})\) is the amount of generated packets at Sat1, \(P_{\textit{send}}(\textit{Sat2})\) is the amount of generated packets at Sat2, \(P_{\textit{received}}(\textit{Sat15})\) is the amount of received packets at Sat15 from Sat1, \(P_{\textit{received}}(\textit{Sat16})\) is the amount of received packets at Sat16 from Sat2, and \(t\) is the simulation time. Both of these can directly affect the network performance in terms of the reliability of the topology control algorithms. It is clear that the lower the average packet loss rate means the better the algorithm performs in the network.