## 5 Numerical Results
### Throughput  
Figure 2 shows the throughput versus the channel load, for the cases of a single satellite with different erasure probabilities (\(\varepsilon=0.01\) and \(\varepsilon=0.9\)), and for the cases of \(K=2\) satellites with erasure probabilities \(\varepsilon_{1}=0.01\) and \(\varepsilon_{2}=0.9\). Moreover, in the figure we also show the sums of the individual throughputs seen by each of the \(K=2\) satellites, as well as the intersections of their throughputs. Recall that the actual throughputs are the differences between the sums of the individual throughputs and their intersections (recall the inclusion-exclusion principle discussed in Section 3). First, considering the case of a single satellite, we can see from Figure 2 that a low erasure probability leads to a larger throughput at low channel loads, and a higher erasure probability is favorable at high channel loads. That is because a high erasure probability limits the collisions, which is desirable at high loads. However, a high erasure probability also leads to an inefficient utilization of the resources of the IoT devices. Interestingly, when we considered \(K=2\) satellites, one with a low and another with a high erasure probability, the benefits were quite large, especially at low to moderate channel loads. Moreover, in the figure we also show simulation results for the case of \(K=2\), validating the analysis in Section 3. Finally, the curves representing the sums and the intersections of the individual throughputs of the two satellites highlight the importance of removing multiplicities when deriving the throughput in Section 3, as the intersection is non-negligible. We extended the analysis for the case of \(K=2\) satellites in Figure 3, in which we fixed the erasure probability \(\varepsilon_{1}\) at the first satellite for each of the subfigures, and we considered different erasure probabilities \(\varepsilon_{2}\) at the second satellite. Thus, we could evaluate the system throughput for a large number of \([\varepsilon_{1},\varepsilon_{2}]\) pairs. Several interesting conclusions can be drawn from these results. First, having low erasure probabilities at both satellites is good only at very low channel loads. Moderate erasure probabilities at one of the satellites, or even at both, are beneficial at low to high channel loads, as they limit the impact of packet collisions. Another important outcome from the results in Figure 3 is that there is an optimum \([\varepsilon_{1},\varepsilon_{2}]\) pair for a given channel load. Finally, it is noteworthy that simulation results agree very well with the theoretical analysis in Section 3. Figure 4 shows the throughput surface as a function of the erasure probabilities \(\varepsilon_{1}\) and \(\varepsilon_{2}\) at the satellites, for the case of \(G=12\) packets per time slot. First, corroborating  
Figure 3: Throughput \(\mathcal{T}\) vs. channel load \(G\) for the case of \(K=2\) satellites. We fix the erasure probability \(\varepsilon_{1}\) at the first satellite at a different value for each of the subfigures, and we consider a set of different erasure probabilities \(\varepsilon_{2}\) at the second satellite in each subfigure. Figure 2: Throughput \(\mathcal{T}\) vs. channel load \(G\), for \(K=1\) with erasure probabilities \(\varepsilon\in\{0.01,0.9\}\) and \(K=2\) with erasure probabilities \(\varepsilon_{1}=0.01\) and \(\varepsilon_{1}=0.9\) (analytical and simulation results). Moreover, we show the sum and intersection of the individual throughputs seen at each satellite. one of the conclusions from Figure 3, there was an optimum pair of erasure probabilities that led to the maximum throughput. Since in this example the channel load was very high, the maximum throughput was obtained when the erasure probabilities were also very high (\(\epsilon_{1}=\epsilon_{2}=0.92\)). However, note that such high erasure probabilities, although favorable in terms of throughput at such high channel load, also led to large packet loss rates.