One key challenge in designing resilient large-scale wireless ad hoc networks is to understand how random node failures affect fundamental network performance. In this work, we show that both network capacity and delay scale as \scalebox{0.65}{$\textstyle Θ\left(\sqrt{\frac{n(1-q)}{\log n}}\right)$}, where $n$ is the total number of nodes and $q$ is the node failure probability. The network capacity degenerates to the classical result given by P. Gupta and P. R. Kumar when $q=0$. Based on these results, we find that even with the same number of non-faulty nodes, a network with $n$ nodes and node failure probability $q$ has lower network capacity than a failure-free network with $n(1-q)$ nodes. To compensate for the network capacity loss caused by random node failures, at least $ε(n,q) nq$ redundant nodes are required, where $ε(n,q)>1$. We further prove that the optimal trade-off between network capacity and delay remains $O(1)$ regardless of node failures, implying that high network capacity and low delay cannot be achieved simultaneously. These results demonstrate robustness against stochastic variations in wireless channels.

翻译：在大规模弹性无线自组网设计中，理解随机节点故障如何影响基础网络性能是一项关键挑战。本文证明，网络容量与时延均按 \scalebox{0.65}{$\textstyle Θ\left(\sqrt{\frac{n(1-q)}{\log n}}\right)$} 规模缩放，其中$n$为节点总数，$q$为节点故障概率。当$q=0$时，网络容量退化至P. Gupta与P. R. Kumar的经典结论。基于这些结果我们发现，即使无故障节点数量相同，含$n$个节点且故障概率为$q$的网络，其容量仍低于含$n(1-q)$个节点的无故障网络。为补偿随机节点故障造成的容量损失，至少需要$ε(n,q)nq$个冗余节点（其中$ε(n,q)>1$）。我们进一步证明，网络容量与时延的最优折衷关系始终维持为$O(1)$且不受节点故障影响，这表明高网络容量与低时延无法同时实现。上述结果证明了无线信道随机波动下的鲁棒性。