Consider a setting where Willie generates a Poisson stream of jobs and routes them to a single server that follows the first-in first-out discipline. Suppose there is an adversary Alice, who desires to receive service without being detected. We ask the question: what is the amount of service that she can receive covertly, i.e. without being detected by Willie? In the case where both Willie and Alice jobs have exponential service times with respective rates $\mu_1$ and $\mu_2$, we demonstrate a phase-transition when Alice adopts the strategy of inserting a single job probabilistically when the server idles : over $n$ busy periods, she can achieve a covert throughput of $\mathcal{O}(\sqrt{n})$ when $\mu_1 < 2\mu_2$, $\mathcal{O}(\sqrt{n/\log n})$ when $\mu_1 = 2\mu_2$, and $\mathcal{O}(n^{\mu_2/\mu_1})$ when $\mu_1 > 2\mu_2$. When both Willie and Alice jobs have general service times we establish an upper bound for the amount of service Alice can get covertly. This bound is related to the Fisher information. Additional upper bounds are obtained for more general insertion policies.

中文翻译：

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单个 FIFO 服务器中的隐蔽循环窃取

考虑一个设置，其中 Willie 生成作业的泊松流并将它们路由到遵循先进先出原则的单个服务器。假设有一个对手 Alice，他希望在不被发现的情况下接受服务。我们问这个问题：她可以秘密接受多少服务，即不被威利发现？在 Willie 和 Alice 作业都具有指数服务时间的情况下，分别为 $\mu_1$ 和 $\mu_2$，我们演示了当 Alice 采用在服务器空闲时概率性地插入单个作业的策略时的相变：超过 $ n$繁忙时段，当$\mu_1 < 2\mu_2$时，她可以实现$\mathcal{O}(\sqrt{n})$的隐蔽吞吐量，$\mathcal{O}(\sqrt{n/\log n})$ 当 $\mu_1 = 2\mu_2$ 时，$\mathcal{O}(n^{\mu_2/\mu_1})$ 当 $\mu_1 > 2\mu_2$ 时。当 Willie 和 Alice 的工作都有一般服务时间时，我们为 Alice 可以秘密获得的服务量建立一个上限。这个界限与Fisher信息有关。为更一般的插入策略获得了额外的上限。