故障注入覆盖率评估:故障抽样与统计分析

Wei Wang, Kishor S. Trivedi
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引用次数: 1

摘要

众所周知,不完全覆盖会大大降低冗余的有效性,但该参数的确定一直难以捉摸。断层注入模拟已被认为是一种有效的数据收集技术,用于统计估计覆盖参数。在故障模拟实验中,从故障种群中采样故障,然后将故障注入系统的仿真模型,以确定系统故障检测机制未发现的故障数量,即导致不安全输出的故障数量。当在大量的模拟运行期间没有发生不安全的故障时(例如,当注入大量的故障时),不可能量化准确的覆盖率,因为唯一可用的信息是系统在所选故障存在的情况下尚未发生故障。然而,在高置信水平下,可以推断覆盖率至少达到某个目标值,即量化安全措施的下界。置信水平和下限都取决于测试持续时间(即样本量)和从故障总体中选择故障的方式。对于简单的随机抽样,其中每个故障都有相同的选择机会,我们提出了一个公式,可以用来预测目标覆盖率下界的最小故障注入数量。对于超可靠和安全关键型系统,期望的覆盖范围应该非常接近于1。在这种情况下,我们的结果表明,最小
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Coverage Evaluation Through Fault Injection: Fault Sampling And Statistical Analysis
Imperfect coverage is known to drastically reduce the effectiveness of redundancy and yet the determination of this parameter has been elusive. Fault injection simulation has been recognized as an effective technique to collect data for the statistical estimation of the coverage parameter. In a fault simulation experiment, faults are sampled from the fault population and then injected into, the system’s simulation model to determine the number of faults that are uncovered by the system’s fault detection mechanisms, i.e., the number of faults that cause unsafe outputs. When no unsafe failure occurs during a large number of simulation runs (i.e., when a large number of faults are injected), it is not possible to quantify the exact coverage since the only available information is that the system has not yet failed in the presence of the chosen faults. However, it is possible, with a high confidence level, to infer that the coverage value reach at least some target value, i.e., to quantify a lower bound on the safety measure. Both the confidence level and the lower bound depend on the test duration (i.e., sample size) and the way in which faults are selected from the fault population. For simple random sampling, where every fault has an equal chance of being selected, we present a formula that can be used to predict the minimum number of fault injections for a target lower bound on coverage. For ultra-reliable and safety-critical systems, the desired coverage is supposedly extremely close to 1. In this case, our result shows that the minimum
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