Dual Failures with General Densities Suppose a system contains two independent widgets, denoted by A and B, in parallel, only one of which is required for the system to be operational, and the system performs a series of missions each lasting D hours. If each widget has a failure rate characterized by the density distribution f(t), and the system is run for T hours or until the first mission when at least one widget fails (whichever comes first), what is the probability that both widgets will fail during a single mission? The joint density function for failure of widget A at time tA and widget B at time tB is just the product f(tA)f(tB). Graphically we can represent the space of possible outcomes as the plane of tA versus tB, and then integrate the joint density function over the region(s) of interest. For the case we have described, the region of interest is the shaded region shown in the figure below. One way of performing the integration is to split up the range into three separate spans, over which the boundaries vary linearly, as shown below. This leads to the following sum of three double-integrals for the total probability However, there are more economical ways of evaluating this overall probability. We could reduce the number of double integrals from three to two by noting the symmetry about the diagonal line, and just integrating the region above the line (and then multiplying by 2). An even better approach is to integrate this same "upper" region diagonally instead of orthogonally, taking the coincidence closeness as the outer parameter, as illustrated below. This leads to an expression for P(T,D) as just one double integral This formulation is particularly nice because it can easily be modified to include "turnback" credit. This relates to cases where the failure of the first widget during a mission causes a turnback that truncates the mission in a time equal to the time from the start of the mission, as indicated in the figure below. To capture this turnback credit we simply replace the "0" lower bound on the inner integral with "u". In comparison, the original orthogonal method requires a sum of four double integrals to give the same result. For any given value of D the probability P(T,D) converges on a constant value as T goes to infinity. This limit represents the probability that the two widgets would fail within D hours of each other assuming we remove the "hard" limit of T hours entirely. Interestingly, if we take the exponential distribution as f(t), then the limit of P(T,D) as T goes to infinity equals the cumulative distribution function F(D) of the exponential distribution. In other words, the coincidence threshold for two exponentially distributed events (with no truncation) is itself exponentially distributed. To illustrate the use of formula (1), suppose we have a system with two widgets in parallel, and the failure characteristic of each widget has a Weibull distribution with scale parameter a = 385 and shape parameter b = 5.55. Recall that the cumulative Weibull distribution has the form This is just a generalization of the exponential distribution, and the parameter a simply scales the time variable, whereas b affects the shape of the distribution. If we take b = 1 we have the exponential distribution (constant failure rate), whereas with b other than 1 we get either an increasing or decreasing failure rate. Typically b is greater than 1, signifying the tendency of the failure rate to increase with time, due to the "wear out" of parts. Differentiating F(t) with respect to t, it follows that the Weibull density distribution is For the stated values of a and b, a plot of this density function is shown below. The density of the dual-widget system, where each widget has the individual failure density f(t), is illustrated in the plot below. To find the probability that both widgets will fail within 12 hours of each other over a period of T, we apply equation (1), which gives the cumulative probability curve shown below This would enable us to determine a suitable exposure time for a dual-widget system to ensure against a complete system failure during any 12-hour mission. Return to MathPages Main Menu