Dual Failures with General Densities

 

Suppose a vehicle has two independent motors, denoted by A and B, only one of which is required for the vehicle to be operational, and the vehicle performs a series of missions each lasting T hours, although if one motor fails the vehicle will either return to its point of origin, divert to an alternate destination that is D hours away, or continue on to its original destination, whichever takes the least amount of time. If each motor has a constant failure rate denoted by λ, small enough so that λT is much less than 1, then the probability of both motors failing in transit can be inferred from the figure below.

 

 

The probability of both motors failing in the next T hours is simply (λT)2, and the desired probability is the fraction of this represented by the shaded area as a fraction of the TxT square. Starting with the area of the whole square and subtracting the areas of the unshaded pairs of triangles, the shaded area is T2 – (T−D)2 – D2 = 2D(T-D), so the desired probability is 2λ2D(T−D). Naturally this applies only for D ≤ T/2.

 

More generally, consider the case where the failure rate of each motor is characterized by the density distribution f(t). In this case the probability density is not uniformly distributed over the square, so we can’t simply work with the ratio of areas. First, suppose we neglect any turn-back, so after the first motor fails the vehicle will proceed to a destination D hours away or its original destination, whichever can be reached sooner. (Assume for the moment that the vehicle’s speed is the same with one or both motors functional.  We will consider the case of different speeds later.) What is the probability that both motors will fail in transit?

 

The joint density function for failure of motor A at time tA and motor 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, without turn-back, the region of interest is the shaded region shown in the figure below.

 

119FIG1

 

One pedestrian way of performing the integration is to split up the range into three separate spans, over which the boundaries vary linearly, as shown below.

 

119fig2

 

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.

 

119fig3

 

This leads to an expression for P(T,D) as just one double integral

 

 

With a constant failure rate, i.e., an exponential density f(t) = λe−λt, with a very small value of λ the density is essentially constant over the mission length T, so we can take f(t) = λ for this mission, in which case this formula gives P(T,D) = λ2D(2T−D). The formulation (1) is particularly nice because it can easily be modified to include "turnback" credit. This relates to cases where the failure of the first motor during a mission causes a turnback that truncates the mission in a time equal to the time from the start of the mission, as depicted in the first figure in this article. To capture this turnback credit we simply replace the "0" lower bound on the inner integral with "u", giving the expression

 

 

In comparison, the original orthogonal method requires a sum of four double integrals to give the same result. Note that, with 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 motors would fail within D hours of each other assuming we remove the "hard" limit of T hours entirely. Interestingly, if f(t) has an exponential distribution, 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.

 

For the case of an exponential distribution that is essentially constant over the mission of duration T, so that f(t) can be taken as the constant λ such that λT is much less than 1, the formula (2) gives P(T,D) = 2λ2D(T−D), in agreement with the simple geometrical analysis discussed previously. Again, as can be seen from the bounds on the inner integration, this applies only for D ≤ T/2.

 

Incidentally, in some cases a mission might be of a specified distance LT, normally traversed at an average speed v1, so we have LT = v1T, but in the event that one motor fails at any point the vehicle moves at a reduced speed v2 back to the point of origin, or on to the destination, or to an alternate location a distance LD = v2D, whichever is closer. For essentially constant individual densities f(t) = λ on the time basis, the rates for the first and second failures on the distance basis are λ/v1 and λ/v2. Hence we can repeat the time-based analysis, but in terms of distances, using v1T in place of T, and v2D in place of D and λ2/(v1v2) in place of λ2. This gives P(T,D) = 2λ2D(T − (v2/v1)D). As before, this applies for v2D ≤ v1T/2. This is discussed further in “Time and Distance for Dual Motors”.

 

To illustrate the use of formula (1), suppose we have a system with two motors in parallel, and the failure characteristic of each motor has a Weibull distribution with scale parameter η = 385 and shape parameter β = 5.55. Recall from Section 6.1 that the cumulative Weibull distribution has the form

 

 

This is just a generalization of the exponential distribution, and the parameter η simply scales the time variable, whereas β affects the shape of the distribution. If we take β = 1 we have the exponential distribution (constant failure rate), whereas with β other than 1 we get either an increasing or decreasing failure rate. Typically β 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, the Weibull density distribution is

 

 

For the stated values of η and β, a plot of this density function is shown below.

 

119fig5

 

The density of the dual-motor system, where each motor has the individual failure density f(t), is illustrated in the plot below.

 

119fig6

 

To find the probability that both motors 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

 

119fig7

 

This would enable us to determine a suitable exposure time for a dual-motor system to ensure against a complete system failure during any 12-hour mission.

 

Return to MathPages Main Menu