Weibull and Lightning 

Suppose a system is designed to be immune to lightning strikes, but the features that provide the protection against lightning can fail, with a timedependent rate l(t). If a system’s protection features are failed, a lightning strike will result in complete failure of the system. Let s denote the rate of occurrence of lightning strikes (of sufficient intensity to cause system failure if the protection features are failed). Also, the system’s protection features are checked periodically, and if found to be failed they are restored to their “new” condition. We will represent this inspection/repair by stipulating that units with failed protection features are repaired at a rate m = 2/t_{I} where t_{I} in the inspection interval. (At the end of this note we evaluate the system using discrete repairs, and compare the results with those obtained using the continuous repair model.) 

If the function l(t) were constant, we could represent this scenario by a trivial Markov model as shown below. 


State “0” has functional lightning protection, and state “V” has failed protection features, so it is vulnerable to a lightning strike. As always, for the steadystate model, the complete system failure (represented by the s transition) feeds back into the fullup state. The rate of complete system failure is sP_{V} where P_{V} is the probability of state V. The system equation is lP_{0} = (m+s)P_{V} with the normalizing equation P_{0} + P_{V} = 1, from which is follows that the rate of complete system failure is 

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However, for our problem the function l(t) is variable, so this simple approach is not applicable. Strictly speaking, a transition with variable rate is not Markovian, but we can represent it as the limit of a sequence of Markov models in which State 0 is split into a timeordered sequence of substates, with each substate containing the systems of a certain age. In these models, the transition rate from each substate is constant, so the models can be solved using ordinary Markov model techniques. We can then evaluate the results in the limit as the sequence of substates becomes infinite, yielding the exact answer for the continuously variable transition rate. (For a related discussion, see the note on Markov Models with Aging Components.) 

To illustrate, we will split State 0 into four substates, representing four spans of time in the overall life T of a system, as shown in the figure below. 


where q = 1/Dt and Dt = T/3. (More generally, if we split up the original State 0 into substates 0 to N, we would set Dt = T/N.) Letting l_{n} denote l(nDt), the steadystate system equations can be written as 

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along with the normalizing relation 

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From the system equations we get 

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We can then insert these expressions into the normalizing equation, solve for P_{0}, and then substitute this value into the expression for P_{V}. This analysis immediately generalizes to substates from 0 to N, leading to the system failure rate 

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Clearly if l(t) is constant, this reduces to equation (1) as expected. This equations can be used to compute the desired result by setting N to an arbitrarily large value (and recalling that Dt = T/N), but we might prefer to have a more explicit expression for the limit as N goes to infinity and Dt goes to zero. To evaluate this limit, we first consider the products 

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This relation becomes exact in the limit as Dt approaches zero. The natural log of this product can be expressed as the summation 

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Taking just the first term of the expansion for the natural logs (which, again, becomes exact in the limit as Dt approaches zero), and converting the summation to an integration, we get 

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Likewise the products from i = 1 to j for j ranging from 1 to N are given by replacing T with a parameter t ranging from 0 to T. The sums of the products can similarly be reduced to integrations, leading to the limiting relations 

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These quantities both go to infinity as N increases, so the additive constants in (2) drop out. Thus, inserting these expressions into (2) and canceling the factors of N/T, we get the system failure rate in the continuous limit: 

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where 
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Equation (3) is formally identical to (1), except that we replace the constant failure rate l with the effective failure rate l^{*}, the latter being a weighted mean of l(t) over the life of the system. 

If the loss of lightning protection has a Weibull failure rate, the function l(t) is of the form 

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where b is the shape parameter and a is the scale parameter. If b is greater than 1, signifying a wearout characteristic such that the failure rate increases with time, we can insert this into the expression for the effective failure rate to give 

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Notice that if b = 1, which signifies a constant failure rate, this expression reduces to l^{*} = 1/a as expected, bearing in mind that the scale factor a corresponds to 1/l for an exponential distribution. On the other hand, if b = 2, corresponding to the case of a linearly increasing failure rate, the expression for the effective failure rate in the limit as T goes to infinity (which is the conservative assumption) can be evaluated in closed form to give 

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Often the values of b for relatively mild wearout characteristics are in the range from 1 to 2, because for values greater than 2 the rate increases exponentially. Therefore, the behavior of many systems can often be bounded between these two cases. As an example, consider the case 

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The figure below shows the system failure rate as a function of system life T for various values of b. 


Just to cite one specific case, with b = 1.6 and T = 7000 we have l_{sys} = (1.244)10^{5}. This agrees with the discrete formula in the limit as N increases. 

Not surprisingly, each curve approaches an asymptotic value as T increases. It’s conservative to assume infinite life, rather than attempting to take credit for the fact that a unit may be removed from service (for some other reason) before it fails. For any value of b between 1 and 2 the asymptotic system failure rates are bounded by the two values 

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In the preceding discussion we represented the periodic inspection/repair at intervals t_{I} by a continuous transition with constant rate 2/t_{I}. This is quite suitable for cases where the rate of entering the failure state doesn’t change much over the duration of any single interval. On the other hand, one might question the suitability of this representation in cases with very long inspection/repair intervals, during which the the rate of entering the failed state may vary significantly. With a Weibull distribution (with b > 1) the failures are more frequent toward the end of an interval, so the mean time to repair of a failed system will be somewhat less than 2/t_{I}. To assess the degree of approximation introduced by the use of the continuous transition model in such cases, consider a system with a life of 96000 hours and an inspection interval of 24000 hours. Suppose the system contains 5 fasteners, and if any one or more of those fasteners fails, the system is vulnerable to a severe lightning strike, and the rate of occurrence of such strikes is s = (8.85)10^{7} per hour. Each fastener has a Weibull failure characteristic with a = 10^{7} and b = 1.6. One very simple and conservative approach would be to assume each fastener has a constant failure rate equal to the value of l(96000), which is the maximum possible failure rate of a fastener over its life. Inserting the values of a and b into equation (4) with t = 96000 we get (9.85)10^{9} per hour, and since there are 5 fasteners, this gives a total rate of l_{max} = (4.93)10^{8} per hour. Using this value for l and using the continuous repair model with m = 2/24000 = (8.33)10^{5}, equation (1) gives l_{sys} = (5.09)10^{10}. For a less simplistic approach, taking credit for the fact that the failure rate is less than the maximum value during most of the system’s life (i.e., accounting for the fact that the Weibull distribution begins with a zero rate and increases with time), we use equation (5) to give the effective failure rate for a single fastener of (6.16)10^{9} per hour, so for the five fasteners we have l^{*} = (3.08)10^{8} per hour. Using this value, equation (3) gives l_{sys} = (3.18)10^{10} per hour. 

Now, to analyze this same system using a discrete representation of the repairs, we first note that the repair rate is two orders of magnitude greater than the rate of lightning strikes, so we can accurately approximate the probability of being in the vulnerable state by neglecting the lightning strikes, and simply considering a system with a Weibull failure distribution and discrete periodic repairs (neglecting lightning strikes). Given a life of 96000 hours with inspection/repair every 24000 hours, we see that the life is divided into four segments, and the probability of a system being failed at the beginning of each segment is zero, as depicted below. 


During each interval, the probability of being vulnerable is simply the integral of the effective density function. Let P_{1} denote the probability at the end of the first interval, P_{2} the probability at the end of the second interval, and so on. We know the density distribution during the first interval is simply the standard Weibull density 

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With this we can compute the probability of being in the vulnerable state at the end of the first interval 

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where t = 24000 is the inspection/repair interval. The density function during the second interval is the weighted average of the density functions for the two possibilities, i.e., either the system was found failed at the end of the first interval or it was not. If it was found failed at time t, then it was restored to the initial condition, whereas if it was not found failed it continues to age using the original density. Thus the density function for the second interval is 

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where the time parameter t is here taken to be zero at the beginning of the second interval. With this we can compute the probability of being in the vulnerable state at the end of the second interval as 

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By the same reasoning as before, the density function for the third interval (now taking t = 0 at the start of the third interval) is 

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from which we can compute the probability of being vulnerable at the end of the third interval as 

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Lastly, the density function for the fourth interval (again taking t = 0 at the start of the interval) is 

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Written explicitly in terms of the d_{1}(t) function, this is 

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where 
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We can now compute the mean probability of being in the vulnerable state over the entire life of the system. It is simply the average of the mean probabilities over each of the four intervals. Letting p_{j}(t) denote the probability in the jth interval (with t = 0 at the start of the interval), we have 

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and the mean probability over that interval is 

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The overall mean is therefore 

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Inserting the values of the parameters for our example, this gives p_{mean} = (6.922)10^{5} for a single fastener, so for one or more of the five fasteners to be in the vulnerable state we multiply this by 5 to give the probability (3.46)10^{4}. Multiplying this by the rate of lightning strikes, s = (8.85)10^{7} we get the rate of total system failure l_{sys} = (3.06)10^{10}. This is quite close to the value of (3.18)10^{10} given by the Markov model based on continuous repairs. 

By the way, since the failure rate is so low in this example, it is very unlikely that any fasteners will be found failed at any inspection, so we could simplify the calculation by simply assuming no failures, and on this basis the mean probability can be given explicitly in terms of the basic Weibull density as 

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This gives the result p_{mean} = (6.923)10^{5}, which is nearly indistinguishable from the exact value.
