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I have a failure rate problem involving structural beams. The scenario is that a set of $n$ structural beams are supporting a weight of $W$ kilograms. For the purpose of simplification, we can assume that all the structural beams equally support the weight. And so, we can say that, if $l$ of the beams break, then the load will then be equally shared by the remaining $n - l$ beams. When we reach the point where $n - 1$ of the beams have broken, new beams are instantly installed to increase the number of working beams back to $n$.

This is an instantaneous failure rate problem. In this case, the rate of failure of any single beam carrying $W^\ast$ kilograms is $\alpha W^\ast$, where $\alpha > 0$ remains constant. Furthermore, this is a Markov process, meaning that the time until failure at time $t$ is independent of the past.

So it seems to me that we can model this problem as a Markov process with $X(t)$ being the number of unbroken beams at time $t$. Furthermore, it seems to me that we have $1 \le l \le n - 1$.

Let's say we're given some failure rate for the structural beams, measured per year per kilogram of weight supported. How do I calculate the probability that some system will survive for some years before replacement beams have to be installed? I'm not so much interested in specific numbers as I am the concept (so that I can make it work for any values I want, based on the specific problem), so feel free to input your own numbers for purposes of illustration.

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Although you could model this as a continuous-time Markov process, there is only one possible set of state transitions, starting from a state with $X=n$ beams and progressing sequentially down to $X=1$, at which time the system magically returns to $X=n$. For me, it was simpler to think about this as a sequence of independent exponentially distributed waiting times,* with a property that surprised me at first.

Start with with $n$ beams supporting a total weight $W$ and rate constant $\alpha$, per year per kilogram supported, for failure of any single beam. Each of the $n$ beams supports $W/n$ weight, so failure of any individual beam in the set of $n$ is Poisson with rate constant $\lambda_n=\alpha W/n$ and a corresponding exponential distribution of waiting times having mean $n/(\alpha W)$ and variance $n^2/(\alpha W)^2$.

If all $n$ beams are assumed independent with the same failure-rate constant $\lambda_n$, the first failure time among the $n$ beams is distributed exponentially with a rate constant of $n \lambda_n= \alpha W$, independent of $n$! Under these assumptions, the fact that the weight is borne by multiple beams is balanced exactly by the faster expected occurrence of the first break with multiple beams.

Thus all the sequential individual beam failures are Poisson with the same rate constant $\alpha W$, regardless of the number of beams (provided there is at least 1). The expected waiting time for the $n-1$ failures needed to reset the system is $\frac{n-1}{\alpha W}$, providing the basis for modeling the distribution of system-reset times.

In a more general case of necessarily sequential states where the transition times might not be so simply described but the state transitions are Markovian, you can model this as an equivalent repeated-events survival model. Then, where $\lambda(t)$ is a time-dependent hazard function, the cumulative hazard $\Lambda(t)=\int_0^t \lambda(\tau) d\tau$ is the expected number of events up to time $t$. So modeling the time for $n-1$ events is just modeling $\Lambda(t)= n-1$, and the Markovian property allows each transition time to be modeled independently.

For example, if the hazard for leaving state $X=m$ to state $X=m-1$ is constant at $\lambda_m$ and the transition is Markovian, the expected time for the transition solves $1=\int_0^t \lambda_m d\tau$ for $t$, or $1/\lambda_m$ As state transitions are assumed independent (except for their necessarily sequential nature), the expected time for $n-1$ transitions just sums up all of those times, with a variance equal to the sum of the individual variances.


*I'd be reluctant to apply this in practice to physical beams, which almost certainly aren't memoryless as posited here.

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  • $\begingroup$ Thanks for the answer. My intention was to model this as a CTMC, rather than some special case. The latter is drastically less useful/robust. $\endgroup$ – The Pointer Apr 23 at 15:31
  • $\begingroup$ @ThePointer a more general CTMC might make sense if each state could make a transition to multiple other states, but all you have here is a defined sequence from state $X=n$ down to state $X=1$ and back again. Such a necessarily sequential CTMC is equivalent to a repeated-event survival model. I'll add a bit to the answer to deal with a more general repeated-event model. $\endgroup$ – EdM Apr 23 at 15:47
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    $\begingroup$ The conundrum is that I'm studying this to learn / better understand CTMC, so the most critical part of this is the pedagogical aspect. An answer that focuses on modelling the problem using special cases, rather than on the conventional CTMC way of modelling the problem, doesn't help me much (although, your information about the special case is obviously still interesting and educational in its own right, just not what I was looking for). $\endgroup$ – The Pointer Apr 23 at 15:52
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    $\begingroup$ @ThePointer this might not be the best pedagogical example for understanding CTMC, as each state only has one possible subsequent state so there is a direct relationship to a repeated-event survival model. You might want to consider a more complicated model in which there's a probability of adding a beam to each state (not just adding $n-1$ back to state $X=1$), instead of just losing one from all states in which $X>1$. $\endgroup$ – EdM Apr 23 at 16:16
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    $\begingroup$ @ThePointer I would have been of no help with a more general CTMC that has more than 1 possible state transition from any given state, of the type I suggested in another comment. I do so much survival modeling that I tend to try to put things into the survival context first, which happened to work in this particular instance. $\endgroup$ – EdM Apr 26 at 15:24

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