I have a load versus capacity problem, and I'm trying to determine the likelihood of failure. Here's a simple example of what I mean.

I have load and capacity discrete pdfs that were determined through a series of computer calculations that analyzed different input values (Monte Carlo simulations to address uncertainties). The figures below are a simplified version of the output:

Load Distribution

Capacity Distribution

As you can see, there is some amount of overlap between the two. I am trying to determine the probability that the system will fail (i.e., the load with exceed the capacity).

Is this as simple as integrating the overlapping area of the empirical cdfs and multiplying, or am I looking for something more complicated, like a Bhattacharyya coefficient?


1 Answer 1


Unfortunately, I don't think you can really answer this question accurately without knowing a bit more about what drives load and capacity variability.

Consider the case of load and capacity in the electrical grid. In the middle of the day in the summertime, as businesses have all their lights turned on and everybody is running their air conditioning, the load (the number of megawatts that people are demanding) soars. On the other hand, at 3:00 a.m. on generally any day of the year, as almost everyone is asleep, load falls. In spite of this fluctuating load problem, in developed countries at least, you almost never see a situation where load exceeds capacity (which would manifest as either a brownout or a rolling blackout). Why is this?

It's because in the electrical grid, load and capacity are tightly correlated over time. Engineers deliberately start up and shut down power plants, or subsections of power plants, in anticipation of time-varying changes in demand.

If your systems behaves in a similar way, it's not enough to simply calculate the overlap between the two overall accumulated static distributions. Instead, you need to Monte Carlo both factors simultaneously, and look for individual instances (i.e., individual "rolls" of the Monte Carlo dice) where load exceeds capacity. This is because, if the random factors in the Monte Carlo lead to correlated variability in both the load and capacity (or for that matter, anti-correlated variability as well) the probability of system failure will be different than just a simple pdf overlap calculation would suggest.

  • $\begingroup$ Thanks stachyra, I understand your comments. However, I think in my case the variables that effect the load are independent of those that influence capacity. I'm actually looking at a power plant, so your comments are well taken, but the problem is a little different. The load (plant temperature) is determined by environmental effects, but the capacity of a material to withstand that temperature is a product of uncertainties related to manufacturing. So uncertainties exist related to both the load and capacity, but they are not correlated. $\endgroup$
    – dnow
    Dec 13, 2013 at 20:22
  • 1
    $\begingroup$ Ah, O.K., I see. Then in that case, I think it is just the overlap in probabilities that you will want. Since you're already doing a numerical simulation anyway, you can simulate this part easily as well, by simply taking random draws from each of the sample distributions that you have pictured, and counting the relative frequency (over many experiments) with which load exceeds capacity. $\endgroup$
    – stachyra
    Dec 13, 2013 at 21:30
  • $\begingroup$ Thanks again, let me see if I understand correctly. Rather than trying to asses the overlap failure probability from the final empirical distributions, I can take a load simulation output and compare it to a capacity simulation output directly. If I repeat this many times (which I'm already doing to develop the distributions), it should approach the actual failure probability with a large enough sample size ("actual" within the confines of my model). That makes sense. $\endgroup$
    – dnow
    Dec 13, 2013 at 21:40
  • $\begingroup$ Yes, I believe the procedure that you have outlined should estimate the failure probability correctly. $\endgroup$
    – stachyra
    Dec 13, 2013 at 22:13

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