I have recently seen a resolved case of components replacement forecast posted in a reliability software company site, whose solution I don't quite agree with, and I would like to know how peers in this forum would solve it – I maybe wrong.

Suppose you have to forecast into the upcoming year, on a quarterly basis, the number of new wheels that must replace the worn ones of a cargo-wagon, after they have attained a certain minimum diameter. Suppose the wagon has 24 wheels and that they all have accumulated 250 days rolling on rails so far (records with mileage just don't exist).

Failure data on the same failure mode (erosion) of similar wheels are abundant. Once statistically treated, the data showed to follow a Weibull distribution with a shape parameter of 1.56 and a scale parameter of 1,520 days.

What would be your forecast for the 1st, 2nd and 3rd quarters of this particular 24 wheeled cargo-wagon?

Thank you in advance.

Rui

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Rui,

And the opurpose of such forecasting would be to ...?

Is it for managing spares and resources? If so, the scale factor gives an average value, which should be adequate for such a prediction, assuming there are many such wagons and that they all start life at different times. With a shape factor so low, the distributuion does not help define time-to-failure well. Add to that using days instead of miles can introduce quite some errors. Is there a point in this exercise if the base data quality is open to large errors?

Hi Vee,

The purpose of the forecast is to provision the funds (capital) necessary for the purchasing of the wheels when they become needed.

Please, take into account that the Weibull distribution has no closed form; therefore you have to use a numerical method or Monte Carlo simulation. I used the latter.

Rui

My point is 1 in the first quarter, 2 in the second quarter and 1 in the third quarter.

Does anybody disagree?

Rui

Rui,

If I were in your place, I would do a history match; any method you use to predict must work for past history too. For these you have records and dont have to hypothise.

Whatever you predict statistically is still just a prediction. Have you done your calculations using Edward's Maximum Likelihood Estimate? Or a variation of Poissan?

Thanks Vee for your comments. Yes it is true that one can always consider history as a basis to forecast, as long as the system doesn't change in the mean time (for example, the mileage per wagon is supposed to be constant). But that is a too general approach and it can't address a particular situation like the one of a particular wagon or automotive. Doing it the way I did – and this is exactly the purpose of the article I mentioned – it becomes more specific, allowing one to estimate in grater detail what the need will be in the case of a particular wagon, locomotive or even a particular bogie.

The parameters of the Weibull distribution that fits the data the best were found by using the Median Ranking Least Square method and not the Maximum Likelihood method. By the way, could you please explain what the variation of Poissan is about? Or did you mean Poisson? If so, allow me to remind you that it applies exclusively to the case of random failures.

The figures I got (1 + 2 + 1 units) were obtained after a few thousand repetitions of a Monte Carlo simulation model. There is an alternative numerical method but it is just an approximation and I don't like this "empirical attempts" to conform a certain result at all.

Perhaps you would like to take a look at the article that brought this subject to light, despite it is in Portuguese (from Brasil) and it addresses 200 wheels instead of just 24 (my version of the case is just a simplification). You can notice that the first table shows the history of these 200 wheels – some failed and some others are still rolling (suspended). The current date is the 31 of December. The last table gives you the number of wheels that are estimated to be replaced from this very moment onwards, on a monthly basis, summing up 16 units in the whole year ahead. This is what I disagree. According to my calculations, it should be 28 units and not 16 and, of course, spreading differently over the next 12 months.

As you can see, any of the traditional analytical methods of forecasting based on the exponential smoothing of historical data such as the Holt or Holt & Winters methods, could never allow you to be so specific and take into consideration the current wheel lives. Those methods would give you just a sense of an average. Suppose that you had recently bought quite a few new wheels – and that is exactly what the case addresses (133 out of the 200 units are brand new) and any average would be spoiled in the short run before the steady state establishes again.

I presented my viewpoint to the editor and I am still waiting for an answer. Because I am really curious to know who is wrong (I maybe am), I decided to pose this question to the community of this forum – people interested in systems reliability. I find the case quite interesting and the solution can be quite informative.

Please see the article attached to this post.

Regards

Rui

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Case_Prevendo_Substituicoes.pdf (37 Kb, 7 downloads)

Do you use history to "validate" if the statistical model is good enough?

Hi Eugene,

It will always be a good management practice to confirm any values that have once been forecasted, for the sake of (eventual) correction of the forecasting method employed or adjustment of the parameters used in a particular method, whenever the magnitude of the difference falls outside the statistical confidence interval accepted.

In the case of the wagon-wheels, as I explained in my last post, one should not extrapolate history as it is not going to work out with sufficient accuracy in case the steady state is not established. This is exactly the situation described in the article: 133 brand new wheels are about to join the bunch of the 67 on service (with different accumulated life times). Only reliability models can take into consideration the inherent degradation of all the individuals at any moment in time.

Regards,

Rui

The editor of the article has just admitted that the result might not be correct and committed himself to come back with the correction. I will then have the chance to confirm my solution of the 24 wagon-wheel problem (1 wheel in the first quarter + 2 wheels in the second quarter + 1 wheel in the third). Does any one have a different solution?