David KrauseThis article summarizes the methodology of my 2009 diploma thesis ("Development and implementation of optimal maintenance strategies based on lifetime distributions", Hochschule Lausitz) and works it through on an example — here not on a foundry molding line as back then, but on the CNC machining center of the reference company that serves as the anchor throughout the book.
1 · The problem with gut feeling
In practice, maintenance intervals arise in three ways: from the manufacturer's documentation, from the maintainers' experience, or — most often — from round numbers. All three can be right. The only way to verify it is to know two things: how the component fails and what the failure costs compared with prevention.
Maintained too late means: unplanned failure, consequential damage, downtime at the worst moment. Maintained too early means: parts replaced that still had half their life left — and every intervention itself creates downtime and assembly risk. The task is a minimization, not a maximization of safety.
2 · The Weibull distribution and its shape parameter
The failure behavior of technical components can be described by the Weibull distribution. It has two parameters: the characteristic life a (the time by which 63.2 % of units have failed) and the shape parameter b. The shape parameter is the truly interesting quantity, because it tells you which failure mechanism is at work:
| b | Failure rate | Phase | Right strategy |
|---|---|---|---|
| b < 1 | decreasing | Early failures (assembly, manufacturing defects) | Run-in inspection, supplier quality — no preventive renewal |
| b = 1 | constant | Random failures (exponential distribution) | Preventive replacement achieves nothing — condition monitoring or run-to-failure |
| b > 1 | increasing | Wear, fatigue, aging | Preventive renewal pays — interval calculable |
A preventive replacement improves the situation only for b > 1. For random failures (b ≈ 1) a new part is just as failure-prone as the installed one — the maintenance then only creates cost and downtime. For early failures (b < 1) it makes things worse, because every new part re-enters the early phase. Whoever sets up maintenance plans without knowing the shape parameter can systematically degrade availability.
Where the parameters come from when no data exists
The practical objection comes at once: "We have no failure statistics." That was the starting point in the thesis too. The solution is expert estimates for lifetime quantiles: maintainers and operators state after which running time roughly 10 % and roughly 90 % of units have failed. From these two points both Weibull parameters can be determined. The method is coarse, but it yields a reliable shape parameter — and that decides the strategy.
3 · The optimization criterion: cost per unit time
The target variable is not availability and not failure probability, but the cost rate — expected cost divided by expected cycle time:
with expected cost = kp · R(τ) + ka · [1 − R(τ)]
and R(τ) = e−(τ/a)b (survival probability)
In words: if the component makes it to the planned date τ without trouble — which happens with probability R(τ) — the cheap preventive cost kp applies. If it fails earlier, the expensive failure cost ka. Both are spread over the running time actually reached.
4 · The cost side — where engineering meets costing
The point where many reliability calculations fail is the valuation of downtime. A failure costs not only the spare part and fitter hours, but lost production. And the right basis for it is not the machine hour rate, but the contribution margin per machine hour: during downtime the variable costs are avoided anyway — what is lost is the contribution margin. For cost center 210 that is €179.00/h (see book, Chapter 13.5).
| Item | planned (kp) | failure (ka) |
|---|---|---|
| Spare bearing set | €2,800 | €2,800 |
| Consequential spindle damage on failure under load | — | €5,500 |
| Assembly labor €65/h | €390 | €910 |
| Downtime loss CM €179.00/h · 6 h resp. 36 h | €1,074 | €6,444 |
| Total per event | €4,264 | €15,654 |
The ratio ka/kp = 3.67 is the second driver alongside the shape parameter. The larger it is, the earlier the preventive intervention pays.
5 · The result
With b = 2.3 and a = 12,000 h (expert estimate: B10 at around 4,500 h), the cost rate as a function of the chosen interval — expressed via the required survival probability — comes out as:
| R required | Interval | €/h | €/year (3,393 h) |
|---|---|---|---|
| 99 % | 1,624 h | 2.704 | €9,175 |
| 95 % | 3,299 h | 1.488 | €5,049 |
| 90 % | 4,511 h | 1.236 | €4,194 |
| 85 % | 5,446 h | 1.151 | €3,904 |
| 80 % | 6,251 h | 1.117 | €3,791 |
| 75 % ★ | 6,981 h | 1.108 | €3,758 |
| 70 % | 7,665 h | 1.110 | €3,768 |
| 60 % | 8,961 h | 1.137 | €3,857 |
| 50 % | 10,232 h | 1.177 | €3,995 |
The optimum is at R = 75 %, not 95 %. The intuitive demand "the component should survive to 95 %" leads to an interval of 3,299 h and costs €5,049/year — about 34 % more than necessary. High survival probability is not an end in itself; it is bought with prematurely discarded remaining life.
Excessive maintenance is more expensive than none. The usual 2,000-hour plan costs €7,582/year. Pure run-to-failure costs €4,996/year. Maintaining too often thus worsens the result more than dropping prevention altogether.
The minimum is flat. Between 6,000 and 8,000 hours the cost rate changes by less than 1 %. That is good news: the calculation need not be exact to hold — it only has to hit the order of magnitude. And it allows the interval to be shifted within that band to fit plant holidays or maintenance windows without losing anything economically.
6 · From single part to strategy
The calculation above concerns one component. In practice a machine has dozens, and each would have its own optimum — which would lead to dozens of separate stoppages. So the second step is coordination: you fix a base interval and assign each component an integer multiple of it. Components are thus shifted slightly away from their own optimum to be renewed together in a single stoppage. Because the minima are flat, this shift costs almost nothing — but saves a complete setup each time.
1. Prioritize components — only units with relevant failure impact. 2. Determine the shape parameter (history or expert estimate over lifetime quantiles). 3. Quantify the cost of both cases, valuing downtime at the contribution margin per hour. 4. Minimize the cost rate, coordinate intervals onto a common base grid.
7 · What this has to do with cost accounting
This methodology intervenes in costing at exactly the point that appears in the book as the maintenance budget within the machine hour rate — for cost center 210 that is €22,000/year, or €6.48/MH. This amount is not a natural constant but the result of the chosen strategy. And it has a second, larger lever: every avoided failure hour raises availability and thus the productive hours in the denominator of the hour rate. Maintenance strategy is therefore directly costing work — it determines both the numerator and the denominator.
Methodological basis: own diploma thesis, Hochschule Lausitz (FH), industrial engineering program, 2009. The worked example was newly built for this article on the fictional reference company "Präzisionsteile Muster GmbH"; the parameters are plausibly chosen, not measured.
