Maintenance Engineering
Changing at 3,000 hours instead of 1,990 costs a fleet of 24 machines ten unplanned stops a year.
— hours
Minimises long-run cost per operating hour
The model assumes replacement restores the component to as-new condition and that the replacement is drawn from the same population, so a part with a different supplier or a different specification invalidates the fit rather than shifting the answer. It assumes a single dominant failure mode: where two modes compete, an early manufacturing defect and a late wear-out for instance, one Weibull fit averages them into a shape that describes neither, and the fit should be split by mode before any interval is taken from it. Failure is treated as detected immediately on occurrence, which is reasonable for a stopping failure and wrong for a hidden one - hidden failures need a failure-finding task and a different calculation entirely. Downtime is priced at a constant rate, so it does not capture a stop that cascades into other machines or one that lands during a bottleneck shift. The integration and search are numerical and converge tightly for beta between 0.5 and 10; results outside that range should be treated with suspicion, as should any answer taken from fewer than eight or ten recorded failure ages, because the confidence interval on beta from a small sample is wide enough to move the optimum substantially. Finally, the annual saving assumes the whole fleet shares one failure distribution and one duty, and comparing it against the cost of actually changing the policy - retraining, spares holding, revised schedules - is a separate decision this tool does not make.
Preventive Maintenance Interval Optimizer (Weibull Age Replacement) — free, with the formula and a worked example, at Textile School.