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Optimal preventive maintenance interval

Enter the component’s Weibull life and the costs of a planned replacement and an in-service failure to find the interval with the lowest cost per hour and the savings over current practice.

Pre-sizing and estimating tool: results do not replace design, calculations or reports signed by a licensed professional, nor verification against the applicable codes and standards.

Preliminary estimate
Component life (Weibull)

Use β and η from a Weibull fit (MTBF calculator) or manufacturer data. Preventive replacement only pays off with β > 1.

Costs
Fleet
Estimate β and η from failure data (paste from Excel)

Paste the operating hours to each failure of the component (one per line). The fit uses median-rank regression (Bernard’s approximation); at least 3 failures are needed, ideally 8 or more.

See what was read
—Optimal preventive replacement interval
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Cost per hour versus interval

Short intervals waste parts; long ones let units fail. The dashed line is run-to-failure.

Tap the chart to see the values.

Preventive and corrective share of the cost

Cost per hour split between planned replacements and in-service failures.

Tap the chart to see the values.

The report gathers this page's inputs, results, equations, assumptions and references, with fields for project, engineer and revision. In the print dialog, choose Save as PDF.

Method, assumptions and references

Equations used

R(t) = e^(−(t/η)^β); F(t) = 1 − R(t)
Cost per hour C(T) = [Cp · R(T) + Cf · F(T)] ÷ ∫₀ᵀ R(t) dt (age replacement)
Run to failure: C∞ = Cf ÷ MTTF, with MTTF = η · Γ(1 + 1/β)
Integral by Simpson’s rule (200 intervals); minimum by grid search and golden-section refinement
Annual savings = (C(current T) − C(T*)) × hours per year × components

Assumptions and limits

  • Replacement leaves the component as good as new; replacement time is negligible compared with the interval (the model minimizes cost, not downtime).
  • Evident failure with economic consequences. For safety or hidden failures, use RCM logic (SAE JA1011) instead of the cost optimum.
  • Validation: the optimal interval matches an independent numerical minimization in Python (SciPy) for the same parameters.

Standards and technical documents

  • IEC 60300-3-11:2009: Dependability management — Reliability centred maintenance
  • SAE JA1011:2009 e JA1012:2011: Evaluation criteria and guide for RCM processes
  • Barlow, R. E.; Proschan, F. Mathematical Theory of Reliability. Wiley, 1965
  • Jardine, A. K. S.; Tsang, A. H. C. Maintenance, Replacement, and Reliability, 2nd ed. CRC, 2013
  • IEC 61649:2008: Weibull analysis

U.S. regulations

  • OSHA 29 CFR 1910.119(j)(4): inspection and testing frequency per manufacturers’ recommendations, good engineering practice and prior experience
  • State boiler and pressure vessel laws (NBIC): mandatory inspection intervals override the economic optimum

Frequently asked questions

How do you set a preventive maintenance interval?

Use the cost per hour of an age replacement policy: add the cost of planned replacements and of failures that still happen and divide by the mean time in use. The interval with the lowest value is optimal.

Why does preventive replacement not help when β ≤ 1?

Because failure risk does not grow with age. Replacing a used part with a new one does not lower the chance of failure and can introduce assembly errors.

Does the optimal interval have to be exact?

No. The bottom of the curve is usually wide, so replacements can be aligned with planned shutdowns at little cost.