KEVOS® MAKEOS · Maintenance
Preventive Replacement Interval Plan (Weibull Reliability and Cost per Hour)
Should I replace this part on a schedule or run it to failure, what interval minimises cost, and what interval gives the reliability I want?
Planned replacement of a wearing component with Weibull failure data: reliability, hazard rate, B10 life and mean life (as the Weibull reliability tool), the interval that gives your target reliability, the age-replacement cost per operating hour (planned cost if it survives, failure cost if not, over the mean running time in the cycle), the interval with the lowest cost, run-to-failure cost per hour, and the yearly saving at the interval used.
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How this tool works
Inputs, units and where to find them
| Input | Unit | Typical source |
|---|---|---|
| Weibull shape β | Weibull shape from failure data (above 1 means wear-out). | |
| Characteristic life η | h | Weibull characteristic life (63.2% have failed by then). |
| Cost of a planned replacement | $ | Total cost of replacing the part as planned. |
| Cost of a failure (repair, downtime, scrap) | $ | Total cost when it fails in service, including downtime and damage. |
| Reliability wanted over the interval | % | Probability of surviving the interval you want. |
| Interval to use (0 = lowest cost) | h | Interval you intend to use; 0 uses the lowest-cost interval. |
| Operating hours per year | h | Operating hours per year. |
Method and formulas
- R(t) = exp[−(t ÷ η)^β]; B10 = η (−ln 0.9)^(1/β); mean life = η Γ(1 + 1/β) (as the Weibull reliability tool).
- Interval for a target R: t = η (−ln R)^(1/β).
- Cost per hour C(T) = [C_p R(T) + C_f (1 − R(T))] ÷ ∫₀ᵀ R dt; run to failure = C_f ÷ mean life; the lowest C(T) is found numerically.
- Example: replacing every 3,140.4 h costs 0.269 $/h against 0.56353 $/h run to failure; 90% reliability needs 3,252.1 h.
Limitations
- Planned replacement restores the part to as-new; no inspection or repair effects.
- Weibull parameters must come from failures of the same part in the same duty.
- Example values are illustrations. Replace them with values for your actual task.
Typical uses
- Set a replacement interval for bearings, belts, seals or tools
- Decide whether planned replacement is worth it
- Show the cost of a chosen interval
Who it is for
- Maintenance engineers
- Reliability engineers
- Maintenance planners
What you get
- Reliability, B10 life and mean life
- Interval for the reliability target and lowest-cost interval
- Cost per hour, run-to-failure cost and yearly saving
Interpreting the result
- Planned replacement only pays when β is above 1 and a failure costs clearly more than a planned change.
References
- Machinery’s Handbook, 29th edition (Industrial Press, 2012), printed pp. 128–133. Topic reference; no source tables or recommended settings are reproduced.
WORKOS tools are screening and decision-support utilities. Verify important financial, engineering, safety, legal or operational decisions using appropriate professional standards and your organisation’s approved data.
Open the Preventive Replacement Interval Plan (Weibull Reliability and Cost per Hour) calculator
