BIZOMAT 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. |
What each input means
Open an input to see what it is, its units, the worked example and the common mistakes, with a small diagram.
What this means: Weibull shape β
- What it is
- The Weibull shape parameter fitted to the part’s failure records. It describes whether failures come early, at random or from wear-out as the part ages.
- Units and format
- Enter a plain number with no unit. The form accepts 0.2 to 10. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 2.5 as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Get it right
- Fit it from your own failure data; planned replacement only reduces failures when it shows wear-out.
What this means: Characteristic life η
- What it is
- The Weibull characteristic life fitted to the failure records: the operating age by which most parts (about 63 percent) have failed.
- Units and format
- Enter a number in h: hours. The form accepts 0.000001 h to 100000000 h. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 8000 h as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Every interval and life the tool gives scales in proportion to it.
- Get it right
- Use operating hours, not calendar hours, matching the hours per year.
What this means: Cost of a planned replacement
- What it is
- The full cost of replacing the part on schedule: the part, labour and any planned downtime.
- Units and format
- Enter a number in $: an amount of money; use one currency for every money input. The form accepts $0 to $100000000. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with $500 as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- A higher planned cost lengthens the lowest-cost interval and reduces the saving.
What this means: Cost of a failure (repair, downtime, scrap)
- What it is
- The full cost when the part fails in service: repair, unplanned downtime, scrap, secondary damage and expediting.
- Units and format
- Enter a number in $: an amount of money; use one currency for every money input. The form accepts $0 or more. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with $4000 as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- A higher failure cost makes planned replacement more attractive and, with wear-out, shortens the lowest-cost interval.
- Get it right
- Include lost production; leaving it out makes run to failure look cheaper than it is.
What this means: Reliability wanted over the interval
- What it is
- The chance you want that a part survives the whole replacement interval without failing. The tool finds the interval that gives it.
- Units and format
- Enter a number in %: percent: the number is read as a percentage, so type 8 for 8 %. The form accepts 1% to 99.999%. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 90% as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- A higher reliability shortens the interval for it.
- Get it right
- Type the percentage itself, not a fraction.
What this means: Interval to use (0 = lowest cost)
- What it is
- The replacement interval, in operating hours, you intend to use. Type 0 to use the interval with the lowest cost per hour.
- Units and format
- Enter a number in h: hours. The form accepts 0 h to 100000000 h. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 0 h as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Get it right
- Use operating hours, the same basis as the characteristic life.
What this means: Operating hours per year
- What it is
- The hours per year the equipment actually runs, used to turn the cost per hour into a yearly saving.
- Units and format
- Enter a number in h: hours. The form accepts 0.000001 h to 8784 h. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 4000 h as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- The yearly saving rises in proportion.
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.
Learn the Concepts Behind This Tool
These BIZOMAT Library readings explain the method this tool uses, with worked examples.
Method explained here · BIZOMAT original book
From model to decision: what to fix first
In Breakdowns and Buffers by the Numbers
What it covers: Pricing a minute of downtime, ranking improvements by the output they add on a buffered line, wear-out parts on a schedule, checking predictions against results, and the Pareto chart that misled.
BIZOMAT 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