A cost curve falling to a single lowest point and rising again, showing that a preventive replacement age has a calculable cost optimum. — Hero

Optimal replacement interval explained

Every planned replacement interval on a maintenance schedule is an answer to a question, and most of those answers were given once and have never been re-examined. The question is precise and it has an arithmetic answer: at what age does replacing this item cost less than continuing to run it. An optimal replacement interval is the age at which total expected cost per operating hour is at its lowest, and it is calculated, not chosen.

This series has built everything the calculation needs. The four functions describe how failure risk behaves over age, the Weibull family puts numbers on that behaviour, censoring and plotting turn maintenance history into a defensible fit, and estimation methods bound the uncertainty around it. This page spends that groundwork on a single output a planner can put into a work order and a manager can defend in a budget review.

It also states the precondition honestly, because the analysis that produces an interval is the same analysis that tells you whether an interval was ever the right instrument. Where no replacement age improves anything, that finding is worth more than a number.

The Decision the Interval Is Actually Making

An age-based replacement policy is a bet placed at a specific age, and the model prices both sides of it.

Under an age-based policy, one of exactly two things happens to every unit installed. Either it survives to the chosen replacement age t_p and is replaced preventively at planned cost, or it fails first and is replaced correctively at the much larger cost unplanned failure carries. Nothing else can occur, so the two outcomes account for the whole population, and their probabilities are R(t_p) and F(t_p), which sum to one by definition.

That structure gives the cost model its shape. Expected cost per cycle is the preventive cost weighted by the probability of reaching t_p plus the failure cost weighted by the probability of not reaching it. Expected cycle length is built the same way, except that a unit which fails contributes its expected failure age rather than t_p. Dividing one by the other gives cost per unit of operating time, the quantity being minimised.

C (t_{p}) = \frac{C _{p m} R ( t _{p} ) + C _{c m} F ( t _{p} )}{t _{p} R ( t _{p} ) + M ( t _{p} ) F ( t _{p} )}

C (t_{p}) C(t_p) = total expected maintenance cost per operating hour if the item is replaced at age t_p t_{p} t_p = the preventive replacement age being tested, in operating hours C_{p m} C_pm = the full cost of one planned preventive replacement C_{c m} C_cm = the full cost of one unplanned failure replacement R (t_{p}) R(t_p) = probability the item survives to age t_p without failing F (t_{p}) F(t_p) = probability the item fails before age t_p, equal to 1 - R(t_p) M (t_{p}) M(t_p) = the expected age at failure, given that failure occurs before t_p

The denominator is where hand-built versions of this model usually go wrong. It is not t_p. A unit that fails at 6,000 hours under a 10,000-hour policy delivered 6,000 hours of service, and averaging cost over the interval intended rather than the life obtained understates the cost of failures. M(t_p) corrects it.

M (t_{p}) = \frac{1}{F ( t _{p} )} \int_{0}^{t_{p}} t f (t) d t

M (t_{p}) M(t_p) = mean age at failure among only those units that fail before t_p f (t) f(t) = the probability density function of the fitted life distribution \int_{0}^{t_{p}} t f (t) d t \int₀^(t_p) t \cdot f(t) dt = the probability-weighted sum of failure ages up to t_p \frac{1}{F ( t _{p} )} 1 / F(t_p) = the term that converts that weighted sum into a conditional average

A diagram showing the two possible outcomes for an item under an age-based replacement policy: surviving to the planned replacement age, or failing before it. — The two cycles an age-based policy can produce

Run the Precondition Before You Open the Model

One test runs before any cost data is gathered, and it decides whether the exercise has an answer at all.

A cost-minimising replacement age exists only when the hazard rate is increasing. Replacing an item early buys a reduction in failure probability over the next interval, and that purchase is only worth making if the item's instantaneous risk of failing is higher at an older age than at a younger one. When the hazard rate is flat, a replaced item is exactly as likely to fail in its first hour as the one it replaced was in its next hour, so the replacement buys nothing and costs C_pm. When the hazard rate is falling, the replacement is worse than nothing, because a new item is at higher risk than the aged one it displaced.

For a Weibull life distribution this test is a single parameter, which is why the fit does the work.

h (t) = \frac{β}{η} (\frac{t}{η})^{β - 1}

h (t) h(t) = hazard rate, the instantaneous failure risk at age t conditional on survival to t β β = the Weibull shape parameter, estimated from the fitted life data η η = the characteristic life, the age by which 63.2% of the population has failed (t / η)^{β - 1} (t / η)^(β-1) = the age-dependent term, which rises with t only when β > 1

The exponent carries the whole result. When β > 1 it is positive and h(t) rises with age, so an age-based task has something to act on (Abernethy 2006). When β = 1 it is zero, h(t) is the constant β/η, and age carries no information. When β < 1 it is negative and h(t) falls with age, so the newest units are the most at risk.

This is not a caveat at the edge of the method, and it is worth running the arithmetic to see how absolute it is. Holding the cost data and the characteristic life of this page's worked example constant and setting β to 0.8, then to 1.0, the total cost per unit time falls at every one of a thousand evaluation points from 200 hours out to 200,000 hours. There is no minimum to find, because the objective function does not have one.

A second condition sits alongside the first. An age limit can only prevent failures that would otherwise have happened after the limit. If most of the population has already failed before reaching the age at which the hazard rate becomes materially higher, moving the interval earlier removes very little risk while adding a great many planned interventions. Both conditions are properties of the fitted distribution, and both are answered before a single cost figure is needed.

A decision diagram showing that a cost-minimising replacement age exists only when the shape parameter is greater than one and the hazard rate rises with age. — The precondition decision gate

What the Cost Terms Must Actually Contain

The model is arithmetically simple, and almost every disappointing result traces back to two numbers filled in too quickly.

C_cm is not the part and the labour. An unplanned failure carries the replacement part, the unplanned labour at whatever rate applies at the hour it happened, expedited freight if the part was not on the shelf, the production lost while the plant was down, the collateral damage to adjacent components, and the safety and environmental exposure the failure created. On a production-limiting asset the lost production term routinely dominates every other term combined, and omitting it produces an interval far longer than the business would want.

C_pm is likewise more than the planned job card. It includes the planned outage, the residual life discarded when a still-serviceable item is thrown away, and the risk the intervention itself introduces, because opening a sound machine creates fresh opportunities for installation error and contamination ingress. That term keeps the model honest in the other direction: it is not a licence to replace everything early.

Build both figures from the same cost taxonomy and the same failure records the reliability fit came from, so the interval and the data behind it can be traced together at review (ISO 14224:2016). Two figures assembled on different bases produce a confidently wrong ratio, and the ratio drives the answer.

Two stacked bars comparing the composition of a planned replacement cost against an unplanned failure cost, with lost production dominating the failure cost. — Anatomy of the two cost terms

Working One Interval End to End

The method is seven steps, and every one produces something a reviewer can check.

Consider the rotor bearings on an induced draught fan handling process off-gas, an arrangement common to smelters, cement plants and boiler installations. Twelve units have run to failure or removal: nine failures at 9,800, 13,600, 17,400, 19,900, 21,300, 24,600, 26,500, 28,900 and 33,700 hours, and three suspensions at 16,200, 23,800 and 31,200 hours, each removed for reasons unrelated to bearing condition.

Fit the life data, including the suspensions, using median rank regression with adjusted ranks.
The three suspensions never failed but each proves survival to its removal age, and excluding them would bias the characteristic life downwards and make the recommended interval shorter than the evidence supports.
Read β from the fit and apply the precondition before going any further.
If β is at or below 1 the cost model has no interior minimum, and every hour spent gathering cost data would be spent producing a number the mathematics cannot support.
Assemble C_pm and C_cm on a consistent basis, including production loss, collateral damage and expedite cost in C_cm.
The ratio between these two figures, not their absolute size, sets where the optimum lands, so an inconsistency between them moves the answer directly.
Evaluate C(t_p) across the full plausible range of replacement ages rather than at a handful of candidate values.
A curve evaluated at three points can hide its own minimum, and the shape of the curve either side of the optimum is itself part of the answer.
Locate the minimum numerically, then confirm it is a genuine interior minimum with cost rising on both sides.
A minimum found at the edge of the scanned range is not a minimum, it is a sign the range was too narrow or the precondition was never met.
Compare the optimum against the practical maintenance windows the site actually has.
An interval that cannot be executed is not an answer, and the cost penalty for moving to a window that can be executed is usually far smaller than engineers expect.
Record the fit, the cost basis, the optimum, the chosen interval and the penalty accepted, as one traceable decision.
The interval will be challenged at the next budget review, and a recorded derivation converts that challenge from an argument into a document review.

Fitting the twelve units by median rank regression with Benard's approximation and Auth adjusted ranks gives β = 2.87 and η = 27,060 hours, with r = 0.998 and r² = 0.996 (Abernethy 2006). β is comfortably above 1, so the precondition is satisfied, and the value is consistent with the progressive fatigue spalling expected of a rolling element bearing under sustained load.

The cost basis: a planned bearing change inside a scheduled outage totals $18,000, covering the bearing set, the labour, the planned downtime slot and the discarded residual life. An unplanned seizure totals $210,000, covering the bearing, the journal damage a seizure causes, expedited freight, the extended unplanned outage and the production lost while the gas path is unavailable. The ratio is 11.7 to 1.

Evaluating the model across the full range gives a minimum at t_p = 9,552 hours, where total cost is $2.913 per operating hour and the probability of reaching that age without failing is 0.951. Running the same bearing to failure costs $8.707 per hour, because the failure cost is then carried by every cycle rather than by 4.9% of them.

A chart of total maintenance cost per operating hour against preventive replacement age, showing a falling preventive cost component, a rising failure cost component, and a total cost curve with its minimum at 9,552 hours. — The total cost curve and its minimum

Does the repair duration change the answer

The fuller form of the model carries the mean duration of each kind of intervention in the denominator, on the reasoning that a cycle occupies the operating time plus the time the item spent being replaced (Siswanto and Kurniati 2018).

C (t_{p}) = \frac{C _{p m} R ( t _{p} ) + C _{c m} F ( t _{p} )}{( t _{p} + T _{p m} ) R ( t _{p} ) + ( M ( t _{p} ) + T _{c m} ) F ( t _{p} )}

T_{p m} T_pm = mean duration of a preventive replacement, in the same time units as t_p T_{c m} T_cm = mean duration of a corrective replacement following failure

Running the worked example through the fuller form with a planned change taking 8 hours and an unplanned one taking 34 hours moves the optimum from 9,552 to 9,549 hours, a shift of 0.03%, and the cost rate by 0.10%. That is the general result, worth establishing by calculation rather than assumption: when repair durations are small relative to the interval, the simplified form is sufficient. Where durations are a material fraction of the interval, the fuller form earns its extra terms.

Why the Curve Is Flat Near Its Minimum

The most useful operational property of this model is that it is forgiving, and knowing exactly how forgiving turns a theoretical optimum into a schedulable task.

Total cost stays within 1% of its minimum from 8,585 to 10,601 operating hours, a window slightly over 2,000 hours wide inside which the choice of interval is, in cost terms, effectively free. The reason is structural: at the minimum, the falling preventive cost per unit time and the rising failure cost per unit time trade off at equal and opposite rates, so first-order changes cancel and only second-order effects remain.

The practical consequence is direct. If this fan sits on an annual planned outage cycle at 8,760 hours, replacing the bearing at that outage rather than at the mathematical optimum costs 0.66% more, which is $170 per year on a task that saves $50,587 per year against running to failure. The interval to put in the plan is 8,760 hours, and the derivation to record is the one showing the optimum at 9,552 hours and the penalty accepted for moving to an executable window. Reporting a near-optimal plateau rather than a single number is both the more useful output and the more defensible one.

How the Answer Moves With the Cost Ratio and With Beta

Two inputs move the answer, and knowing how each one moves it is what lets an engineer sense-check a result rather than accept it.

The cost ratio pulls the interval in hard. Holding the same fit and varying only the ratio of failure cost to preventive cost, the optimum moves from 22,300 hours at a ratio of 2 to 1, to 12,475 hours at 6 to 1, to 9,552 hours at the worked example's 11.7 to 1, to 6,066 hours at 40 to 1. Against the characteristic life, that is a move from 0.82η down to 0.22η. A high ratio makes each avoided failure worth many preventive replacements, so the model buys risk reduction earlier and more often.

The shape parameter behaves less obviously. Holding the cost ratio at 11.7 and varying β alone, the optimum falls from 15,815 hours at β = 1.2 to 8,351 hours at β = 2.0, then rises again to 11,387 hours at β = 4.0 and 20,749 hours at β = 20. The optimum is not monotonic in β. A very high β makes the failure age nearly deterministic, so the best policy is to replace just short of the characteristic life and the interval migrates back towards η. A β only slightly above 1 offers so little wear-out signal that replacing early mostly discards useful life. The tightest intervals sit between the two.

A second reading of that sweep matters more than the interval itself, because existence and value are different questions. At β = 1.2 the model returns a finite optimum, but acting on it saves only 7.3% against running to failure. At β = 2.0 the same comparison saves 50%, and at β = 4.0 it saves 75%. The precondition tells you whether an answer exists, and the saving tells you whether it is worth executing.

Two charts, one showing the optimal replacement age shortening as the failure-to-preventive cost ratio rises, and one showing that the total cost curve has no minimum when the shape parameter is at or below one. — Sensitivity of the optimum to the cost ratio and to beta

When the Right Answer Is Not an Interval

A model that returns no interval has not failed. It has told you the failure mechanism does not respond to age, which is a finding, and acting on it correctly is worth more than any number.

Where β is at or below 1, or where the saving against run-to-failure is too small to justify the intervention, three alternatives each address the situation the arithmetic described. Condition monitoring acts on evidence of the failure developing rather than on elapsed age, which suits failures that arrive without regard to age but do give warning. Redesign or specification change acts on the mechanism itself, which suits a decreasing hazard pointing to installation, commissioning or manufacturing quality as the driver. Run to failure with the spare positioned and the response planned suits a tolerable consequence that no age-based or condition-based task would improve (SAE JA1012).

Deciding not to schedule a replacement is a legitimate engineering decision, and it should be documented as carefully as a decision to schedule one: the fit, the β, the precondition result, the alternative selected and the reason. An undocumented absence of a task looks identical to an oversight at audit, while a documented one is evidence the question was asked and answered.

This is also where the intuition that shortening an interval always reduces risk comes apart. Shortening an interval reduces exposure to the aged part of the distribution, which only helps when the aged part is where the risk lives. When the hazard rate is flat or falling, a shorter interval adds interventions and adds exposure to installation-related early failures while buying no reduction in the underlying risk. The interval is a lever connected to a specific mechanism, and pulling it harder only works if the lever is connected.

Closing the Loop: An Interval Is a Reliability Statement

This series opened by insisting that a reliability statement is only actionable when it names a required function, the conditions, a time interval and a probability. The interval just calculated is exactly that statement, written in the form a planner can execute.

Replacing that fan bearing at 8,760 operating hours is the claim that the bearing will maintain the fan's required function of sustaining rated gas flow at rated pressure, under continuous off-gas duty at the stated load and temperature, for 8,760 operating hours, with a probability of 0.961. All four elements are present, each traceable to the fit and the cost basis behind it, and the number carries its own justification into whatever review it faces. That is the difference between an interval that is defended and an interval that is merely inherited.

The mathematics that produced it is not exotic. It is a general reliability function, a life distribution fitted honestly to real maintenance history including its suspensions, a hazard rate whose shape decides whether age is a lever at all, and a cost comparison priced on both sides (MIL-HDBK-338B 1998). Each of those is a page in this series, and this one is where they combine into a decision. An interval derived this way can be challenged on its inputs, and it cannot be dismissed as a guess.

If you would like a replacement interval derived, reviewed or defended for a specific critical asset, we would be glad to work through it with you.

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Optimal Replacement Interval: Turning Weibull Output into a Cost-Justified Task Frequency