Advanced Category: Maintenance Planning & Strategy • ~6 min read

Preventive Maintenance Interval Optimizer

Find the mathematically optimum PM interval that minimizes total maintenance cost per unit time.

Q:

When is Preventive Maintenance (PM) mathematically justified?

Preventive maintenance is mathematically viable only if two conditions hold simultaneously: 1) The component exhibits an increasing failure rate (Weibull shape β > 1), and 2) The cost of an unplanned failure exceeds the planned PM cost (C_f > C_p). If β ≤ 1, time-based PM increases total plant costs.

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Weibull & Cost Parameters

Must be > 1 for age-based replacement (β > 1 indicates mechanical wear-out).
hours
$
$
Optimal PM Interval
776 hours
Lowest cost point
Optimal Cost Rate
$1.73 /hr
Expected hourly cost
Cost Savings vs RTF
52.7%
RTF: $3.66/hr
Expected Cost Rate Curve C(T)Min = 776h
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The Age Replacement Optimization Model

Under the classical age replacement policy, equipment is replaced preventively at operating age T_p or upon failure, whichever comes first. By balancing the low cost of planned replacement against the high cost of emergency downtime, this model computes the exact inflection point that minimizes total life-cycle cost rate.

Formulas

Expected Cost Rate: C(T_p) = [ C_p · R(T_p) + C_f · (1 - R(T_p)) ] / ∫₀T_p R(t) dt
Run-to-Failure Rate: CRTF = C_f / MTTF
Cost Savings: [ (CRTF - C(T_p^*)) / CRTF ] × 100%

Frequently Asked Questions

What happens if failure cost equals planned cost (Cf = Cp)?

If Cf = Cp, the optimal policy is run-to-failure (Tp* = ∞). There is no financial penalty for failing in service, so replacing components before they fail prematurely discards remaining useful life.

Engineering & Regulatory Notice: The calculations, projections, and failure models provided in this suite are for engineering estimation and educational purposes. Verify all calculations and assumptions independently before making operational, equipment replacement, safety-critical, or regulatory compliance decisions.

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