Corrective Maintenance Downtime and a PM Interval Trade
A plant runs 14 identical packaging lines under a service contract that specifies corrective maintenance mean downtime no greater than 2.0 hours and a 90th-percentile downtime no greater than 4.0 hours per repair. Last quarter the CMMS logged 40 corrective work orders and 96 hours of line downtime. Twenty-two were mechanical repairs averaging 1.5 hours of active work; 12 required a spare part that was not stocked, so the line sat idle an average of 3.5 hours before the same 1.5-hour repair began; and 6 were cleared by a control reset averaging 25 minutes. The maintenance manager reports the mean as 1.35 hours by excluding waiting time, and the customer's auditor has rejected that calculation. Separately, preventive maintenance is performed every 250 hours at a cost of $600 per event. The line's failure history fits a Weibull with beta = 2.1 and eta = 1,200 hours, and each unplanned failure costs about $6,300 in repair and lost production. To reduce cost, the manager proposes moving PM to a 400-hour interval.
1. Recompute the mean and 90th-percentile downtime with and without logistics delay. Which figure is contractual, and what is the real root cause of any noncompliance?
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Recompute from the work orders, not from the summary.
- Total downtime = 22 x 1.5 + 12 x (3.5 + 1.5) + 6 x 0.42 = 33 + 60 + 2.5 = 95.5 hours, giving a mean of 2.39 hours against a 2.0-hour limit.
- Sorted, the 12 part-wait repairs all sit near 5.0 hours, and the 90th percentile lands at roughly 5.0 hours against a 4.0-hour limit.
- Excluding logistics delay: 53.5 hours / 40 = 1.34 hours, with a 90th percentile near 2.6 hours. Both pass.
Which number is contractual? The contract says downtime per repair, and line downtime is elapsed time the asset is unavailable, so the waiting hours count. The auditor is right, and the plant's own availability calculation uses total downtime, so excluding it would make the two reports internally inconsistent.
The maintenance manager is not wrong about maintainability engineering, though. MTTR as a design metric measures active repair time and legitimately excludes administrative and logistics delay, because those are supply problems rather than design problems. Report both, labelled clearly: active MTTR of 1.34 h for maintainability improvement, maintenance downtime of 2.39 h for the contract.
The root cause is visible in the decomposition: 12 of 40 events (30%) produced 42 of 96 downtime hours (44%). Stocking those parts, or kitting them at the line, removes essentially the whole gap and brings both metrics inside their limits with margin. Shortening the repairs themselves would barely move the number.
2. Evaluate the proposal to move PM from 250 to 400 hours using the Weibull and cost data. Show the calculation and state a recommendation.
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Do the arithmetic before rejecting the proposal. With beta = 2.1 the hazard is increasing, but that alone does not justify the current interval. What matters is the cost ratio between a PM event ($600) and a failure ($6,300).
Use the age-replacement cost rate CR(T) = [Cp + Cf x F(T)] / integral of R(t) dt, with F(T) = 1 - exp[-(T/1200)^2.1].
| T (h) | F(T) | Mean cycle (h) | Cost/cycle | Cost/hour |
|---|---|---|---|---|
| 250 | 0.036 | ~246 | $827 | $3.36 |
| 400 | 0.095 | ~386 | $1,199 | $3.11 |
| 500 | 0.147 | ~472 | $1,526 | $3.23 |
| 600 | 0.208 | ~552 | $1,910 | $3.46 |
The closed form T* = eta x [Cp/(Cf(beta-1))]^(1/beta) = 1,200 x [600/(6,300 x 1.1)]^0.476 = about 375 hours, consistent with the table. The proposal is defensible: 400 hours is near the optimum and roughly 7-8% cheaper per operating hour than 250 hours.
Two observations. At 250 hours the probability of failure before PM is only 3.6%, so PM is being performed far more often than the failure economics require; the interval was probably set by habit rather than analysis. And the conditional probability of failure in the extended 150-hour window is [R(250) - R(400)]/R(250) = about 6%, which is acceptable only if those failures are detectable and benign.
3. What conditions or evidence would overturn the PM interval recommendation, and how would you verify the chosen interval in practice?
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What would reverse the conclusion:
- Severity of failure. The $6,300 figure prices repair and lost production only. If an unplanned failure can injure someone, release material, or scrap a batch, Cf is far higher, the optimum moves sharply toward shorter intervals, and eventually toward redesign. The cost model is valid only while failures are economically bounded.
- Restoration quality. Age replacement assumes PM returns the unit to as-good-as-new. If PM is imperfect, the effective beta is lower than 2.1 and the modeled benefit of running longer evaporates; the post-PM hazard can sit above the Weibull prediction for several cycles.
- Maintenance-induced failures. Every PM event is an opportunity to introduce a defect - missed fasteners, wrong lubricant, disturbed connections. If the PM-induced failure rate is not negligible, more frequent PM is worse than the model says.
- Heterogeneity. Pooling data from 14 lines can hide a subpopulation with a wear-out cliff. A mixture would show a steeper late hazard, and the pooled beta understates risk for the affected units. Check lines individually before applying one interval fleet-wide.
Verification: document the model and its assumptions, then run a controlled trial - half the lines at 400 hours for one quarter, half unchanged - and compare failures per operating hour and total cost per hour. Predefine the decision criterion and the stopping rule so the result is not reinterpreted after the fact.
Absent any of the reversing conditions, 400 hours is the better economic choice, and the decision record should say so explicitly with the cost ratio that drove it.