ModelRisk example · Manufacturing
Spare Parts and Maintenance Strategy
Run to failure or replace on a schedule — and how many spare kits to hold? Bearing wear-out on eight spindles.
From the model’s saved run
Which strategy, and how many spare kits?
Mean two-year cost (US$k) for the eight spindles: replacements, failures, waits for a kit and holding the spare kits.
| Strategy | Textbook long run | 0 kits simulated | 1 kit simulated | 2 kits simulated | 3 kits simulated |
|---|---|---|---|---|---|
| 1 Run to failure | 324.6 | 697.0 | 541.0 | 376.5 | 291.8 |
| 2 Replace at 2,500 hours | 205.9 | 280.8 | 201.8 | 186.0 | 185.3 |
| 3 Replace at 3,500 hours | 194.6 | 327.1 | 220.5 | 183.4 | 177.3 |
| 4 Replace at 5,000 hours | 214.5 | 428.4 | 292.2 | 216.4 | 194.3 |
Textbook: the renewal-reward cost once the bearings' ages have spread out, with a kit always on the shelf. Green: the lowest mean cost.
S-curve (cumulative probability) of Run to failure, Early replacement, Mid-life replacement, Late replacement from the saved simulation.
The business question
A machine shop runs eight machining centres on two shifts, about 6,000 hours a year. Their spindle bearings wear out: half last beyond about 6,000 hours, but a few fail before 2,000. A bearing that fails in service damages the spindle and stops the machine for twelve hours — about US$21,000 in all.
Replaced on schedule, in a weekend maintenance window, it costs US$4,000: a kit and the labour. All eight spindles were rebuilt with new bearings in this year's shutdown.
The maintenance manager must choose a strategy for the next two years — run the bearings to failure, or replace each at a fixed age — and decide how many spare kits to keep on the shelf: a kit ordered after a failure takes eight weeks to arrive, and a failure that finds no kit waits 36 hours for an expedited one.
- Who it is for
- Maintenance and reliability engineers, plant managers, spare-parts and MRO planners.
- Main outputs
- The two-year cost of each strategy with 0 to 3 spare kits, beside the textbook long-run cost · failures, planned replacements and kits used · the chance that a failure finds no kit on the shelf · failures by quarter.
How the model works
Three sheets after this one, read from left to right. Each one feeds the next.
Inputs sheet
Operating hours a year; the bearing's Weibull shape and scale; three replacement ages to compare with running to failure. The cost of a kit, the labour, the damage and the downtime of a failure; the spare kits held, their resupply time and the cost of holding one.
Spindles sheet
VoseWeibull draws the lives of ten successive bearings on each spindle. For each strategy a bearing comes out at its replacement age or when it fails, and the next one starts: a renewal process. Failures and planned replacements in the two years are counted, and for each failure the failures in the resupply time up to it show whether a spare kit was left.
The textbook long-run cost is calculated alongside.
Results sheet
The mean two-year cost of each strategy with 0 to 3 spare kits. Failures, kits used and the chance of running out of kits; failures by quarter.
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The results in detail
Failures, kits and the risk of running out
In the two years, for the eight spindles.
Line chart: Run to failure, Early replacement, Mid-life replacement, Late replacement.
| Strategy | Failures mean | Failures 90th pct | Planned mean | Kits used 90th pct | No kit chance* |
|---|---|---|---|---|---|
| 1 Run to failure | 12.2 | 15 | – | 15 | 96% |
| 2 Replace at 2,500 hours | 2.8 | 5 | 30.9 | 35 | 5% |
| 3 Replace at 3,500 hours | 4.3 | 7 | 20.7 | 26 | 17% |
| 4 Replace at 5,000 hours | 6.7 | 10 | 11.0 | 19 | 46% |
* The chance that at least one failure finds no kit, with the spare kits held on the Inputs sheet.
| Chance that a failure finds no kit | 0 kits | 1 kit | 2 kits | 3 kits |
|---|---|---|---|---|
| 1 Run to failure | 100% | 100% | 96% | 50% |
| 2 Replace at 2,500 hours | 94% | 37% | 5% | 0% |
| 3 Replace at 3,500 hours | 99% | 67% | 17% | 2% |
| 4 Replace at 5,000 hours | 100% | 91% | 46% | 10% |
How to read these results
Running the bearings to failure costs the most, whatever the spare kits: about 12 failures in two years at US$21,000 each. Even with three spare kits, half of the futures have a failure that finds the shelf empty.
All the bearings were new after the shutdown, so they wear out together: failures climb through the first year and stay high in the second (see the chart).
- Replacing each bearing at 3,500 hours, with three spare kits, costs the least: about US$177,000 over two years, against about US$292,000 for running to failure with the same kits. About 21 bearings are replaced in weekend windows and about 4 fail in service. At 2,500 hours fewer fail (about 3) but about 31 are replaced. At 5,000 hours about 7 fail first.
- The spare kits matter as much as the replacement age. A kit held for two years costs US$1,250. A failure that waits for one costs US$36,000. With two kits, replacing at 3,500 hours runs out in 17% of futures and at 2,500 hours in 5%, and both cost about US$183,000–186,000: the earlier the replacements, the fewer the failures that draw on the shelf.
- The textbook formula — the renewal-reward cost once the bearings' ages have spread out — ranks the strategies the same way and puts the best age near 3,500 hours. It cannot see the spare kits, nor the years just after a shutdown, when young bearings fail less often than in the long run and then fail together.
- For your own machines, fit the Weibull to your failure records (MF-03 shows how, with parameter uncertainty), then try replacement ages around the best one and the spare kits you can hold.
See it move. Open the workbook, change an input on the Inputs sheet and run the simulation: every table and chart on this page updates in Excel.
Inputs and assumptions
Every assumption in the model, with its uncertainty and where it comes from. The figures are illustrative.
The spindles 1 input
Eight machining centres, all with new spindle bearings after this year's shutdown.
| Assumption | Unit | Value |
|---|---|---|
| Operating hours a yearTwo shifts, five days a week. | hours | 6,000 |
Bearing life 2 inputs
A Weibull distribution of operating hours to failure.
| Assumption | Unit | Value |
|---|---|---|
| Weibull shapeAbove 1: the bearings wear out — the older, the likelier to fail. | 2.50 | |
| Weibull scale63% of bearings fail by this age; fitted to the shop's records. | hours | 7,000 |
Strategies 3 inputs
Replace each bearing at this age, or when it fails. Running to failure is compared with all three.
| Assumption | Unit | Value |
|---|---|---|
| Early replacement at | hours | 2,500 |
| Mid-life replacement at | hours | 3,500 |
| Late replacement at | hours | 5,000 |
Costs 6 inputs
| Assumption | Unit | Value |
|---|---|---|
| Bearing kit | US$k | 2.5 |
| Labour to replace the bearingsPlanned or not. | US$k | 1.5 |
| A failure: damage to the shaft and housing | US$k | 5.0 |
| A failure: hours the machine standsStrip, rebuild and test the spindle. | hours | 12 |
| Contribution lost per hour a machine stands | US$k | 1.00 |
| No kit on the shelf: extra hours waitingFor an expedited kit from the maker. | hours | 36 |
Spare kits 3 inputs
| Assumption | Unit | Value |
|---|---|---|
| Spare kits heldThe tables on the Results sheet show 0 to 3. | kits | 2 |
| Resupply time of a kitOrdered as soon as a failure uses one. | weeks | 8 |
| Cost of holding a kitCapital, storage and obsolescence. | per year | 25% |
Units and conventions
- Hours
- Operating hours since this year's shutdown: 6,000 a year, 12,000 in the two years.
- Money
- US$ thousand (US$k), not discounted over the two years.
- Strategy
- Replace each bearing when it reaches the replacement age or when it fails, whichever comes first; running to failure has no replacement age.
- Stock-out
- A failure that finds no spare kit on the shelf. Every kit used by a failure is reordered at once and arrives one resupply time later.
Key assumptions and limitations
- Bearings
- Lives independent and Weibull; a new bearing is as good as new; all eight start new after the shutdown.
- Planned work
- Done in weekend maintenance windows: no lost production; its kits are ordered ahead and never wait.
- Spare kits
- One kit reordered for each failure; a failure that finds no kit waits for an expedited one.
- Data
- The shop and all figures are invented for this example.
How to use it
- Enter your operating hours and the bearing's Weibull life — fitted to your failure records (see MF-03).
- Enter the costs of a planned replacement and of a failure, and the resupply time of a kit.
- Run a simulation from the ModelRisk ribbon (10,000 samples with a fixed seed are set).
- Read the best strategy and number of kits from the cost table; then try other replacement ages.
Simulate this model with ModelRisk’s classic engine: it does not run in Turbo mode yet.
ModelRisk functions used
- VoseWeibull
- The life of each bearing, in operating hours.
- VoseWeibullObject, VoseProb
- The chance that a bearing fails before its replacement age, for the long-run formula.
- VoseOutput
- Failures, kits, stock-outs and costs, recorded at every sample.
- VoseSimMean, VoseSimPercentile, VoseSimProbability
- Statistics of the simulation on the Results sheet.
Further reading
- Monte Carlo simulation explained (Vose guide)
- Jardine, A. K. S. & Tsang, A. H. C. (2013). Maintenance, Replacement, and Reliability, 2nd edition. CRC Press.
- Barlow, R. E. & Proschan, F. (1965). Mathematical Theory of Reliability. Wiley — age replacement.
- Sherbrooke, C. C. (2004). Optimal Inventory Modeling of Systems, 2nd edition. Kluwer.
Download the workbook and run it yourself. It opens in Excel with ModelRisk. The free trial runs every example model.
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An illustrative example: the figures are invented for teaching and do not describe a real company, project or asset. Last updated 9 October 2026.