Spare Parts Optimization Calculator
Spare holdings are a bet: too few and you extend downtime, too many and you tie up capital in parts that may never be used. When failures are random and independent, demand over a lead time follows a Poisson distribution, which turns the decision into an explicit trade between stock level and service level.
Formula
Worked example
12 identical pumps, each with MTBF = 8760 hours. Spare lead time is 90 days (2160 hours).
- λ = 12 × 2160 / 8760 = 25920 / 8760 = 2.959
- P(0) = e^(−2.959) = 0.0519
- P(≤1) = 0.0519 × (1 + 2.959) = 0.2055
- P(≤2) = 0.2055 + e^(−2.959)·2.959²/2 = 0.2055 + 0.2273 = 0.4328
- P(≤5) = 0.9166; P(≤6) = 0.9666; P(≤7) = 0.9877
Six spares give a 96.7% service level; seven give 98.8%. The seventh spare buys 2.1 percentage points — whether that is worth its purchase and carrying cost depends entirely on the cost of an hour of downtime. This is the calculation that turns a stocking argument into an arithmetic one.
Common mistakes
- Using the Poisson model for parts with a strong wear-out pattern. Poisson assumes random independent failures; wear-out produces clustered demand that this model badly underestimates.
- Ignoring lead time variability. A supplier whose 90-day lead time is really 90 ± 40 days needs additional safety stock beyond the Poisson result.
- Optimising each part in isolation when the budget is shared. Across a portfolio, marginal-service-level-per-dollar allocation beats per-part targets.
- Setting a blanket service level for all parts regardless of criticality or downtime cost.
Related tools
About Reliability Pro
Spare Parts Optimizer is part of a suite of 126 reliability, maintenance and quality engineering tools covering life data analysis, accelerated testing, system reliability, FMEA and root cause, SPC, and design for reliability. It runs in the browser and as native iOS and Android apps, so the same calculation is available at a desk or in front of the asset.