Mean Time Between Failures (MTBF) is the expected operating time between failures for a repairable system. It is not a guarantee — it is a statistical expectation derived from failure rate data. For automation cells, MTBF drives maintenance intervals, spare parts inventory, and whether a given uptime target is physically achievable.
Core definitions
| Term | Symbol | Relationship |
|---|---|---|
| Failure rate | λ | Failures per operating hour |
| Mean time between failures | MTBF | 1 / λ |
| Mean time to repair | MTTR | Average hours from failure to restored operation |
| Availability | A | MTBF / (MTBF + MTTR) |
A component with MTBF = 80,000h has λ = 0.0000125 failures/hour. Over one year (8,760h), the expected number of failures is 0.0875 — approximately one failure every 9.1 years under stated conditions.
Series system: no redundancy
In a series configuration, any single component failure stops the cell. The system failure rate is the sum of all component failure rates:
λ_system = λ₁ + λ₂ + λ₃ + ... + λₙ
MTBF_system = 1 / λ_systemExample cell:
| Component | MTBF (h) | λ (failures/h) |
|---|---|---|
| PLC (S7-1500) | 200,000 | 0.00000500 |
| HMI panel | 50,000 | 0.00002000 |
| Servo drive | 80,000 | 0.00001250 |
| Servo motor | 60,000 | 0.00001670 |
| Safety relay | 150,000 | 0.00000670 |
| System | — | 0.00006090 |
MTBF_system = 1 / 0.0000609 ≈ 16,420 hours (~1.87 years)The HMI and servo motor dominate. They are the weakest links — the most cost-effective targets for replacement or redundancy.
Redundant pair: active standby
For a component with an active standby, both units must fail for the system to fail. The effective failure rate of the pair:
λ_pair = (λ₁ × λ₂) / (λ₁ + λ₂)For two identical servo drives (λ = 0.0000125 each):
λ_pair = (0.0000125 × 0.0000125) / (0.0000125 + 0.0000125)
= 0.000000000015625 / 0.000025
= 0.00000000625 failures/hThat is a ~2,000× reduction in the drive's failure rate contribution. Redundancy is only worth its cost when the component is a dominant contributor to λ_system and the standby can take over without manual intervention.
Availability calculation
For the series cell above, with MTTR = 4 hours:
A = MTBF / (MTBF + MTTR)
= 16,420 / (16,420 + 4)
≈ 99.976%To determine the MTBF required to hit a specific availability target, rearrange:
MTBF = (A × MTTR) / (1 - A)Target: 99.99% availability at MTTR = 4h:
MTBF = (0.9999 × 4) / (1 - 0.9999)
= 3.9996 / 0.0001
= 39,996 hours (~4.6 years)The current cell MTBF of 16,420h falls short. To reach 99.99%, either reduce MTTR (faster diagnosis, local spares, trained technicians) or raise MTBF by replacing the two weakest components.
Effect of MTTR on availability
MTBF improvements are expensive — they require better components or redundancy. MTTR improvements are operational — spares on-site, documented procedures, trained staff.
| MTBF (h) | MTTR = 8h | MTTR = 4h | MTTR = 1h |
|---|---|---|---|
| 16,420 | 99.951% | 99.976% | 99.994% |
| 39,996 | 99.980% | 99.990% | 99.997% |
| 100,000 | 99.992% | 99.996% | 99.999% |
Halving MTTR delivers a larger availability gain than doubling MTBF at these ranges. For most automation cells, the fastest path to higher availability is reducing repair time, not purchasing better hardware.
MTBF data sources
| Source | Basis | Use case |
|---|---|---|
| Manufacturer datasheets | Vendor testing, controlled conditions | First estimate |
| MIL-HDBK-217F | US military component-level models | Conservative baseline |
| IEC 61709 | International component failure rates | Reference standard |
| CMMS field data | Your actual operating conditions | Most accurate |
Manufacturer MTBF figures assume 25°C and low vibration. Apply a derating factor of 0.5–0.7 for high-temperature enclosures or high-vibration environments. A drive rated for 80,000h in a lab may deliver 40,000–56,000h on a stamping press.
Build a CMMS-backed failure log from day one. Field data from your specific environment will diverge from handbook values within two to three years — and that divergence is the most useful reliability signal you have.