# What is the MTBF?

MTBF calculator: mean time between failures from total operating hours and failures, with the failure rate, reliability over a mission time, and chi-square confidence bounds, including the zero-failure case.

- Page: https://www.acalculator.org/physics/mtbf-calculator
- JSON spec: https://www.acalculator.org/physics/mtbf-calculator.json
- Version: 2d7231ddd8e1

## Default answer

Example with the default inputs (Number of units 1, Operating time per unit 800 h, Failures 2, Confidence level 90%, Mission time 100 h): MTBF: 400 hours, from 800 hours of operation.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| n | Number of units | How many units ran (each for the time below). |
| t | Operating time per unit | How long each unit ran. |
| r | Failures | The number of failures seen in that time; 0 if none. |
| c | Confidence level | The confidence for the MTBF bounds, for example 90%. |
| m | Mission time | Optional: a time to find the chance of running that long without a failure. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| result | MTBF | Total operating time ÷ failures, in hours; with no failures, the lower confidence bound. |
| mtbf | MTBF (hours) | Total operating time ÷ failures. None when there are no failures. |
| totalHours | Total operating time (hours) | Units × operating time per unit. |
| rate | Failure rate (per hour) | λ = failures ÷ total operating time = 1 ÷ MTBF. |
| perMillion | Failures per million hours | λ × 10⁶. |
| fit | Failures per billion hours (FIT) | λ × 10⁹, the FIT rate. |
| lower | Lower confidence bound (hours) | One-sided: the MTBF is at least this at the chosen confidence, 2T ÷ χ²(c; 2r + 2). |
| intervalLow | Two-sided interval, low (hours) | 2T ÷ χ²(1 − α/2; 2r + 2), with α = 1 − confidence. |
| intervalHigh | Two-sided interval, high (hours) | 2T ÷ χ²(α/2; 2r). Needs at least one failure. |
| reliability | Reliability over the mission (%) | R = e^(−mission ÷ MTBF), as a percent: the chance of no failure in the mission time. |
| failureChance | Chance of a failure in the mission (%) | The chance of at least one failure in the mission time, 1 − R, as a percent. |

## Method

T = units × time per unit; MTBF = T ÷ r; λ = r ÷ T; lower bound 2T ÷ χ²(c; 2r + 2); two-sided interval [2T ÷ χ²(1 − α/2; 2r + 2), 2T ÷ χ²(α/2; 2r)] with α = 1 − c; R(t) = e^(−t ÷ MTBF).

## Assumptions

- A constant failure rate (the exponential model): failures arrive at random, and failed units are repaired or replaced at once.
- The test runs for a fixed time (Type I censoring), as in NIST §8.4.5.1; χ²(p; k) is the value a chi-square variable with k degrees of freedom stays below with probability p.

## Worked examples

1. n = 1, t = 2,880,000, r = 2, c = 90%, m = 360,000 gives result = 400 hours, mtbf = 400, rate = 0.0025, perMillion = 2,500, intervalLow = 127.068968, intervalHigh = 2,251.228611, reliability = 77.880078. Source: NIST/SEMATECH e-Handbook of Statistical Methods, §8.4.5.1 Constant repair rate (HPP/exponential) model (MTBF = total time ÷ total failures; 2T ÷ χ² bounds; zero failures: T ÷ (−ln α); example: 800 h and 2 failures, MTBF 400 h, 90% interval (127, 2251), 127 a 95% lower limit), https://www.itl.nist.gov/div898/handbook/apr/section4/apr451.htm (retrieved 2026-10-18); NIST/SEMATECH e-Handbook of Statistical Methods, §8.1.6.1 Exponential (R(t) = e^(−λt)), https://www.itl.nist.gov/div898/handbook/apr/section1/apr161.htm (retrieved 2026-10-18).
2. n = 1, t = 2,880,000, r = 2, c = 95% gives lower = 127.068968. Source: NIST/SEMATECH e-Handbook of Statistical Methods, §8.4.5.1 Constant repair rate (HPP/exponential) model (MTBF = total time ÷ total failures; 2T ÷ χ² bounds; zero failures: T ÷ (−ln α); example: 800 h and 2 failures, MTBF 400 h, 90% interval (127, 2251), 127 a 95% lower limit), https://www.itl.nist.gov/div898/handbook/apr/section4/apr451.htm (retrieved 2026-10-18): 127 is a 95% lower limit for the true MTBF.
3. n = 1, t = 2,880,000, r = 0, c = 95% gives lower = 267.046561, result = At least 267.047 hours (95% lower bound, no failures yet), rate = 0. Source: NIST/SEMATECH e-Handbook of Statistical Methods, §8.4.5.1 Constant repair rate (HPP/exponential) model (MTBF = total time ÷ total failures; 2T ÷ χ² bounds; zero failures: T ÷ (−ln α); example: 800 h and 2 failures, MTBF 400 h, 90% interval (127, 2251), 127 a 95% lower limit), https://www.itl.nist.gov/div898/handbook/apr/section4/apr451.htm (retrieved 2026-10-18): zero failures, MTBF lower bound T ÷ (−ln α) = 800 ÷ (−ln 0.05) = 267.
4. n = 50, t = 3,600,000, r = 4, c = 90%, m = 31,536,000 gives mtbf = 12,500, fit = 80,000, reliability = 49.618819, failureChance = 50.381181. Source: NIST/SEMATECH e-Handbook of Statistical Methods, §8.4.5.1 Constant repair rate (HPP/exponential) model (MTBF = total time ÷ total failures; 2T ÷ χ² bounds; zero failures: T ÷ (−ln α); example: 800 h and 2 failures, MTBF 400 h, 90% interval (127, 2251), 127 a 95% lower limit), https://www.itl.nist.gov/div898/handbook/apr/section4/apr451.htm (retrieved 2026-10-18); NIST/SEMATECH e-Handbook of Statistical Methods, §8.1.6.1 Exponential (R(t) = e^(−λt)), https://www.itl.nist.gov/div898/handbook/apr/section1/apr161.htm (retrieved 2026-10-18).

## FAQ

### How do you calculate MTBF?

Divide the total operating time by the number of failures. A system that ran 800 hours and failed twice has an MTBF of 800 ÷ 2 = 400 hours. For many units, add their hours first: 50 units × 1,000 hours with 4 failures gives 50,000 ÷ 4 = 12,500 hours.

### How do you convert MTBF to a failure rate?

The failure rate is 1 ÷ MTBF. An MTBF of 12,500 hours is 0.00008 failures per hour, 80 per million hours, or 80,000 per billion hours (80,000 FIT).

### What is the chance of a failure during a mission?

With a constant failure rate, the chance of no failure in a time t is R = e^(−t ÷ MTBF). With an MTBF of 400 hours, a 100-hour mission succeeds e^(−0.25) = 77.9% of the time, so a failure happens 22.1% of the time.

### Does an MTBF of 10,000 hours mean a unit lasts 10,000 hours?

No. MTBF is an average over many failures. With a constant failure rate, the chance of running for one MTBF without a failure is e^(−1) = 36.8%, so a failure comes first 63.2% of the time.

### What if there were no failures?

You cannot divide by 0, but you can bound the MTBF. NIST gives the lower bound T ÷ (−ln α): with 800 hours and no failures, the MTBF is at least 800 ÷ 2.996 = 267 hours at 95% confidence.

### How are the confidence bounds found?

From the chi-square distribution, as in the NIST/SEMATECH e-Handbook. For 800 hours and 2 failures, the 90% two-sided interval is 127 to 2,251 hours, and 127 hours is also a 95% one-sided lower bound.

## Sources

- NIST/SEMATECH e-Handbook of Statistical Methods, §8.4.5.1 Constant repair rate (HPP/exponential) model: the MTBF estimate is total time ÷ total failures; chi-square confidence bounds for Type I (time) censoring; the zero-failure bound T ÷ (−ln α); the 800-hour example, retrieved 2026-10-18. https://www.itl.nist.gov/div898/handbook/apr/section4/apr451.htm
- NIST/SEMATECH e-Handbook of Statistical Methods, §8.1.6.1 Exponential: R(t) = e^(−λt), constant failure rate λ, mean 1 ÷ λ, retrieved 2026-10-18. https://www.itl.nist.gov/div898/handbook/apr/section1/apr161.htm
- NIST/SEMATECH e-Handbook of Statistical Methods, §8.1.7.1 Homogeneous Poisson Process: times between failures are exponential and MTBF = 1 ÷ λ, retrieved 2026-10-18. https://www.itl.nist.gov/div898/handbook/apr/section1/apr171.htm
- Wikipedia, Failure rate, section Units: FIT, failures in one billion (10⁹) device-hours, retrieved 2026-10-18. https://en.wikipedia.org/wiki/Failure_rate