What is the MTBF?
Type how many units ran, for how long, and how many failures you saw. The MTBF calculator gives the mean time between failures, the failure rate, the chance of getting through a mission without a failure, and confidence bounds, even with no failures yet.
- MTBF
- 400 hours
MTBF: 400 hours, from 800 hours of operation.
- MTBF (hours)
- 400
- Total operating time (hours)
- 800
- Failure rate (per hour)
- 0.0025
- Failures per million hours
- 2,500
- Failures per billion hours (FIT)
- 2,500,000
- Lower confidence bound (hours)
- 150.31
- Two-sided interval, low (hours)
- 127.069
- Two-sided interval, high (hours)
- 2,251.23
- Reliability over the mission (%)
- 77.8801
- Chance of a failure in the mission (%)
- 22.1199
MTBF: 400 hours. MTBF: 400 hours, from 800 hours of operation.
How to calculate
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.
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.
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).
- 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
Each example is checked against the calculator on every build.
- Number of units 1, Operating time per unit 800 h, Failures 2, Confidence level 90%, Mission time 100 h gives MTBF 400 hours, MTBF (hours) 400, Failure rate (per hour) 0.0025, Failures per million hours 2,500, Two-sided interval, low (hours) 127.068968, Two-sided interval, high (hours) 2,251.228611, Reliability over the mission (%) 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)
- Number of units 1, Operating time per unit 800 h, Failures 2, Confidence level 95% gives Lower confidence bound (hours) 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
- Number of units 1, Operating time per unit 800 h, Failures 0, Confidence level 95% gives Lower confidence bound (hours) 267.046561, MTBF At least 267.047 hours (95% lower bound, no failures yet), Failure rate (per hour) 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
- Number of units 50, Operating time per unit 1,000 h, Failures 4, Confidence level 90%, Mission time 8,760 h gives MTBF (hours) 12,500, Failures per billion hours (FIT) 80,000, Reliability over the mission (%) 49.618819, Chance of a failure in the mission (%) 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)
How it works
- Total operating time T = number of units × operating time per unit, in hours (1 d = 24 h, 1 wk = 168 h, 1 yr = 365 days = 8,760 h).
- MTBF = T ÷ r, with r the number of failures (NIST §8.4.5.1). It needs at least one failure.
- Failure rate λ = r ÷ T per hour (1 ÷ MTBF), also given per million hours (λ × 10⁶) and per billion hours (λ × 10⁹, FIT).
- Reliability over a mission time t: R = e^(−t ÷ MTBF), and the chance of a failure is 1 − R, computed as −expm1(−t ÷ MTBF) so small chances keep their digits (NIST §8.1.6.1). Both need at least one failure.
- Confidence bounds at confidence c (α = 1 − c), for a test that runs a fixed time (Type I censoring), with χ²(p; k) the value a chi-square variable with k degrees of freedom stays below with probability p:
- one-sided lower bound = 2T ÷ χ²(c; 2r + 2). With r = 0 this is T ÷ (−ln α), the NIST zero-failure bound;
- two-sided interval = 2T ÷ χ²(1 − α/2; 2r + 2) to 2T ÷ χ²(α/2; 2r), for r ≥ 1.
The headline is the MTBF in hours, or with no failures "At least L hours (c% lower bound, no failures yet)".
Rules
- Units from 1 to 10⁹ (whole); operating time per unit from 1 s to 10¹³ s; failures from 0 to 100,000 (whole); confidence from 1% to 99.99%; mission time from 0 to 10¹³ s, optional.
- Reliability is left out when e^(−t ÷ MTBF) is below 10⁻³⁰⁰, which is when t is over 300 ln 10 = 690.776 MTBFs; the chance of a failure then shows 100%.
Output format
The headline and the numbers show to 6 significant figures with thousands commas, and below 10⁻⁶ or from 10¹⁵ up as a × 10ⁿ. The reliability and the chance of a failure are percents (their labels say %), so a reliability of 5.9034 × 10⁻²⁹⁸ % shows as 5.9034 × 10⁻²⁹⁸. T, the MTBF and λ are exact fractions of the typed values, rounded once; the bounds and the reliability use the chi-square quantile and e^x, so they are 64-bit floats.
Worked examples by hand
800 hours, 2 failures. MTBF = 800 ÷ 2 = 400 hours; λ = 0.0025 per hour (2,500 per million hours). At 90%, α = 0.10: the interval is 1,600 ÷ χ²(0.95; 6) = 1,600 ÷ 12.5916 = 127.07 hours to 1,600 ÷ χ²(0.05; 4) = 1,600 ÷ 0.71072 = 2,251.2 hours. A 100-hour mission: R = e^(−0.25) = 77.8801%.
The 95% lower bound. 1,600 ÷ χ²(0.95; 6) = 127.07 hours, the same 127 as the low end of the 90% interval.
No failures. 800 hours, 95%: 800 ÷ (−ln 0.05) = 800 ÷ 2.99573 = 267.047 hours.
A fleet. 50 units × 1,000 hours = 50,000 hours with 4 failures: MTBF = 12,500 hours; λ = 8 × 10⁻⁵ per hour = 80,000 FIT. A one-year mission (8,760 hours): R = e^(−0.7008) = 49.6188%, so the chance of a failure is 50.3812%.
Other questions people ask
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.