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MTBF and MTTR calculator

Mean time between failures and mean time to repair from totals or a failure log, with chi-square confidence bounds, availability, failure rate and reliability per machine.

Calculate from
Confidence and mission time (both)

Intervals and R(t) assume a constant failure rate (exponential times between failures). A period of records that simply ends on a date is “at a fixed time” (time-terminated).

Failure log input

Example dataReplace with your own failure log

Paste, open a CSV, TSV, TXT or XLSX file (up to 20 MB), or drop it on the field. Files are read in this browser only.

Columns

Start, End and Failures are found by their header names; without Start and End the number column is the repair duration.

Machine  Start             End
CNC-1    2026-03-02 06:40  2026-03-02 08:10
Long format
1 row = 1 failure

Columns: tab, ; or ,
Decimals: 12,5 or 12.5

Read as: tab-separated · whole numbers · 12 rows · header row skipped

Data as understood — first 10 of 12 rows
#MachineStartEnd
1CNC-12026-03-02 06:402026-03-02 08:10
2CNC-12026-03-09 14:052026-03-09 15:00
3CNC-12026-03-17 22:302026-03-18 01:15
4CNC-12026-03-26 10:002026-03-26 10:40
5CNC-22026-03-04 09:152026-03-04 13:45
6CNC-22026-03-20 18:202026-03-20 19:05
7CNC-32026-03-03 07:002026-03-03 07:35
8CNC-32026-03-06 12:102026-03-06 13:20
9CNC-32026-03-11 16:452026-03-11 18:00
10CNC-32026-03-15 08:302026-03-15 09:10

+2 more rows

One row per failure: the machine and the start and end of the repair (2026-03-02 14:05 or 02.03.2026 14:05), or the repair duration. A Failures column lets one row stand for several failures. Operating time = scheduled time − repair time.

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Failure log results

118.5 hMTBF (all machines)
1.49 hMTTR
98.76%Inherent availability
0.935Reliability R(t)

All machines: 12 failures in 1422.17 operating hours · MTBF 90 % interval 73.147 to 205.39 h · λ 0.008438 per hour

  • The fleet MTBF adds the operating time of all machines: it is the MTBF of one machine only if all of them fail at the same rate.
MTBF of each machine with its confidence interval: CNC-1: 4 failures, MTBF 118.5 h (51.8 h to 347 h), MTTR 1.458 h; CNC-2: 2 failures, MTBF 237.4 h (75.41 h to 1336 h), MTTR 2.625 h; CNC-3: 6 failures, MTBF 78.88 h (39.96 h to 181.1 h), MTTR 1.125 h; All machines: 12 failures, MTBF 118.5 h (73.15 h to 205.4 h), MTTR 1.486 h.05001000MTBF (hours)CNC-1118.5CNC-2237.4CNC-378.88All machines118.5
MTBF of a machineAll machinesConfidence intervalLower bound
Reliability R(t) = exp(−t / MTBF) of All machines, MTBF 118.5 h, lower bound 73.15 h, assuming a constant failure rate. At the mission time of 8 h: R = 0.9347, 0.8964 at the lower bound.0.000.250.500.751.00050100150200Operating hours tR = 0.935
R(t) at the MTBFR(t) at the lower boundMission time
MTBF and MTTR (hours)
QuantityCNC-1CNC-2CNC-3All machines
Operating time474.167474.75473.251422.17
Failures42612
Repair time5.833335.256.7517.8333
MTBF118.54237.3878.875118.51
MTBF lower (90 %)51.80275.40739.96273.147
MTBF upper347.041336181.11205.39
MTTR1.45832.6251.1251.4861
Failure rate λ (per 1000 h)8.4364.21312.688.438
Inherent availability98.78%98.91%98.59%98.76%
R(8 h)0.93470.96690.90350.9347
R at the lower MTBF0.85690.89930.81860.8964

Totals input

Operating time is the time the equipment was running or ready to run, without the repair time. Repairs default to the number of failures.

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Totals results

120.8 hMTBF
1.96 hMTTR
98.41%Inherent availability
0.936Reliability R(t)

Totals: 12 failures in 1450 operating hours · MTBF 90 % interval 74.579 to 209.41 h · λ 0.008276 per hour

    Reliability R(t) = exp(−t / MTBF) of Totals, MTBF 120.8 h, lower bound 74.58 h, assuming a constant failure rate. At the mission time of 8 h: R = 0.9359, 0.8983 at the lower bound.0.000.250.500.751.00050100150200Operating hours tR = 0.936
    R(t) at the MTBFR(t) at the lower boundMission time
    MTBF and MTTR (hours)
    QuantityTotals
    Operating time1450
    Failures12
    Repair time23.5
    MTBF120.83
    MTBF lower (90 %)74.579
    MTBF upper209.41
    MTTR1.9583
    Failure rate λ (per 1000 h)8.276
    Inherent availability98.41%
    R(8 h)0.9359
    R at the lower MTBF0.8983

    Download offline versionSingle HTML file · v1.5 · 567 KB · works without internet

    How MTBF and MTTR are calculated

    Mean time between failures (MTBF) says how long repairable equipment runs, on average, between two failures; mean time to repair (MTTR) how long it takes to get it running again. Together they give the inherent availability and the failure rate, the figures behind the availability loss of OEE and behind maintenance planning.

    Formulas

    Worked example (the tool's example)

    1. Three CNC machines were scheduled for 480 h each in March. The log has 12 failures: 4 on CNC-1, 2 on CNC-2, 6 on CNC-3.
    2. CNC-1 was down 90 + 55 + 165 + 40 = 350 min = 5.83 h, so it operated 474.17 h: MTBF = 474.17 / 4 = 118.5 h and MTTR = 5.83 / 4 = 1.46 h.
    3. All machines: 1422.17 operating hours and 17.83 repair hours for 12 failures: MTBF = 118.5 h, MTTR = 1.49 h, Ai = 118.5 / (118.5 + 1.49) = 98.76 %.
    4. 90 % interval, time-terminated: 2 × 1422.17 / χ²(0.95; 26) = 2844.33 / 38.885 = 73.1 h to 2 × 1422.17 / χ²(0.05; 24) = 2844.33 / 13.848 = 205.4 h. Twelve failures leave a wide range.
    5. Reliability over an 8-hour shift: R(8) = exp(−8 / 118.5) = 0.935; at the lower bound 73.1 h it is 0.896.

    Common mistakes

    Input and files

    Questions

    Is MTBF the time a machine runs before it fails?
    No. MTBF is an average over many failures, not a guaranteed failure-free time. With a constant failure rate only about 37 % of the units are still running at a time equal to the MTBF; most fail before it.
    Should the repair time be taken off the operating time?
    Yes. MTBF counts the time the equipment was running or ready to run, so time spent down for repair is not part of it. The calculator subtracts each machine’s repair time from its scheduled time.
    Why use 2r + 2 degrees of freedom for the lower bound?
    When the records end at a fixed date rather than at a failure (a time-terminated test), the next failure has not happened yet, so the lower bound allows for one more failure: chi-square with 2r + 2 degrees of freedom. A test stopped at a failure (failure-terminated) uses 2r.
    When is R(t) = exp(−t / MTBF) wrong?
    When the failure rate is not constant: early-life failures make the rate fall, wear-out makes it rise. Plot the failure times or fit a Weibull distribution; a Weibull shape clearly different from 1 means the exponential formulas understate or overstate the risk.