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MTBF, MTTR & Availability Calculator

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Understanding the MTBF, MTTR and availability calculator — concept and principle

MTBF (mean time between failures), MTTR (mean time to repair) and availability are the three basic numbers of maintenance and reliability engineering. From the total operating time of a repairable system, the number of failures observed in that time and the average repair time, the calculator gives the MTBF, the failure rate, the steady-state availability and the expected downtime over one year (8,760 hours). Engineers use them to compare equipment, set service targets and judge where spares or redundancy pay off.

Uptime and repair cycles over one yearFour failures in 8,760 hours give MTBF 2,190 h; with 3 h repairs availability is 2,190 ÷ 2,193 = 99.86%, about 12 hours of downtime a year.One year: run, fail, repair, repeat8,760 h of operation, 4 failures2,190 h2,190 h2,190 h2,190 hred = 3 h repair (drawn wider)Availability = up ÷ (up + down)2,190÷(2,190+3)=99.86%MTBF = 8,760 ÷ 4 = 2,190 h · λ = 1 ÷ 2,190 = 0.00046/hDowntime ≈ 12 h a year8,760 × (1 − 0.99863) = 11.98 h
Each cycle is 2,190 h running plus 3 h repair; availability is the running share of the cycle.

How it is calculated

Operating time should count only running hours (uptime), not calendar time including repairs.

The principle

A repairable system alternates between an "up" period, which on average lasts one MTBF, and a "down" period, which on average lasts one MTTR. Over a long run the fraction of time spent up is simply the average up period divided by the average full cycle, MTBF ÷ (MTBF + MTTR). This is inherent availability: it counts only repair time and ignores waiting for parts and planned maintenance. λ = 1/MTBF describes a constant failure rate (exponential model), which fits random failures rather than wear-out.

Worked example

  1. Inputs: operating time 8,760 h, 4 failures, MTTR 3 h.
  2. MTBF = 8,760 ÷ 4 = 2,190 h.
  3. Failure rate = 4 ÷ 8,760 = 0.000456621 per hour.
  4. Availability = 2,190 ÷ (2,190 + 3) = 2,190 ÷ 2,193 = 99.86%.
  5. Downtime per year = 8,760 × (1 − 0.998632) ≈ 11.98 h.
How MTBF (h) changes with FailuresHow MTBF changes with Failures02,0004,0006,0008,00010,0001234568Example: 2,190Failures
Each point is the calculator re-run with only “Failures” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.
FailuresMTBF (h)
18,760
24,380
32,920
42,190
51,752
61,460
81,095

How to read the result

Availability is often expressed in "nines": 99% allows about 87.6 h of downtime per year, 99.9% about 8.8 h, 99.99% about 53 minutes. Raise availability by making failures rarer (higher MTBF) or repairs faster (lower MTTR); in the example, cutting MTTR to 1.5 h halves yearly downtime, often more cheaply than redesign.

Common mistakes

Keep in mind: These formulas assume a steady state, a roughly constant failure rate and an MTTR that covers all downtime. Wear-out, early failures and logistic delays change the result; use it for planning, not as a guarantee.

Frequently asked questions

What is the difference between MTBF and MTTF?

MTTF is the mean time to the first (and only) failure of a non-repairable item. MTBF applies to repairable systems that return to service after each failure.

What is a good availability figure?

It depends on the service; set the target from the cost of an hour of downtime. Critical IT services often aim for 99.9% or more.

Sources: IEC 60050-192, International Electrotechnical Vocabulary – Part 192: Dependability · MIL-HDBK-338B, Electronic Reliability Design Handbook, U.S. Department of Defense

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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