Library / Reliability / REL-03
Reliability
Availability
Fraction of time a system is ready for use, from the mean time between failures and the mean time to repair.
- Ref
- REL-03
- Rev
- v1
- Field
- Reliability
- Result
- A
MTBF is the mean time between failures and MTTR the mean time to repair, in the same unit (hours here). A is the long-run fraction of time the system is ready to be used. It depends only on the ratio of MTTR to MTBF, since A = 1/(1 + MTTR/MTBF).
Fraction of time a system is ready for use, from the mean time between failures and the mean time to repair.
Calculator
- Downtime per year (8760 h)86.73 h
- Availability if the repair time is halved0.995
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- An availability of 99 % sounds high, but it still means about 87 hours of downtime a year. Halving the repair time to 2.5 h gives 99.5 %.
- Improving repair speed (spares, training, easy access) is often cheaper than improving reliability, because A depends on the ratio MTTR/MTBF.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing () | Local step | Change in result |
|---|---|---|---|---|---|---|
| MTBF Mean time between failures | 1 to 100,000 h | 0.1667 | 1 | 0.8333 | +10% | +0.0901 % |
| MTTR Mean time to repair | 0.1 to 1,000 h | 0.3333 | 0.9998 | 0.6665 | +10% | −0.0989 % |
Sweep of Mean time to repair
| Mean time to repair (h) | Availability |
|---|---|
| 0.1 | 0.9998 |
| 0.3162 | 0.9994 |
| 1 | 0.998 |
| 3.162 | 0.9937 |
| 10 | 0.9804 |
| 31.62 | 0.9405 |
| 100 | 0.8333 |
| 316.2 | 0.6126 |
| 1,000 | 0.3333 |
Sweep one variable
Variables and how each one contributes
Availability
What the equation gives you. Everything else feeds it.
Mean time between failures
The mean time the equipment runs between failures, in hours. Raises availability with diminishing returns. At the default a 10 % rise in MTBF raises A by only about 0.1 %, because A is already above 99 %.
Mean time to repair
The mean time to restore the equipment to service after a failure, in hours. Lowers availability: halving the repair time has the same effect as doubling the MTBF.
Worked example with the values above
- Total cycle: MTBF + MTTR = 500 + 5 = 505 h.
- Fraction of the cycle spent working: 500 / 505.
- A = 0.9901, or 99.01 %.
Assumptions and limits
- Steady-state operation with constant failure and repair rates (exponential times).
- Repair restores the equipment to as-good-as-new, with no waiting for parts or staff counted in MTTR unless it is included.
Common mistakes
- Counting planned maintenance as downtime in one place and not in the other.
- Leaving logistic delay (waiting for spares) out of MTTR.
Statistics and uncertainty
For high availability A ≈ 1 − MTTR/MTBF. MTBF estimated from a small number of failures is uncertain by about 1/√n (30 % with 10 failures), and since the unavailability 1 − A is proportional to 1/MTBF, the quoted downtime is uncertain by the same proportion.
Where it is used
- Readiness of vehicles, radars and other equipment.
- Comparing a reliability improvement with a repair-time improvement.
Maths you need
- Ratios and fractions: time working over total time
- Mean time: average of many cycles