Tool
Enter failure, repair, and system data
Core formulas: Availability = MTBF / (MTBF + MTTR); Reliability R(t) = e−t/MTBF
Calculator Library / Reliability
Turn MTBF and MTTR into availability, downtime hours and cost per year, mission reliability, and the effect of series or parallel redundancy.
Tool
Core formulas: Availability = MTBF / (MTBF + MTTR); Reliability R(t) = e−t/MTBF
System results
Series means every unit must work. Parallel means the system works if at least one unit works, with repair capacity for every unit.
Comparison
Table
| Units | Series availability | Series downtime (h/yr) | Parallel availability | Parallel downtime (h/yr) |
|---|
Instructions
This calculator turns two numbers from your maintenance records, mean time between failures (MTBF) and mean time to repair (MTTR), into availability, expected downtime, the probability of surviving a mission, and the effect of redundancy.
Use it to compare options such as adding a spare unit, reducing repair time, or improving component reliability, and to put a dollar value on downtime for a business case.
| Measure | Formula | Meaning |
|---|---|---|
| Inherent availability | MTBF / (MTBF + MTTR) | The fraction of time an item is up, counting only failures and repairs. |
| Reliability over time t | e−t / MTBF | Probability of no failure over t hours, assuming a constant failure rate. |
| Series system of n units | An and Rn | Every unit must work; more units lower availability. |
| Parallel system (1 needed) | 1 − (1 − A)n | At least one unit works; redundancy raises availability. |
| Downtime per year | Operating hours × (1 − A) | Expected hours down in the required operating period. |
A pump has an MTBF of 2,000 hours and an MTTR of 8 hours. Availability is 2,000 / 2,008 = 99.60%. Over an 8,760-hour year that is about 4.4 failures and 34.9 hours of downtime, which at $5,000 per hour costs about $175,000. The chance the pump runs a 720-hour month without failing is e−0.36 = 69.8%.
With two identical pumps, needing both (series) gives 99.20% availability, or about 69.7 hours of downtime. Needing only one (parallel, with each unit repairable independently) gives 99.998%, about 0.14 hours. The second pump turns roughly 35 expected hours of downtime into a fraction of an hour, which shows why redundancy is used for critical duties.
Reliability is the probability that an item runs without failing for a stated time, so it depends on failure behavior only. Availability is the fraction of time an item is ready to work, so it depends on both how often it fails and how long repair takes. A unit can have modest reliability and high availability if repairs are quick.
MTBF is total operating time divided by the number of failures over that time, and MTTR is total repair time divided by the number of repairs. Use consistent units, and include only the failures and repairs relevant to the failure mode or asset you are studying.
A parallel arrangement improves availability when failures are independent and the standby unit switches in reliably. Common-cause failures, shared components, and failed switchover can erase much of the benefit, so redundancy should be designed and tested, not assumed.
The exponential model is simple and reasonable for many electronic and random failures, and it needs only MTBF. Equipment with wear-out behavior follows a different pattern, so use Weibull analysis when you have failure-time data.