How to use the reliability, mtbf and availability calculator
- Choose how the parts work together: in series (all must work), in parallel (any one is enough) or k of n (at least k must work).
- Enter the mission time, the number of hours the system must run without failing.
- List the parts with the mean time between failures (MTBF) for each. Add the mean time to repair (MTTR) if you want availability.
- Read the chance the system survives the mission, its availability and its mean time to failure, with the same figures for each part.
Formula
part reliability over time t = e^(-t / MTBF)
part availability = MTBF / (MTBF + MTTR)
series: product of the reliabilities; parallel: 1 - product of (1 - reliability)
k of n: the chance that at least k parts work, summed over every combination
mean time to failure = area under the system reliability curve
All parts are assumed to fail at a constant rate, which is what makes the reliability an exponential. The mean time to failure is worked out exactly by expanding the system reliability into exponential terms.
A controller, a motor and a sensor in series
A controller with an MTBF of 20,000 hours, a motor with 8,000 and a sensor with 15,000 are all needed. Over a 1,000 hour mission they survive with probabilities 95.12, 88.25 and 93.55 percent. All three survive with probability 78.53 percent.
With repair times of 4, 12 and 2 hours, the availabilities are 99.980, 99.850 and 99.987 percent, and the system availability is 99.82 percent. The mean time to failure of the series is 1 / (1/20,000 + 1/8,000 + 1/15,000) = 4,137.93 hours. Two identical pumps of MTBF 8,000 hours in parallel survive 1,000 hours with probability 98.62 percent, and their mean time to failure is 12,000 hours.
Series, parallel and k of n
In series, every part is a single point of failure, so adding parts always lowers reliability. In parallel, a spare covers for a failure, so adding parts raises reliability, with diminishing returns. A k of n system sits between them, such as a vote of three sensors where two must agree.
The page computes the k of n case exactly by working through how many parts survive, treating the parts as independent. Series is the case k = n and parallel is the case k = 1.
Reliability versus availability
Reliability is the chance of getting through a mission without a failure. Availability is the share of time a repairable part is working. A part that fails often but is fixed in minutes can have high availability and low reliability. The availability shown here is the long-run figure and treats repairs of different parts as independent.
For equipment effectiveness in a production line, use the OEE Calculator. For counts of failures in a period, there is the Poisson Distribution Calculator, and for queues and work in progress, see Little's Law Calculator.
Assumptions and limits
- Every part has a constant failure rate, so its life is exponential. Wear-out and early failures are not modeled.
- Parts fail independently. Shared causes such as a power loss are not modeled.
- Availability is the steady-state figure with independent repair of each part, and it needs a repair time for every part.
- The mean time to failure is for a system that is not repaired while running.
- Up to 12 parts are accepted, and nothing typed here leaves your browser.
Frequently asked questions
What is the difference between MTBF and the chance of surviving?
MTBF is an average. With a constant failure rate, a part survives a time equal to its MTBF with probability about 36.8 percent, not 50 percent, because the failure times are spread out. The page converts MTBF into a chance for the mission time you enter.
Why does availability need repair times?
Availability is the share of time a part is up, which depends on how long each repair takes. Without a repair time for every part, the page leaves the system availability out.
Can I mix different kinds of parts in a k of n system?
Yes. Each part has its own MTBF. The page works out the chance that at least k of them work, using each part's own reliability.
What is the mean time to failure of a parallel system?
It is longer than the longest part, because a spare keeps the system going. For two equal parts it is 1.5 times a single part's MTBF, and the page computes it exactly for any mix.