Compute availability from MTBF (mean time between failures) and MTTR (mean time to repair), and convert it to downtime per year. You can also compute the availability of a whole system when units are connected in series, duplicated in parallel for redundancy, or arranged so that k of n units must work. It is handy for checking availability problems from IT certification exams after solving them by hand.
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Whole system
System availability
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Typical downtime
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Number of nines
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Units are assumed to fail independently, and availability is treated as a long-term average. A year is taken as 8760 hours.
See it as a graph
Other inputs stay at their current values while the chosen input varies from 1/4 to 4 times its current value. The red dot is the current value.
MTBF, MTTR and availability
MTBF (mean time between failures) is the average operating time from one failure to the next, and MTTR (mean time to repair) is the average time to restore service after a failure. Availability is the fraction of time the system is working: A = MTBF ÷ (MTBF + MTTR). With the default MTBF of 480 hours and MTTR of 20 hours, it is 480 ÷ 500 = 0.96 (96%). To raise availability, make failures rarer to lengthen the MTBF, or make repairs faster to shorten the MTTR.
Series systems: multiply the availabilities
When every unit must work for the system to work, the units are in series, and the system availability is the product of the unit availabilities. Two units of 0.9 in series give 0.9 × 0.9 = 0.81, so adding units in series lowers availability. Network devices, servers and storage on the processing path count as series elements, because the whole system stops if any one of them stops. This tool connects the row groups in series to compute the total.
Parallel systems (redundancy): 1 − probability that all fail
When several units do the same job and any one working is enough, they are in parallel. The system stops only when all units are down at once, so the availability is 1 − (1 − a)ⁿ. Two units of 0.9 in parallel give 1 − 0.1 × 0.1 = 0.99. Connecting the default first row (90%, 1 of 2 units) and second row (95%, 1 unit) in series gives 0.99 × 0.95 = 0.9405 (94.05%). Redundancy is the basic way to eliminate single points of failure, components whose failure stops the whole system.
k-out-of-n systems
For configurations such as "2 of 3 units are enough to keep up", add up the probabilities that k or more units are working. The probability that at least k of n units with availability a are up is Σ C(n, j)·aʲ·(1 − a)ⁿ⁻ʲ for j = k to n. With three units of 0.9 and two needed, it is 3 × 0.9² × 0.1 + 0.9³ = 0.243 + 0.729 = 0.972. Because k = 1 gives the parallel formula and k = n the series formula, this tool treats every row as a k-of-n group.
Availability, downtime and nines
Differences in availability are easier to grasp as downtime. 99% means about 87.6 hours of downtime per year, 99.9% about 8.76 hours, 99.99% about 53 minutes and 99.999% about 5 minutes. From the number of 9s, 99.9% is called "three nines" and 99.999% "five nines". Cloud and other services often state availability targets in an SLA (service level agreement). In practice, check whether planned maintenance is included and whether the assumption of independent failures really holds.
AThe average time from a failure until repair is completed
BThe shortest time until the first failure occurs
CFor a repairable system, the average operating time from one failure to the next
Show answer and explanation
Answer: C. For a repairable system, the average operating time from one failure to the next
MTBF (mean time between failures) is the average operating time between failures of a system that is repaired and kept in service; higher means more reliable. Mean time from failure to completed repair is MTTR, and mean time to failure for non-repaired parts is MTTF. Because MTBF is an average, it does not mean nothing fails before then, nor is it the shortest time to first failure.
Add the probabilities that at least two of the three servers are running. The probability that exactly two are running is 3 × 0.95² × 0.05 = 0.135375, and the probability that all three are running is 0.95³ = 0.857375, so the availability is 0.135375 + 0.857375 = 0.99275. 0.9025 is the value with the two required servers in series and leaves out the effect of the spare. 0.999875 is the value for a parallel system in which any one of the three is enough.
FAQ
How is availability calculated?
As MTBF ÷ (MTBF + MTTR). With an MTBF of 480 hours and an MTTR of 20 hours, it is 480 ÷ 500 = 0.96 (96%). A longer MTBF and a shorter MTTR both raise availability.
What are the formulas for series and parallel availability?
Series is the product a₁ × a₂ × …. Parallel is 1 − (1 − a₁)(1 − a₂)…, one minus the probability that all units are down at once. With two units of 0.9, series gives 0.81 and parallel 0.99.
How do I handle a mix of series and parallel?
First compute each parallel part as one group, then connect the groups and the remaining units in series and multiply. In this tool, enter each group as a row (k of n units), and the rows are connected in series.
How much downtime does 99.99% availability mean per year?
0.01% of 8760 hours, about 0.876 hours or roughly 53 minutes. 99.9% is about 8.76 hours, and 99.999% about 5 minutes.
What is the difference between MTBF and MTTF?
MTTF (mean time to failure) is the average time until a non-repairable part fails. MTBF is the average interval between failures of a repairable system, and it is what you use to compute the availability of repairable equipment.