MTBF, Repair and Availability

Availability is simply the fraction of time a repairable machine is actually up and working. The formula $A = \frac{\text{MTBF}}{\text{MTBF} + \text{MTTR}}$ is a plain ratio: MTBF (the average stretch of running between failures) over MTBF plus MTTR (the average time to get it fixed and running again). So if a pump runs 2,000 hours and then sits 20 hours waiting for repair, it's up 2,000 of every 2,020 hours. The maintenance part of the lesson asks one more question: when is it cheaper to replace a part before it breaks?

A sawtooth life. A cooling-water pump runs about 2,000 hours between failures, and each repair takes about 10 hours. Its life is a sawtooth: up, down, up, down. Over a year, what fraction of the time is it running? And would it be smarter to replace its seals every few hundred hours instead of waiting for them to fail?

Availability from one cycle. Each failure ends a cycle made of one up period (mean MTBF) and one repair (mean MTTR). Over many cycles, the law of large numbers says the fraction of time up approaches

For the pump, $A = 2000/2010 = 99.50\%$, about 44 hours down a year. Availability is often quoted in nines: 99.9% is 8.8 hours of downtime a year, 99.99% is 53 minutes.

The same answer from a differential equation. Let $P(t)$ be the probability the pump is up at time $t$, with constant failure rate $\lambda = 1/\text{MTBF}$ and repair rate $\mu = 1/\text{MTTR}$. Up pumps leave at rate $\lambda$ and down pumps return at rate $\mu$:

This is a linear first-order equation. Starting up, $P(t) = \frac{\mu}{\lambda + \mu} + \frac{\lambda}{\lambda + \mu}e^{-(\lambda + \mu)t}$, which settles within a few repair times to $\frac{\mu}{\lambda + \mu} = \frac{\text{MTBF}}{\text{MTBF} + \text{MTTR}}$.

MTBF is a rate in disguise. An MTBF of a million hours doesn't mean a drive lasts 114 years. It means a fleet of 1,000 drives sees about $1000 \times 8760/10^6 \approx 9$ failures a year during its useful life. MTBF belongs to repairable systems and MTTF to parts that are thrown away; for a constant failure rate both equal $1/\lambda$.

Systems. With independent failures and separate repair crews, availabilities combine exactly like reliabilities in Series, Parallel and k-out-of-n Systems: multiply for series, $1 - \prod(1 - A_i)$ for parallel. Repair makes redundancy enormously powerful. Two servers that are each down 1% of the time are both down only 0.01% of the time, because a failed server is fixed long before its partner is likely to fail.

Spares. MTTR includes waiting for parts, so the stockroom matters. With $n$ units at a constant failure rate, failures form a Poisson process (Exponential Lifetimes), so the number in a period $T$ is Poisson with mean $m = nT/\text{MTBF}$. Stock the smallest $s$ with $P(N \le s)$ at your target. For 12 motors with MTBF 20,000 h over six months (4,380 h), $m = 2.63$: five spares cover 94.9% of half-years and six cover 98.2%.

Preventive replacement. Replace a part at age $T$ or at failure, whichever comes first. A cycle then lasts $\min(\text{life}, T)$, whose mean is $\int_0^T R(t)\,dt$ (the area-under-$R$ idea from Reliability and Hazard Functions), and costs $c_p$ with probability $R(T)$ or $c_f$ with probability $1 - R(T)$. Mean cost per cycle over mean cycle length gives the long-run cost rate $C(T)$ above, the same renewal argument that gave $A$. Then find the $T$ that minimizes it (Optimization).

It only pays for wear-out. With a constant hazard, $R = e^{-\lambda T}$ and

which falls steadily toward $\lambda c_f$ as $T \to \infty$: never replace early. A memoryless part isn't worn, so a new one is no better. A finite best age appears only when the hazard rises ($\beta > 1$ in the Weibull model) and failures cost more than planned swaps.

A common mistake. It's tempting to think scheduled replacement always helps, since fresh parts feel safer. For random-failure electronics ($\beta = 1$) it doesn't: a new unit is no less likely to fail than the old one, and every swap adds cost and a chance of an installation error. Other slips: treating MTBF as a guaranteed life, forgetting that MTTR includes the wait for parts, and combining parallel availabilities as if one crew could fix both units at once.

Availability targets are written into contracts: cloud providers promise 99.9% or 99.99% uptime, and power plants, telecom networks and hospital equipment are run to availability goals. Maintenance planners use exactly these tools: Poisson spares calculations to stock warehouses, MTTR reduction (spares on site, modular swaps) to raise uptime, and age-based replacement for wear-out parts like rail wheels, pump seals and aircraft engine components, while random-failure electronics are left to run until they fail.