Dear Johanno,

"repair time" hints to MTBF while "failure rate" relates to statistical failure, not typically related to "end of life".

So please do not mix up these two things.

If you double the amount of identic components in a given system, the failure rate (FIT) doubles as well purely for statistical reasons. The MTBF remains the same.

If you have two systems that can independently do the same job - even if it takes some time to switch from one to another - the time to repair doubles. Also by pure math.

Systems having 2x100% installed but working one system only to provide 100% leaves the second redundant system without operation, not adding to the operating hours.

The main unit thus does a number of hours of operation while the second unit usually has 0 hours or just the few hours needed to replace the first unit.

Therefore, the main unit will reach it's end of life after X hours while the second unit has X-(hours operated) left.

Cheers - Martin

Hi Johanno,

First, MTBF = 1 / Failure Rate thus MTBF is the inverse of failure rate. MTBF has nothing to do with repair time. MTTR, or mean time to repair, is commonly used to describe the average repair time. See NoMTBf.com for more on the many common confusions surrounding MTBF. In short, do not use MTBF for anything.

With a redundant system, the availability improves, meaning the ability of the system to operate increases as it can continue to operate after the primary element fails and the backup unit takes over the operation.

The failure rate of a redundant system tends to increase given the increased complexity. In some cases, if the elements in parallel share the load, they may actually have lower failure rates. Really depends on the design and use.

At home I have a backup battery system for the office equipment (an UPS), it has it's own set of failure mechanisms, yet does protect both my work and equipment when we experience power drops or sags. My office equipment is working well while the UPS takes the damage caused by the power drops.

Repair time - here again the answer is highly dependent on your system and logicistics system

The modeling of standby systems is typically done with state-based thinking. You need to distinguish three system states, assuming that the backup unit is a warm spare:

• Both units are ok.
• Only the backup unit is working.
• Both units are broken.

You can now create something like a Markow chain and model the different state transition rates based on your given input. You will end up with probabilities for being in some of the states, expressible as reliability function, but no longer with an overall failure rate or repair rate for the whole system. As the other authors pointed out, it doesn't make sense in most cases. Talking in 'rates' assumes an exponential distribution of failure and repair times - but your standby system does not fail in a 'memoryless' fashion. The failure probability of the overall system at any given point in time depends on the question if and how long the primary unit already failed . There is some in-build dependency in your system behavior, so you can not seriously talk in failure and repair rates about it.

Shooman discussed a simplified example of your case in his book (978-0-471-29342-2) and showed how to derive a reliability function for the resulting system. May this is good enough for your purpose. You can also read papers by K. Trivedi, who investigated a lot of similar cases in the past. Dynamic fault trees have a spare gate for exactly such cases, so any reasonable reliability analysis tool will give you a reliability function for the resulting system. 

The failure rate depends fundamentally on the type of component/system such as IFR, CFR, etc. Moreover, the standby component need not be identical to the initial one. Otherwise, the lifetime doubles but not the failure rate. I guess that the answer by Martin was intended to mean the double life and not the double rate. The repair time depends on many other factors such as the diagnosis time, the response time, the availability of spares, the actual repair etc

If Ls is the failure rate of your system, Ls.dt is the probability that the system fails between t and t+dt given no failure has occurred before t.

When the running unit fails, the system fails at once as the second unit is still in standby. Therefore, rigorously speaking and if if no other assumption is made, the equivalent failure rate of the system is simply the same as the failure rate Lu of the operating unit.

If the switching is instantaneous, the duration of the state (one unit failed, one unit in standby) is equal to 0. The system doesn't fail when the 1st unit fails as the second unit starts at once. It fails when the 2nd unit are fails. In this case the equivalent failure rate Ls converge to an asymptotic  value equal to about Lu*[Lu/(Lu+Mu)] after about 3 time 1/Mu where Mu is the repair rate. (See the theory of critical running states in Gondran-Pagès)

If the switching time is not negligible but that you make the assumption that the functioning interruption  is acceptable  - ie that the state (one unit failed, one unit in standby) is a running state - then Ls is also going to converge toward about Lu*[Nu/(Nu+Mu)]*[Lu/(Lu+Mu)] where Nu is the switching rate (1/time to switch)

Note that when the failure rates are constant (Markov case), 1/Ls is the MTTF of the system (not the MTBF).

Note also that the acronym MTBF is confusing as it means "mean time between failures" for some peoples and "mean operating time between failures" for others. Therefore it is better to use METBF (mean elapsed time between failure) and MOTBF (mean operating time between failures) to avoid any confusion (see IEC 61703:2016)