# Standby Redundancy, Equal Failure Rates, Perfect Switching

First off, switches are not perfect, so this situation is hypothetical. Yet, when you are exploring adding standby redundancy and haven’t sorted out the switching mechanism, you may be purely curious about the benefits of the redundancy.

Second, in practice, we use different equipment or components for the standby redundancy.

For example, we may use a diesel generator as a backup for grid power. Or use a refurbished pump to back up the main pump. Consider this situation for standby redundancy with equal failure rates (same type and age equipment) and a perfect ability to detect and turn on the backup unit as an exercise that leads to other realistic cases.

Let’s explore the math to estimate the system reliability given standby redundancy with an example.

## Two power supply systems with one supply in standby

The two supplies have the same expected failure rate, 0.0005 failures per hour, over the duration we expect the system to operate, 730 hours. Only one operates at a time. And, we’re going to ignore how the second supply is turned on when the first one fails for now. We’re assuming a perfect switch.

Here’s a simple diagram of the situation.

Where R_{1} is the reliability function of the power supplies. So, what is the reliability expectation of this standby redundancy configuration?

The formula for we can use in this situation is

$$ \large\displaystyle R\left( t \right)={{e}^{-\lambda t}}\left( 1+\lambda t \right)$$

This is slightly different then the k-out-of-n formula with a 1-out-of-2 situation as the first unit operates alone while the standby unit is dormant and assumed to have not dormant failure period failures.

Adding the values for t and $- \lambda -$ we find

$$ \large\displaystyle \begin{array}{l}R\left( 730 \right)={{e}^{-\left( 0.0005 \right)730}}\left( 1+\left( 0.0005 \right)730 \right)\\R\left( 730 \right)={{e}^{-0.365}}\left( 1.365 \right)\\R\left( 730 \right)=0.948\end{array}$$

## A general formula

In the case with n standby units each with a perfect switch when the previous elements have failed, the general formula is

$$ \large\displaystyle R\left( t \right)={{e}^{-\lambda t}}\sum\limits_{i=0}^{n-1}{\frac{{{\left( \lambda t \right)}^{i}}}{i!}}$$

Here’s a simple diagram of a system with n=3. Note the second standby units doesn’t become active until the primary and first standby units fail. After the unit brought online by the first switch fails, the second switch closes allowing the system to operate with the second standby unit.

You may envision n such standby switches and units in a configuration. Now you can calculate the system reliability for these standby parallel situations.

Related:

Standby Redundancy, Equal Failure Rates, Imperfect Switching (article)

Parallel Systems (article)

RBD and Design Process (article)

sameer.sheeba@gmail.com says

Yes, this is for perfect switching. In case of imperfect switching you need to just multiply the probability of failure of that imperfect switch with the equation you have provided. But, have you experienced it by iterating 1000 times through Monte-carlo simulation. I mean the result.What results did you get.

Fred Schenkelberg says

Thanks for the comment Sameer,

I’ve not tried the Monte Carlo approach and will have to explore it later. For some cases it is possible to analytically solve for system reliability including imperfect switching, which is the topic of another post.

Cheers,

Fred

Dick Somes says

I’m trying to understand the rationale behind the computation. Can you provide a derivation of the formula? Fifty years ago I was a reliability engineer in my first job. At the time we focused on active redundancy because of the state of failure detection and switching technology.

Fred Schenkelberg says

HI Dick,

I found a reasonable explanation in

Reliability in Engineering Design” target=”_blank”>Reliability in Engineering Design by Kailash C. Kapur. Wiley, 1977 pg 218-221 [affiliate link to Amazon]

Here are the scanned pages from my copy. You will need to be logged into Accendo Reliability to download.

Cheers,

Fred

Aaron McDonald says

Hello,

Is it possible to be directed to a derivation for this? I generally understand how to set-up the problem for a duty/stand-by system but can’t get over the hump of fully solving it.

Regards,

Aaron

Fred Schenkelberg says

Hi Aaron,

good question – most of my references just include the formulas for the various situations with perfect or imperfect switches. A general derivation might start with thinking through the probabilities involved and how to combine them. An imperfect switch would be in series with the second unit, I suppose.

Cheers,

Fred

Farid Jalali says

Hello Fred,

is there a formula or referenced that you can point me to for when active and standby units have different failure rates?

Thank you,

Farid

Fred Schenkelberg says

Hi Farid,

I haven’t written about such a case and a quick check of a couple of the reference books I have hasn’t turned up such a case. It is a good idea, so will add to my list of topics to research and write a short tutorial.

cheers,

Fred

ayca says

I found this article while i search generator reliability. In case of any power failure, the generator is activated, in this case is always generator redundant a system? When the generator is used in this way, is it a subsystem that always increases the reliability of the system?

Thank for all informations.

ayca

Fred Schenkelberg says

Hi Ayca,

The reliability of the various elements of the system don’t really change – the redundant elements improve the availability of the overall system.

At the overall system level, with the generator, the ability (probability) of the system to provide power (function) over time in the specified environment has an improved reliability performance. The tradeoff comes with a more complex system and an increased cost of maintenance.

cheers,

Fred