The design of a product includes the arrangement of all of the product elements. When considering the reliability of a system, the arrangement matters. Many systems are arranged serially. This means that with the failure of any one element, the system will not work. See the article on Series Systems for more details.

Another approach is the arrangement of the elements in a parallel structure. This means that the elements share the load of providing the function. The basic analysis of this type of system is to assume that any one element of all that work in parallel can operate alone for the system to work. This increases system reliability, regardless of the reliability of each individual element in parallel. See the post on parallel systems for more.

A third approach is to arrange the elements in parallel, where one or more elements must function for the system to work. If there are, n elements in parallel, the system may require k elements as a minimum to operate. K must be less than or equal to n. When k equals n, it is a series system, as all have to work. When k = 1 and n is greater than 1 than it is a simple parallel system. When k = 2 and n= 4, the system operates if any two of the four elements are working. Thus we get some benefit of the parallel structure, yet it is not as expensive or reliable as a simple redundancy system.

The general formula is similar to the binomial distribution reliability function in construction. The first term with factorials determines the number of ways the n units could combine in working and not working elements. The second term is the probability of the product successfully working, for the k minimum elements for successful operation of the system. The later term is the probability of failure for the n-k remaining units. In this case pay attention to the summation, otherwise, the formula is pretty flexible.

$$ \large\displaystyle \begin{array}{l}R(t)={{e}^{-\lambda t}}\\R(1)=0.995={{e}^{-\lambda (1)}}\end{array}$$ $$ \large\displaystyle \begin{array}{l}\ln (0.995)=-\lambda \\\lambda =0.005\end{array}$$

Î» is the failure rate and single parameter for the exponential distribution.

n is the number of elements in parallel

k is the minimum number of elements required for successful operation of the system.

## Example Problem

The autonomous parking system of a car contains three computers and sensor set to determine the appropriate parking maneuver for aÂ given situation. The computers consider the information and plan the steering and acceleration to successfully park the car. Then the computers compare results before attempting the parking maneuver. When two computers agree, and one does not, the car parks with the plan created by the two computers, and warns the driving that maintenance is required for the faulty computer.

If each computer has a 0.995 probability of success, what is the probability of a successfully parked car?

## Example Solution

The car has a 2 out of 3 active redundancy system. Thus we can use the above formula where k = 2 and n = 3.

First we need the failure rate, and given only the reliability for the parking maneuver, we assume the exponential distribution and solve forÂ Î».

$$ \large\displaystyle \begin{array}{l}R(t)={{e}^{-\lambda t}}\\R(1)=0.995={{e}^{-\lambda (1)}}\end{array}$$ $$ \large\displaystyle \begin{array}{l}\ln (0.995)=-\lambda \\\lambda =0.005\end{array}$$

Using the formula from above, k = 2, n = 3,Â Î» = 0.005, and t = 1 to find

$$ \large\displaystyleÂ \begin{array}{l}R(t)=\sum\limits_{k}^{n}{\frac{n!}{k!(n-k)!}{{\left( {{e}^{-\lambda t}} \right)}^{k}}}{{\left( 1-{{e}^{-\lambda t}} \right)}^{n-k}}\\R(1)=\sum\limits_{2}^{3}{\frac{3!}{2!(3-2)!}{{\left( {{e}^{-0.005}} \right)}^{2}}}{{\left( 1-{{e}^{-0.005}} \right)}^{3-2}}\\R(1)=\text{0.999926}\end{array}$$

Now, I’m not sure if the new cars with the fancy self parking feature really have triple redundancy, yet if the onboard computers work as well as my old PC at home, then I would hope it does have the active redundancy. Still, its always important to consider system arrangement in product design, weighing all aspects of reliability, costs and benefits.

Related:

Reliability Block Diagrams Overview and ValueÂ (article)

Parallel SystemsÂ (article)

Reliability ApportionmentÂ (article)

Tarapada Pyne says

Very easy to understand. Thank you so much Er Fred. Keeping complex things to a Simple, is always your free-flowing interactions and sharing . I learn a lot from you.

Paul Thomas says

Some text books this concept is not clearly explained.

I am able to understand very clearly your explanation.

Thanks Fred.

Fred Schenkelberg says

Thanks for the note Paul. Glad to help. Cheers, Fred

Cristian Rosa says

First of all, I would like to thank you for this post. It is completely useful and understandable. I only have a question regarding the RBD k-out-n. How can I perform this calculation if I have different Reliability block figures?

Fred Schenkelberg says

Hi Cristian,

I’m not entirely sure. My first thought would to use a monte carlo type simulation of the k-out-n structure with the various different blocks in order to get a single expected distribution to use in the larger model. Just a thought.

cheers,

Fred