# k out of n system

The design of a product includes the arrangement of all of the product elements. When considering the reliability of 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 element 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 .

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

λ 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 3 computers and sensor set to determine the appropriate parking maneuver for a 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 for λ.

$\displaystyle \begin{array}{l}R(t)={{e}^{-\lambda t}}\\R(1)=0.995.995={{e}^{-\lambda (1)}}\end{array}$$\displaystyle \begin{array}{l}\ln (0.995)=-\lambda \\\lambda =0.005\end{array}$

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

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

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.

# Exponential Reliability

Down to the last week of preparation for the exam on March 2nd. Good luck to all those signed up for that exam date. Time to focus on preparing your notes, organizing your references and doing a final run though of practice exams. Continue reading

# RBD and Design Process

A reliability block diagram (RBD) for a product that has no redundancy or complex use profile, is often very simple. A series system (reliability wise) implies that any one part or element of the product that fails the entire product fails. One might ask if an RBD is even necessary. Continue reading

# Series System

During design and development, Reliability Engineers often receive reliability parameters in many forms. The most common reliability parameter is the mean time to failure (MTTF), which can also be specified as the failure rate (this is expressed as a frequency or Conditional Probability Density Function (PDF)) or the number of failures during a given period. Continue reading

# Parallel Systems

Speaking reliability-wise, parallel, means any of the elements in parallel structure permit the system to function. This does not mean they are physically parallel (in all cases), as capacitors in parallel provide a specific behavior in the circuit and if one capacitor fails that system might fail.
In this simple drowning there are n components in parallel and any one component is needed for the system to function. Continue reading