The k-out-of-n configuration is a special case of parallel redundancy. This type of configuration requires that at least k components succeed out of the total n parallel components for the system to succeed. For example, consider an airplane that has four engines. Furthermore, suppose that the design of the aircraft is such that at least two engines are required to function for the aircraft to remain airborne. This means that the engines are reliability-wise in a k-out-of-n configuration, where k = 2 and n = 4. More specifically, they are in a 2-out-of-4 configuration.
Figure 4.9: 2-out-of-4 configuration.
Even though we classified the k-out-of-n configuration as a special case of parallel redundancy, it can also be viewed as a general configuration type. As the number of units required to keep the system functioning approaches the total number of units is the system, the system's behavior tends towards that of a series system. If the number of units required is equal to the number of units in the system, it is a series system. In other words, a series system of statistically independent components is an n-out-of-n system and a parallel system of statistically independent components is a 1-out-of-n system.
The simplest case of components in a k-out-of-n configuration is when the components are independent and identical. In other words, all the components have the same failure distribution and whenever a failure occurs, the remaining components are not affected. In this case, the reliability of the system with such a configuration can be evaluated using the binomial distribution, or:
(6)
Where:
n is the total number of units in parallel.
k is the minimum number of units required for system success.
R is the reliability of each unit.
Consider a system of six pumps of which at least four must function properly for system success. Each pump has an 85% reliability for the mission duration. What is the probability of success of the system for the same mission duration?
Using Eqn. (6) for k = 4 and n = 6:
One can examine the effect of increasing the number of units required for system success while the total number of units remains constant (in this example, six units). In Figure 4.9, the reliability of the k-out-of-6 configuration was plotted versus different numbers of required units.
Note that the system configuration becomes a simple parallel configuration for k = 1 and the system is a six unit series configuration ((0.85)6 = 0.377) for k = 6.
Figure 4.10: Reliability of a k-out-of-6 configuration for k values.
In the case where the k-out-of-n components are not identical, the reliability must be calculated in a different way. One approach, described in detail later in this chapter, is to use the event space method. In this method, all possible operational combinations are considered in order to obtain the system's reliability. The method is illustrated with the following example.
Three hard drives in a computer system are configured reliability-wise in parallel. At least two of them must function in order for the computer to work properly. Each hard drive is of the same size and speed, but they are made by different manufacturers and have different reliabilities. The reliability of HD #1 is 0.9, HD #2 is 0.88 and HD #3 is 0.85, all at the same mission time.
Since at least two hard drives must be functioning at all times, only one failure is allowed. This is a 2-out-of-3 configuration.
The following operational combinations are possible for a system success:
All 3 hard drives operate.
HD #1 fails, while HDs #2 and #3 continue to operate.
HD #2 fails, while HDs #1 and #3 continue to operate.
HD #3 fails, while HDs #1 and #2 continue to operate.
The probability of success for the system (reliability) can now be expressed as:
This equation for the reliability of the system can be reduced to:
Or:
If all three hard drives had the same reliability, R, then the equation for the reliability of the system could be further reduced to:
Or, using the binomial approach:
The example can be repeated using BlockSim. The following graphic demonstrates the RBD for the system.
The RBD is analyzed and the system reliability equation is returned. Figure 4.11 shows the equation returned by BlockSim.
Figure 4.11: System equation results for Example 9.
Using the Analytical Quick Calculation Pad, the reliability can be calculated to be 0.9586. Figure 4.12 shows the returned result.
Figure 4.12: Reliability results for Example 19.
Note that you are not required to enter a Mission End Time for this system into the Analytical QCP because all of the components are static and thus the reliability results are independent of time.
Consider the four engine aircraft discussed previously. If we were to change the problem statement to "two out of four engines are required, however no two engines on the same side may fail," then the block diagram would change to the configuration shown in Figure 4.13.
Figure 4.13: Block diagram for Example 10.
Note that this is the same as having two engines in parallel on each wing and then putting the two wings in series.
There are other multiple redundancy types and multiple industry terms. One such example is what is referred to as a consecutive k-out-of-n : F system. To illustrate this configuration type, consider a telecommunications system that consists of a transmitter and receiver with six relay stations to connect them. The relays are situated so that the signal originating from one station can be picked up by the next two stations down the line. For example, a signal from the transmitter can be received by Relay 1 and Relay 2, a signal from Relay 1 can be received by Relay 2 and Relay 3, and so forth. Thus, this arrangement would require two consecutive relays to fail for the system to fail. A diagram of this configuration is shown in Figure 4.14.
Figure 4.14: RBD for the consecutive k-out-of-n : F system
This type of a configuration is also referred to as a complex system. Complex systems are discussed in the next section.
See Also:
RBDs and Analytical System Reliability
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