The fault tree below visualizes how probabilities propagate through a fault tree.
- The sliders can be moved to set the basic event probabilitie.
- The Sankey diagrams show the failure probabilities for the intermediate events.
The Fault Tree is made up out of Basic Events (BEs, the boxes), AND gates, OR gates, and k/N (k out of N) gates. Basic Events represent elemental failures of components of the overall system. Gates are used to describe how these elemental failures interact with each other, leading to the eventual failure of the overall system. Every gate has 2 or more connections to other gates or Basic Events. The OR gate models something has failed if any of the connected gates/BEs has failed. The AND gate describes something has failed if all the connected gates/BEs have failed. The k/N gate describes something has failed if k out of the N connected gates have failed.
The example below illustrates the potential total failure of a road trip:
- The road trip fails if the car fails and the phone fails, since road side assistance can’t be called.
- The car fails if the engine fails, or the tires fail.
- The tires fail if 2 out of 5 of the individual, identical tires (4 used normally and 1 spare) fail.
- The cell phone fails if its power fails, or the connection fails.
- The cell phone runs out of power if its battery fails, or the car engine fails, since it can’t be charged by the alternator.
The failure probability of the Basic Events and gates is described with a value from 0 to 1 (0% to 100%). The probabilities of the Basic events in the model can be adjusted with the sliders under the Basic Events.
An and gate fails if all its children fail, so the failure probability of AND gates is calculated by multiplying the failure probability of all its children: Pg = Pc1 * Pc2.
The failure probability of OR gates is calculated through the inverse. An OR gate fails if ANY of its children fails, so it has not failed only if ALL its children have not failed, which is analogous to the AND gate. The probability an OR gate has failed is 1 minus the probability it has not failed, which is the probabity for each child to not have failed multiplied with each other, which in turn is 1 minus the probability the child has failed: Pg = 1 – [ (1-Pc1) * (1-Pc2) ].
Calculating the failure probability of the k/N gate is significantly more involved, and usually includes calculating and adding the probability of every unique comination of states for the children that would result in the gate failing, or not failing and taking the 1 – not failing probability.
In the below example a 2 out of 5 gate is used, and there are only 6 combinations of child states where the gate has not failed: All children have not failed, or a single child has failed and the others have not failed, once for each of the 5 children. Thus, the failure probability is 1 – the probability of not failing, being the probability of each child not failing plus the probability of a single child failing for each distinct child:
Pg = 1 – [ (1-Pc1) * (1-Pc2) * (1-Pc3) * (1-Pc4) * (1-Pc5)
+ (Pc1) * (1-Pc2) * (1-Pc3) * (1-Pc4) * (1-Pc5)
+ (1-Pc1) * (Pc2) * (1-Pc3) * (1-Pc4) * (1-Pc5)
+ (1-Pc1) * (1-Pc2) * (Pc3) * (1-Pc4) * (1-Pc5)
+ (1-Pc1) * (1-Pc2) * (1-Pc3) * (Pc4) * (1-Pc5)
+ (1-Pc1) * (1-Pc2) * (1-Pc3) * (1-Pc4) * (Pc5) ]
In the below example, the failure probability of each of the children of the k/N gate is identical, so this calculation can be further simplified:
Pg = 1 – [ (1-Pc)^5 + 5 * (Pc) * (1-Pc)^4 ]