Criticality Analysis

Failure mode criticality, as used in FMECA can be generically described as the joint probability of three events:
 * 1) The probability of failure of the component;
 * 2) the probability that the mode under consideration was the culprit; and
 * 3) the probability that this mode would result in a system failure ( which for the case of non-redundancy or if in series =1).

This number is then used to provide relative ranking for the different failure modes. This general approach is described in Kececioglu and expanded further here.

General Criticality Number Computation
Then mathematically and for a specific failure mode,


 * $${{C}_{\bmod e}}={{\beta }_{\bmod e}}\cdot {{\alpha }_{\bmod e}}\cdot Q{{(t)}_{component}}$$

Where
 * $${{C}_{component}}=$$Criticality number for the failure mode.
 * $${{\beta }_{\bmod e}}=$$Conditional probability of mission loss.
 * $${{\alpha }_{\bmod e}}=$$Failure mode ratio.
 * $$Q{{(t)}_{\bmod e}}=$$Unreliability or probability of failure due to that mode at time t.

In the case that probability of occurrence of each mode is defined, and the modes are assumed to be in series, then a better way to obtain a criticality value is:


 * $${{C}_{\bmod {{e}_{i}}}}={{\beta }_{\bmod {{e}_{i}}}}\cdot \frac{\frac{Q{{(t)}_{\bmod {{e}_{i}}}}}{Q{{(t)}_{Component}}}}{\sum\limits_{i=1}^{M}{\frac{Q{{(t)}_{\bmod {{e}_{i}}}}}{Q{{(t)}_{Component}}}}}$$

Note that this differs from the specialized definition given in Task 102 of MIL-STD-1629A, Section 3.2.1.6.

Specifically a procedure is given for computing the “Failure mode criticality number” when an assumption of a constant failure mode (i.e. Exponential distribution) is used and for small values of lambda:

This is given by:
 * $${{C}_{m}}=\beta \alpha {{\lambda }_{P}}t$$

where
 * $${{C}_{m}}=$$Criticality number for a failure mode.
 * $$\beta =$$Conditional probability of mission loss.
 * $$\alpha =$$Failure mode ratio.
 * $${{\lambda }_{P}}=$$Part failure rate.
 * $$t=$$Mission Duration.

It is important to note that in this formulation, and in addition to the constant failure rate assumption, assumes that part failure rate values for a mode are relatively small and thus the following simplification is made:


 * $$Q(t)=1-{{e}^{-{{\lambda }_{P}}\cdot t}}\approx {{\lambda }_{P}}\cdot t$$

This approximation holds true when using an exponential distribution and small values of lambda (<1E-4), but not for all values. ReliaSoft chose to use the general formulation instead of the specialized and limited 1629A formulation.

Example
To illustrate this assume the above Fault Tree with 3 failure modes, FM1, FM2 and FM3. The unreliability is defined by the failure model of each mode at t=1,000 hrs (in this case FM1 is Weibull (3, 3000 hrs) and FM2 and 3 are Exponential with MTTF=10,000 hrs). Then

$$\begin{align} & \sum\limits_{i=1}^{M}{\frac{Q{{(t)}_{\bmod {{e}_{i}}}}}{Q{{(t)}_{Component}}}}=\frac{0.036360+0.095163+0.095163}{0.211038} \\ & =\frac{0.226686}{0.211038} \\ & =1.074148 \end{align}$$

Derivation
$$\begin{align} & {{C}_{FM1}}={{\beta }_{FM1}}\cdot \frac{\frac{Q{{(t)}_{FM1}}}{Q{{(t)}_{Component}}}}{\sum\limits_{i=1}^{M}{\frac{Q{{(t)}_{\bmod {{e}_{i}}}}}{Q{{(t)}_{Component}}}}} \\ & =1\cdot \frac{\frac{0.36360}{0.211038}}{1.074148} \\ & =0.160398 \end{align}$$

$$\begin{align} & {{C}_{FM2}}={{\beta }_{FM2}}\cdot \frac{\frac{Q{{(t)}_{FM2}}}{Q{{(t)}_{Component}}}}{\sum\limits_{i=1}^{M}{\frac{Q{{(t)}_{\bmod {{e}_{i}}}}}{Q{{(t)}_{Component}}}}} \\ & =1\cdot \frac{\frac{0.095163}{0.211038}}{1.074148} \\ & =0.419801 \end{align}$$

and

$$\begin{align} & {{C}_{FM3}}={{\beta }_{FM3}}\cdot \frac{\frac{Q{{(t)}_{FM3}}}{Q{{(t)}_{Component}}}}{\sum\limits_{i=1}^{M}{\frac{Q{{(t)}_{\bmod {{e}_{i}}}}}{Q{{(t)}_{Component}}}}} \\ & =1\cdot \frac{\frac{0.095163}{0.211038}}{1.074148} \\ & =0.419801 \end{align}$$

Derivation
Given component failed before time T, the probability that it is caused by mode B is: The Exact Solution is:

$$$$			(1) We can get the upper bound of P1 $${{P}_{1}}=\frac{\int_{0}^{T}{{{f}_{B}}(t){{R}_{A}}(t)dt}}<\frac{\int_{0}^{T}{{{f}_{B}}(t)dt}}=\frac=\frac{{{Q}_{A}}+{{Q}_{B}}-{{Q}_{AB}}}$$ We can get the lower bound of P1 $$\begin{align} & {{P}_{1}}=\frac{\int_{0}^{T}{{{f}_{B}}(t){{R}_{A}}(t)dt}}>\frac{\int_{0}^{T}{{{f}_{B}}(t){{R}_{A}}(T)dt}}=\frac{{{Q}_{B}}{{R}_{A}}(T)}=\frac{{{Q}_{B}}(1-{{Q}_{A}})}{{{Q}_{A}}+{{Q}_{B}}-{{Q}_{A}}{{Q}_{B}}} \\ & =\frac{{{Q}_{B}}-{{Q}_{B}}{{Q}_{A}}}{{{Q}_{A}}+{{Q}_{B}}-{{Q}_{A}}{{Q}_{B}}} \\ \end{align}$$ So the probability that the failure is caused by mode B is $$\frac{{{Q}_{B}}-{{Q}_{B}}{{Q}_{A}}}{{{Q}_{A}}+{{Q}_{B}}-{{Q}_{A}}{{Q}_{B}}}<{{P}_{1}}<\frac{{{Q}_{A}}+{{Q}_{B}}-{{Q}_{AB}}}$$ We can use the approximated value: ... It is a value with the bounds. Similarly, given component failed before time T, the probability that it is caused by mode A is: The Exact Solution is: (2) and the sum of P1 and P2 is: $$\begin{align} & {{P}_{1}}+{{P}_{2}}=\frac{\int_{0}^{T}{{{f}_{A}}(t){{R}_{B}}(t)dt}+\int_{0}^{T}{{{f}_{B}}(t){{R}_{A}}(t)dt}}{1-{{R}_{A}}(T){{R}_{B}}(T)} \\ & =\frac{{{F}_{A}}(T)-\int_{0}^{T}{{{f}_{A}}(t){{F}_{B}}(t)dt}+{{F}_{B}}(T)-\left[ {{F}_{A}}(T){{F}_{B}}(T)\left| \begin{align} & T \\ & 0 \\ \end{align} \right.-\int_{0}^{T}{{{F}_{B}}(T){{f}_{A}}(T)dt} \right]}{1-{{R}_{A}}(T){{R}_{B}}(T)} \\ & =\frac{{{F}_{A}}(T)+{{F}_{B}}(T)-{{F}_{A}}(T){{F}_{B}}(T)}{1-{{R}_{A}}(T){{R}_{B}}(T)} \\ & =1 \\ \end{align}$$  			(3) We can use the approximated value: . It is a value with the bounds. So $${{\tilde{P}}_{1}}+{{\tilde{P}}_{2}}=1$$.