Template:Competing failure modes

Competing Failure Modes
Often, a group of products will fail due to more than one failure mode. One can take the view that the products could have failed due to any one of the possible failure modes, but since an item cannot fail more than one time, there can only be one failure mode for each failed product. In this view, the failure modes compete as to which causes the failure for each particular item. This can be viewed as a series system reliability model, with each failure mode composing a block of the series system. Competing failure modes analysis segregates the analyses of failure modes and then combines the results to provide an overall model for the product in question.

In order to begin analyzing data sets with more than one competing failure mode, one must perform a separate analysis for each failure mode. During each of these analyses, the failure times for all other failure modes not being analyzed are considered to be suspensions. This is because the units under test would have failed at some time in the future due to the failure mode being analyzed, had the unrelated (not analyzed) mode not occurred. Thus, in this case, the information available is that the mode under consideration did not occur and the unit under consideration accumulated test time without a failure due to the mode under consideration (or a suspension due to that mode).

Once the analysis for each separate failure mode has been completed (using the same principles as before), the resulting reliability equation for all modes is the product of the reliability equation for each mode, or:


 * $$R(t)={{R}_{1}}(t)\cdot {{R}_{2}}(t)\cdot ...\cdot {{R}_{n}}(t)$$

where $$n$$  is the total number of failure modes considered. This is the product rule for the reliability of series systems with statistically independent components, which states that the reliability for a series system is equal to the product of the reliability values of the components comprising the system. Do note that in Eqn. (sysrel)    is the reliability function based on any assumed life distribution. In Weibull++ this life distribution can be either the 2-parameter Weibull, lognormal, normal or the 1-parameter exponential.

Example 13
From Meeker & Escobar [27], the following table gives failure times for an electric component that has two failure modes.

One failure mode is due to random voltage spikes which cause failure by overloading the system (denoted as a $$V$$  in the table). The other failure mode is due to wear-out failures which usually happen only after the system has run for many cycles (this failure mode is denoted as a $$W$$  in the table).

Considering that these are competing failure modes, determine the overall reliability for the component at 100,000 cycles.

$$\begin{matrix} Number & Failure & Failure & Number & Failure & Failure \\ in State & Time* & Mode & in State & Time* & Mode \\ \text{1} & \text{2} & \text{V} & \text{1} & \text{147} & \text{W} \\ \text{1} & \text{10} & \text{V} & \text{1} & \text{173} & \text{V} \\ \text{1} & \text{13} & \text{V} & \text{1} & \text{181} & \text{W} \\ \text{2} & \text{23} & \text{V} & \text{1} & \text{212} & \text{W} \\ \text{1} & \text{28} & \text{V} & \text{1} & \text{245} & \text{W} \\ \text{1} & \text{30} & \text{V} & \text{1} & \text{247} & \text{V} \\ \text{1} & \text{65} & \text{V} & \text{1} & \text{261} & \text{V} \\ \text{1} & \text{80} & \text{V} & \text{1} & \text{266} & \text{W} \\ \text{1} & \text{88} & \text{V} & \text{1} & \text{275} & \text{W} \\ \text{1} & \text{106} & \text{V} & \text{1} & \text{293} & \text{W} \\ \text{1} & \text{143} & \text{V} & \text{8} & \text{300} & \text{suspended} \\ \end{matrix}$$


 * Failure times given are in thousands of cycles.

Solution to Example 13
We will begin by performing a Weibull analysis of the voltage spike ( $$V$$ ) failure mode. In order to do this, we must consider all of the failures for the wear-out mode to be suspensions. The input data for the analysis are shown next:



Analyzing this data set using the maximum likelihood method (recommended due to the number of suspensions in the data) and assuming a Weibull distribution, we obtain the parameters $${{\beta }_{V}}=0.6711$$  and  $${{\eta }_{V}}=449.4$$. The reliability for this failure mode at $$t=100$$  is  $${{R}_{V}}(100)=0.694$$.

We follow an identical procedure for the wear-out failure mode, counting only the $$W$$  entries as failures and assuming the  $$V$$  entries are suspensions. This is shown next.



Once again, analyzing with a Weibull distribution with maximum likelihood estimators, we obtain the parameters $${{\beta }_{W}}=4.337$$  and  $${{\eta }_{W}}=340.4$$. The reliability for this failure mode at $$t=100$$  is  $${{R}_{W}}(100)=0.995$$.

We can now use Eqn. (sysrel) to determine the overall system reliability at 100,000 cycles:


 * $$\begin{align}

& {{R}_{sys}}(100)= & {{R}_{V}}(100)\cdot {{R}_{W}}(100) \\ & = & 0.694\cdot 0.995 \\ & = & 0.69053 \end{align}$$

Or the reliability of the unit (or system) under both modes is $${{R}_{sys}}(100)=69.053%.$$

Note that Weibull++ can perform this analysis for you automatically. To accomplish this, the data would be entered in a single data sheet and competing failure modes chosen as the analysis method. This is shown in the next graphic.



Plotting Competing Failure Modes
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When plotting competing failure modes in Weibull++, your plots can contain the combined mode line as well as the individual mode lines. The User's Guide describes how these options can be turned on or off.