Weibull++ Example 10 - Competing Failure Modes Analysis

Download Example File for Version 10 (*.rsgz10) or Version 9 (*.rsr9)

 

Background

An electronic component has two competing failure modes. One failure mode is due to random voltage spikes, which cause failure by overloading the system. The other failure mode is due to wear-out failures, which usually happen only after the system has run for many cycles. The objective of this test is to determine the overall reliability for the component at 100,000 cycles.

Experiment and Data

(This example has been abstracted from Example 15.6 from the Meeker and Escobar textbook Statistical Methods for Reliability Data, published by John Wiley and Sons.)

30 units are tested and the failure times are recorded in the following tables. The failures that are due to the random voltage spikes are denoted by a V in the table. The failures that are due to wearout failures are denoted by a W.

Number in State Failure Time* Failure Mode   Number in State Failure Time* Failure Mode
1 2 V   1 147 W
1 10 V   1 173 V
1 13 V   1 181 W
2 23 V   1 212 W
1 28 V   1 245 W
1 30 V   1 247 V
1 65 V   1 261 V
1 80 V   1 266 W
1 88 V   1 275 W
1 106 V   1 293 W
1 143 V   8 300 suspended

* Failure times given in thousands of cycles.

Analysis

Step 1: Create a new Weibull++ standard folio data sheet that holds grouped times-to-failure data with suspensions.

Step 2: Enter the data into the folio, using the Subset ID column to identify the failure mode. On the control panel, choose Competing Failure Modes > CFM-Weibull. Then select the MLE analysis method, as shown next.

Figure 1: Standard folio with data entered.
Figure 1: Standard folio with data entered.

Step 3: Click Calculate. In the window that appears, identify Mode 1 as the V failure mode and Mode 2 as the W failure mode, as shown next.

Figure 2: Select Subsets window for associating subset IDs with failure modes.
Figure 2: Select Subsets window for associating subset IDs with failure modes.

Step 4: Open the Quick Calculation Pad and determine the overall reliability of the component at 100,000 cycles. The result is estimated to be 69.1%, as shown next.

Figure 3: Calculating reliability at 100,000 cycles.
Figure 3: Calculating reliability at 100,000 cycles.

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