Expected Failure Time Plot
Expected Failure Time Plot
When a reliability life test is planned it is useful to visualize the expected outcome of the experiment. The Expected Failure Time Plot (introduced by ReliaSoft in Weibull++ 8)provides such a visual. Figure 1 below shows such a plot for h a sample size of 5 and an assumed Weibull distribution with β = 2 and η = 2,000 hrs and at a 90% confidence.
Fig. 1: Expected Failure Time Plot with a sample size of 5, an assumed Weibull distribution with β = 2 and η = 2,000 hrs and at a 90% confidence. |
Interpreting the EFT Plot
The EFT plot in Figure 1 is based on the assumption that the time-to-failure of the units follows a Weibull distribution with the given parameters. Based on that assumption the median time to failure of each unit would be as follows: 558; 921; 1,249; 1,615 and 2,155 hrs. In other words with a 50% probability we expect the first failure to occur at 558 hours and the last at 2,155 hours. Obviously, and since this is a probabilistic event, there is a 50% probability that the number is lower or higher than the stated median. While this is informative, in practical situations wjhat would be of more interest is the range of plausibale values, or a range of values that we are likely to observe X% of the time. As an example a more useful approach would be to come up with a range of times that a failure would occure with a 90% probability (or 90% 2-sided confidence interval). Table 1
Failure Number |
Expected Time 5% probability that it will be less. |
Expected Time Median (hrs) |
Expected Time 5% probability that it will be higher. |
---|---|---|---|
1 | 152 | 558 | 1161 |
2 | 423 | 921 | 1552 |
3 | 687 | 1249 | 1935 |
4 | 971 | 1615 | 2405 |
5 | 1,339 | 2155 | 3211 |
Background & Calculations
Using the cumulative binomial, for a defined sample size, one can compute a rank (Median Rank if at 50% probability) for each ordered failure. As an example and for a sample size of 6 the 5%, 50% and 95% ranks would be as follows:
Order Number | 5% | 50% | 95% |
---|---|---|---|
1 | 0.85% | 10.91% | 39.30% |
2 | 6.29% | 26.45% | 58.18% |
3 | 15.32% | 42.14% | 72.87% |
4 | 27.13% | 57.86% | 84.68% |
5 | 41.82% | 73.55% | 93.71% |
6 | 60.70% |
89.09% |
99.15% |
Furthermore, consider that for the units to be tested the underlying reliability model assumption is given by a Weibull distribution with β = 2, and η = 100 hr. Then the median time to failure of the first unit on test can be determined by solving the Weibull reliability equation for t, at each probability,
or [math]\displaystyle{ R(t)=e^{\big({t \over \eta}\big)^\beta} }[/math]
then for 0.85%,
[math]\displaystyle{ 1-0.0085=e^{\big({t \over 100}\big)^2} }[/math]
and so forths as shown in the table below:
Order Number | Lowest Expected Time-to-failure (hr) | Median Expected Time-to-failure (hr) | Highest Expected Time-to-failure (hr) |
---|---|---|---|
1 | 9.25 | 33.99 | 70.66 |
2 | 25.48 | 55.42 | 93.37 |
3 | 40.77 | 73.97 | 114.21 |
4 | 56.26 | 92.96 | 136.98 |
5 | 73.60 | 115.33 | 166.34 |
6 |
96.64 |
148.84 | 218.32 |
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