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=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 visual.  
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 visual.  


= Background & Calculations  =
== 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:  
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:  


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Revision as of 18:11, 2 March 2011

UNDER CONSTRUCTION
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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 visual.

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:


Table 1: 5%, 50% and 95% Ranks for a sample size of 6. 
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:


Failure 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