Expected Failure Time Plot: Difference between revisions

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#REDIRECT [[Reliability_Test_Design#Expected_Failure_Times_Plots]]
 
 
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:
 
<br>
 
{| border="1" cellspacing="1" cellpadding="1" width="400" align="center"
|+ Table 1: 5%, 50% and 95% Ranks for a sample size of 6.&nbsp;
|-
! bgcolor="#cccccc" valign="middle" scope="col" align="center" | Order Number
! bgcolor="#cccccc" valign="middle" scope="col" align="center" | 5%
! bgcolor="#cccccc" valign="middle" scope="col" align="center" | 50%
! bgcolor="#cccccc" valign="middle" scope="col" align="center" | 95%
|-
| valign="middle" align="center" | 1
| valign="middle" align="center" | 0.85%
| valign="middle" align="center" | 10.91%
| valign="middle" align="center" | 39.30%
|-
| valign="middle" align="center" | 2
| valign="middle" align="center" | 6.29%
| valign="middle" align="center" | 26.45%
| valign="middle" align="center" | 58.18%
|-
| valign="middle" align="center" | 3
| valign="middle" align="center" | 15.32%
| valign="middle" align="center" | 42.14%
| valign="middle" align="center" | 72.87%
|-
| valign="middle" align="center" | 4
| valign="middle" align="center" | 27.13%
| valign="middle" align="center" | 57.86%
| valign="middle" align="center" | 84.68%
|-
| valign="middle" align="center" | 5
| valign="middle" align="center" | 41.82%
| valign="middle" align="center" | 73.55%
| valign="middle" align="center" | 93.71%
|-
| valign="middle" align="center" | 6
| valign="middle" align="center" | 60.70%
| valign="middle" align="center" |
89.09%
 
| valign="middle" align="center" |
99.15%
 
|}
 
<br>
 
Furthermore, consider that for the units to be tested the underlying reliability model assumption is given by a Weibull distribution with <span class="texhtml">β = 2</span>, and <span class="texhtml">η = 100</span> 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>R(t)=e^{\big({t \over \eta}\big)^\beta}</math>
 
then for 0.85%,
 
<br><math>1-0.0085=e^{\big({t \over 100}\big)^2}</math>
 
and so forths as shown in the table below:
 
<br>
 
{| border="1" cellspacing="1" cellpadding="1" width="400" align="center"
|-
! bgcolor="#cccccc" scope="col" | Failure Order Number
! bgcolor="#cccccc" scope="col" | Lowest Expected Time-to-failure (hr)
! bgcolor="#cccccc" scope="col" | Median Expected Time-to-failure (hr)
! bgcolor="#cccccc" scope="col" | Highest Expected Time-to-failure (hr)
|-
| valign="middle" align="center" | 1
| valign="middle" align="center" | 9.25
| valign="middle" align="center" | 33.99
| valign="middle" align="center" | 70.66
|-
| valign="middle" align="center" | 2
| valign="middle" align="center" | 25.48
| valign="middle" align="center" | 55.42
| valign="middle" align="center" | 93.37
|-
| valign="middle" align="center" | 3
| valign="middle" align="center" | 40.77
| valign="middle" align="center" | 73.97
| valign="middle" align="center" | 114.21
|-
| valign="middle" align="center" | 4
| valign="middle" align="center" | 56.26
| valign="middle" align="center" | 92.96
| valign="middle" align="center" | 136.98
|-
| valign="middle" align="center" | 5
| valign="middle" align="center" | 73.60
| valign="middle" align="center" | 115.33
| valign="middle" align="center" | 166.34
|-
| valign="middle" align="center" | 6
| valign="middle" align="center" |
96.64
 
| valign="middle" align="center" | 148.84
| valign="middle" align="center" | 218.32
|}
 
<br><br>
 
<br>
 
<br>
 
 
 
<br><br>
 
[[Category:Weibull++]] [[Category:Test_Design]] [[Category:Special_Tools]]

Latest revision as of 22:51, 21 August 2012

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