Expected Failure Time Plot: Difference between revisions
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{{UConstruction}} | <p><span class="fck_mw_template">{{UConstruction}}</span> | ||
</p> | |||
<h1>Expected Failure Time Plot</h1> | |||
<p>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. | |||
</p> | |||
<h2> Background & Calculations </h2> | |||
<p>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: | |||
</p><p><br /> | |||
</p> | |||
<table border="1" cellspacing="1" cellpadding="1" width="400" align="center"> | |||
<caption> Table 1: 5%, 50% and 95% Ranks for a sample size of 6. | |||
</caption> | |||
<tr> | |||
<th bgcolor="#cccccc" valign="middle" scope="col" align="center"> Order Number | |||
</th><th bgcolor="#cccccc" valign="middle" scope="col" align="center"> 5% | |||
</th><th bgcolor="#cccccc" valign="middle" scope="col" align="center"> 50% | |||
</th><th bgcolor="#cccccc" valign="middle" scope="col" align="center"> 95% | |||
</th></tr> | |||
<tr> | |||
<td valign="middle" align="center"> 1 | |||
</td><td valign="middle" align="center"> 0.85% | |||
</td><td valign="middle" align="center"> 10.91% | |||
</td><td valign="middle" align="center"> 39.30% | |||
</td></tr> | |||
<tr> | |||
<td valign="middle" align="center"> 2 | |||
</td><td valign="middle" align="center"> 6.29% | |||
</td><td valign="middle" align="center"> 26.45% | |||
</td><td valign="middle" align="center"> 58.18% | |||
</td></tr> | |||
<tr> | |||
<td valign="middle" align="center"> 3 | |||
</td><td valign="middle" align="center"> 15.32% | |||
</td><td valign="middle" align="center"> 42.14% | |||
</td><td valign="middle" align="center"> 72.87% | |||
</td></tr> | |||
<tr> | |||
<td valign="middle" align="center"> 4 | |||
</td><td valign="middle" align="center"> 27.13% | |||
</td><td valign="middle" align="center"> 57.86% | |||
</td><td valign="middle" align="center"> 84.68% | |||
</td></tr> | |||
<tr> | |||
<td valign="middle" align="center"> 5 | |||
</td><td valign="middle" align="center"> 41.82% | |||
</td><td valign="middle" align="center"> 73.55% | |||
</td><td valign="middle" align="center"> 93.71% | |||
</td></tr> | |||
<tr> | |||
<td valign="middle" align="center"> 6 | |||
</td><td valign="middle" align="center"> 60.70% | |||
</td><td valign="middle" align="center"> | |||
<p>89.09% | |||
</p> | |||
</td><td valign="middle" align="center"> | |||
<p>99.15% | |||
</p> | |||
</td></tr></table> | |||
<p><br /> | |||
</p><p>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, | |||
</p><p>or | |||
</p><p><img _fckfakelement="true" _fck_mw_math="R(t)=e^{\big({t \over \eta}\big)^\beta}" src="/images/math/9/b/2/9b21aed609d5cefddaae485bbfbc3a2f.png" /> | |||
</p><p>then for 0.85%, | |||
</p><p><br /><img _fckfakelement="true" _fck_mw_math="1-0.0085=e^{\big({t \over 100}\big)^2}" src="/images/math/d/b/e/dbe99885cf4bd0ea65638a820287544a.png" /> | |||
</p><p>and so forths as shown in the table below: | |||
</p><p><br /> | |||
</p> | |||
<table border="1" cellspacing="1" cellpadding="1" width="400" align="center"> | |||
<tr> | |||
<th bgcolor="#cccccc" scope="col"> Failure Order Number | |||
</th><th bgcolor="#cccccc" scope="col"> Lowest Expected Time-to-failure (hr) | |||
</th><th bgcolor="#cccccc" scope="col"> Median Expected Time-to-failure (hr) | |||
</th><th bgcolor="#cccccc" scope="col"> Highest Expected Time-to-failure (hr) | |||
</th></tr> | |||
< | <tr> | ||
<td valign="middle" align="center"> 1 | |||
</td><td valign="middle" align="center"> 9.25 | |||
</td><td valign="middle" align="center"> 33.99 | |||
</td><td valign="middle" align="center"> 70.66 | |||
</td></tr> | |||
<tr> | |||
<td valign="middle" align="center"> 2 | |||
</td><td valign="middle" align="center"> 25.48 | |||
</td><td valign="middle" align="center"> 55.42 | |||
</td><td valign="middle" align="center"> 93.37 | |||
</td></tr> | |||
<tr> | |||
<td valign="middle" align="center"> 3 | |||
</td><td valign="middle" align="center"> 40.77 | |||
</td><td valign="middle" align="center"> 73.97 | |||
</td><td valign="middle" align="center"> 114.21 | |||
</td></tr> | |||
<tr> | |||
<td valign="middle" align="center"> 4 | |||
</td><td valign="middle" align="center"> 56.26 | |||
</td><td valign="middle" align="center"> 92.96 | |||
</td><td valign="middle" align="center"> 136.98 | |||
</td></tr> | |||
<tr> | |||
<td valign="middle" align="center"> 5 | |||
</td><td valign="middle" align="center"> 73.60 | |||
</td><td valign="middle" align="center"> 115.33 | |||
</td><td valign="middle" align="center"> 166.34 | |||
</td></tr> | |||
<tr> | |||
<td valign="middle" align="center"> 6 | |||
</td><td valign="middle" align="center"> | |||
<p>96.64 | |||
</p> | |||
</td><td valign="middle" align="center"> 148.84 | |||
</td><td valign="middle" align="center"> 218.32 | |||
</td></tr></table> | |||
<p><br /><br /> | |||
</p><p><br /> | |||
</p><p><br /> | |||
</p><p><br /> | |||
</p><p><br /><br /> | |||
</p><a _fcknotitle="true" href="Category:Weibull++">Weibull++</a> <a _fcknotitle="true" href="Category:Test_Design">Test_Design</a> <a _fcknotitle="true" href="Category:Special_Tools">Special_Tools</a> | |||
96.64 | |||
<br><br> | |||
<br> | |||
<br> | |||
<br><br> | |||
Revision as of 18:24, 2 March 2011
UNDER CONSTRUCTION |
Please come back later. |
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:
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
<img _fckfakelement="true" _fck_mw_math="R(t)=e^{\big({t \over \eta}\big)^\beta}" src="/images/math/9/b/2/9b21aed609d5cefddaae485bbfbc3a2f.png" />
then for 0.85%,
<img _fckfakelement="true" _fck_mw_math="1-0.0085=e^{\big({t \over 100}\big)^2}" src="/images/math/d/b/e/dbe99885cf4bd0ea65638a820287544a.png" />
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 |
<a _fcknotitle="true" href="Category:Weibull++">Weibull++</a> <a _fcknotitle="true" href="Category:Test_Design">Test_Design</a> <a _fcknotitle="true" href="Category:Special_Tools">Special_Tools</a>