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= {{Pending}}Expected Failure Time Plot  =
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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 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.  
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.  


{| width="200" border="0" cellpadding="1" cellspacing="1" align="center" siber__q92dpb7seovvtbh5__vptr="717b7a0" sourceindex="11"
{| width="200" border="0" cellpadding="1" cellspacing="1" align="center" siber__q92dpb7seovvtbh5__vptr="71cd140" sourceindex="13"
|- siber__q92dpb7seovvtbh5__vptr="71914d0" sourceindex="13"
|- siber__q92dpb7seovvtbh5__vptr="710e950" sourceindex="15"
| siber__q92dpb7seovvtbh5__vptr="7213f80" sourceindex="14" |  
| siber__q92dpb7seovvtbh5__vptr="71a1840" sourceindex="16" |  
[[Image:EFTP1.png|border|center|700px|Expected Failure Time Plot with a sample size of 5, an assumed Weibull distribution with b=2 and h-1,500 hrs and at a 90% confidence.]]  
[[Image:EFTP1.png|border|center|700px|Expected Failure Time Plot with a sample size of 5, an assumed Weibull distribution with b=2 and h-1,500 hrs and at a 90% confidence.]]  


'''Fig. 1:''' Expected Failure Time Plot with a sample size of 5, an assumed Weibull distribution with <span class="texhtml" siber__q92dpb7seovvtbh5__vptr="7191eb0" sourceindex="19">β = 2</span> and&nbsp;<span class="texhtml" siber__q92dpb7seovvtbh5__vptr="7103640" sourceindex="20">η = 2,000</span> hrs and at a 90% confidence.  
'''Fig. 1:''' Expected Failure Time Plot with a sample size of 5, an assumed Weibull distribution with <span class="texhtml" siber__q92dpb7seovvtbh5__vptr="71cd650" sourceindex="21">β = 2</span> and&nbsp;<span class="texhtml" siber__q92dpb7seovvtbh5__vptr="71cd040" sourceindex="22">η = 2,000</span> hrs and at a 90% confidence.  


|- siber__q92dpb7seovvtbh5__vptr="710e6e0" sourceindex="21"
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{| border="1" cellspacing="1" cellpadding="1" width="400" align="center" siber__q92dpb7seovvtbh5__vptr="71cdbc0" sourceindex="45"
|+ '''Table 1: 5%, 50% and 95% Ranks for a sample size of 6.&nbsp;'''  
|+ '''Table 1: 5%, 50% and 95% Ranks for a sample size of 6.&nbsp;'''  
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89.09%  
89.09%  


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Furthermore, consider that for the units to be tested the underlying reliability model assumption is given by a Weibull distribution with <span class="texhtml" siber__q92dpb7seovvtbh5__vptr="71de090" sourceindex="87">β = 2</span>, and <span class="texhtml" siber__q92dpb7seovvtbh5__vptr="7113f10" sourceindex="88">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,  
Furthermore, consider that for the units to be tested the underlying reliability model assumption is given by a Weibull distribution with <span class="texhtml" siber__q92dpb7seovvtbh5__vptr="7182ae0" sourceindex="89">β = 2</span>, and <span class="texhtml" siber__q92dpb7seovvtbh5__vptr="401970" sourceindex="90">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  
or  
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{| border="1" cellspacing="1" cellpadding="1" width="400" align="center" siber__q92dpb7seovvtbh5__vptr="3fa660" sourceindex="99"
{| border="1" cellspacing="1" cellpadding="1" width="400" align="center" siber__q92dpb7seovvtbh5__vptr="7171dd0" sourceindex="101"
|+ '''Table 2: Times corresponding to the 5%, 50% and 95% Ranks for a sample size of 6. and assuming Weibull distribution with <span class="texhtml" siber__q92dpb7seovvtbh5__vptr="71717c0" sourceindex="102">β = 2</span>, and <span class="texhtml" siber__q92dpb7seovvtbh5__vptr="71d3c50" sourceindex="103">η = 100</span> hr.'''  
|+ '''Table 2: Times corresponding to the 5%, 50% and 95% Ranks for a sample size of 6. and assuming Weibull distribution with <span class="texhtml" siber__q92dpb7seovvtbh5__vptr="7171e60" sourceindex="104">β = 2</span>, and <span class="texhtml" siber__q92dpb7seovvtbh5__vptr="711de50" sourceindex="105">η = 100</span> hr.'''  
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! bgcolor="#cccccc" scope="col" siber__q92dpb7seovvtbh5__vptr="71c32c0" sourceindex="107" | Lowest Expected Time-to-failure (hr)  
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96.64  
96.64  


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Revision as of 11:53, 10 March 2011


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

This EFT plot is based on an 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.  



1 151.9291 558.4953 1161.0655
2 422.9902 920.5177 1552.4924
3 687.0619 1248.8311 1935.3269
4 971.4768 1614.8296 2405.2631
5 1339.0443 2144.7716 3211.8056


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

R(t)=e^{\big({t \over \eta}\big)^\beta}

then for 0.85%,


1-0.0085=e^{\big({t \over 100}\big)^2}


and so forths as shown in the table below:


Table 2: Times corresponding to the 5%, 50% and 95% Ranks for a sample size of 6. and assuming Weibull distribution with β = 2, and η = 100 hr.
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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