Expected Failure Time Plot: Difference between revisions

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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="71918b0" sourceindex="13"
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[[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="71a66f0" sourceindex="21">β = 2</span> and&nbsp;<span class="texhtml" siber__q92dpb7seovvtbh5__vptr="71d3220" sourceindex="22">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">β = 2</span> and&nbsp;<span class="texhtml">η = 2,000</span> hrs and at a 90% confidence.  


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== Interpreting the EFT Plot  ==
== 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. &nbsp;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. &nbsp;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. &nbsp;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&nbsp;
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. &nbsp;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. &nbsp;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. &nbsp;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&nbsp;  


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<br>  


{| border="1" align="center" cellpadding="1" cellspacing="1" siber__q92dpb7seovvtbh5__vptr="79adb60" sourceindex="31"
{| border="1" align="center" cellpadding="1" cellspacing="1"
|+ '''Table1'''. Expected times-to-failure with a 90% 2-sided confidence interval.
|+ '''Table1'''. Expected times-to-failure with a 90% 2-sided confidence interval.  
|- siber__q92dpb7seovvtbh5__vptr="79a7d20" sourceindex="34"
|-
! scope="col" siber__q92dpb7seovvtbh5__vptr="799dd70" sourceindex="35" align="center" valign="middle" | Failure<br> Number
! scope="col" align="center" valign="middle" | Failure<br> Number  
! scope="col" siber__q92dpb7seovvtbh5__vptr="79a7520" sourceindex="37" |  
! scope="col" |  
Expected Time <br>Low (hrs)
Expected Time <br>Low (hrs)  


'''5% probability that it will be less.'''
'''5% probability that it will be less.'''  


! scope="col" siber__q92dpb7seovvtbh5__vptr="79a2260" sourceindex="42" | Expected Time<br>Median (hrs)
! scope="col" | Expected Time<br>Median (hrs)  
! scope="col" siber__q92dpb7seovvtbh5__vptr="79a7b70" sourceindex="44" |  
! scope="col" |  
Expected Time<br>High (hrs)
Expected Time<br>High (hrs)  


5% probability that it will be higher.
5% probability that it will be higher.  


|- siber__q92dpb7seovvtbh5__vptr="79a7ce0" sourceindex="49"
|-
| siber__q92dpb7seovvtbh5__vptr="79a2320" sourceindex="50" | 1
| 1  
| siber__q92dpb7seovvtbh5__vptr="79adc40" sourceindex="51" | 152
| 152  
| siber__q92dpb7seovvtbh5__vptr="79a2290" sourceindex="52" | 558
| 558  
| siber__q92dpb7seovvtbh5__vptr="799d210" sourceindex="53" | 1161
| 1161
|- siber__q92dpb7seovvtbh5__vptr="79a2460" sourceindex="54"
|-
| siber__q92dpb7seovvtbh5__vptr="79a25f0" sourceindex="55" | 2
| 2  
| siber__q92dpb7seovvtbh5__vptr="799dfa0" sourceindex="56" | 423
| 423  
| siber__q92dpb7seovvtbh5__vptr="79a25e0" sourceindex="57" | 921
| 921  
| siber__q92dpb7seovvtbh5__vptr="79a24d0" sourceindex="58" | 1552
| 1552
|- siber__q92dpb7seovvtbh5__vptr="799dd40" sourceindex="59"
|-
| siber__q92dpb7seovvtbh5__vptr="79a23a0" sourceindex="60" | 3
| 3  
| siber__q92dpb7seovvtbh5__vptr="79a2590" sourceindex="61" | 687
| 687  
| siber__q92dpb7seovvtbh5__vptr="79a2740" sourceindex="62" | 1249
| 1249  
| siber__q92dpb7seovvtbh5__vptr="79a2600" sourceindex="63" | 1935
| 1935
|- siber__q92dpb7seovvtbh5__vptr="799db00" sourceindex="64"
|-
| siber__q92dpb7seovvtbh5__vptr="79a2610" sourceindex="65" | 4
| 4  
| siber__q92dpb7seovvtbh5__vptr="79a2780" sourceindex="66" | 971
| 971  
| siber__q92dpb7seovvtbh5__vptr="79a23d0" sourceindex="67" | 1615
| 1615  
| siber__q92dpb7seovvtbh5__vptr="79a20b0" sourceindex="68" | 2405
| 2405
|- siber__q92dpb7seovvtbh5__vptr="79a2770" sourceindex="69"
|-
| siber__q92dpb7seovvtbh5__vptr="79a2510" sourceindex="70" | 5
| 5  
| siber__q92dpb7seovvtbh5__vptr="79a2430" sourceindex="71" | 1,339
| 1,339  
| siber__q92dpb7seovvtbh5__vptr="79a2670" sourceindex="72" | 2155
| 2155  
| siber__q92dpb7seovvtbh5__vptr="79a2980" sourceindex="73" align="center" valign="middle" | 3211
| align="center" valign="middle" | 3211
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{| border="1" cellspacing="1" cellpadding="1" width="400" align="center" siber__q92dpb7seovvtbh5__vptr="71bd7e0" sourceindex="90"
{| border="1" cellspacing="1" cellpadding="1" width="400" align="center"
|+ '''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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! bgcolor="#cccccc" valign="middle" scope="col" align="center" siber__q92dpb7seovvtbh5__vptr="7103c80" sourceindex="95" | Order Number  
! bgcolor="#cccccc" valign="middle" scope="col" align="center" | Order Number  
! bgcolor="#cccccc" valign="middle" scope="col" align="center" siber__q92dpb7seovvtbh5__vptr="3add00" sourceindex="96" | 5%  
! bgcolor="#cccccc" valign="middle" scope="col" align="center" | 5%  
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| valign="middle" align="center" siber__q92dpb7seovvtbh5__vptr="71084f0" sourceindex="116" | 27.13%  
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| valign="middle" align="center" siber__q92dpb7seovvtbh5__vptr="7108390" sourceindex="117" | 57.86%  
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| valign="middle" align="center" siber__q92dpb7seovvtbh5__vptr="718bf60" sourceindex="118" | 84.68%
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89.09%  
89.09%  


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99.15%  
99.15%  


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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="71d9380" sourceindex="134">β = 2</span>, and <span class="texhtml" siber__q92dpb7seovvtbh5__vptr="718b6c0" sourceindex="135">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">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  
or  
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{| border="1" cellspacing="1" cellpadding="1" width="400" align="center" siber__q92dpb7seovvtbh5__vptr="71a1110" sourceindex="146"
{| border="1" cellspacing="1" cellpadding="1" width="400" align="center"
|+ '''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="7135690" sourceindex="149">β = 2</span>, and <span class="texhtml" siber__q92dpb7seovvtbh5__vptr="71d9740" sourceindex="150">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">β = 2</span>, and <span class="texhtml">η = 100</span> hr.'''  
|- siber__q92dpb7seovvtbh5__vptr="71bdcc0" sourceindex="152"
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! bgcolor="#cccccc" scope="col" siber__q92dpb7seovvtbh5__vptr="719bf50" sourceindex="153" | Order Number  
! bgcolor="#cccccc" scope="col" | Order Number  
! bgcolor="#cccccc" scope="col" siber__q92dpb7seovvtbh5__vptr="7191fe0" sourceindex="154" | Lowest Expected Time-to-failure (hr)  
! bgcolor="#cccccc" scope="col" | Lowest Expected Time-to-failure (hr)  
! bgcolor="#cccccc" scope="col" siber__q92dpb7seovvtbh5__vptr="7185a20" sourceindex="155" | Median Expected Time-to-failure (hr)  
! bgcolor="#cccccc" scope="col" | Median Expected Time-to-failure (hr)  
! bgcolor="#cccccc" scope="col" siber__q92dpb7seovvtbh5__vptr="71cd090" sourceindex="156" | Highest Expected Time-to-failure (hr)
! bgcolor="#cccccc" scope="col" | Highest Expected Time-to-failure (hr)
|- siber__q92dpb7seovvtbh5__vptr="7191a20" sourceindex="157"
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| valign="middle" align="center" siber__q92dpb7seovvtbh5__vptr="71039a0" sourceindex="159" | 9.25  
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|- siber__q92dpb7seovvtbh5__vptr="3fa3f0" sourceindex="167"
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96.64  
96.64  


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Revision as of 12:13, 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

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 


Table1. Expected times-to-failure with a 90% 2-sided confidence interval.
Failure
Number

Expected Time
Low (hrs)

5% probability that it will be less.

Expected Time
Median (hrs)

Expected Time
High (hrs)

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:


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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