Reliability Test Design - Revision history

Lisa Hacker: /* Expected Failure Times Plots */

2015-12-10T21:22:35Z

Expected Failure Times Plots

← Older revision		Revision as of 21:22, 10 December 2015
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	Using the above equation, for a given ''Q'', we can get the corresponding time ''t''. The above calculation gives the '''median''' of each failure time for ''CL'' = 0.5. If we set ''CL'' at different values, the confidence bounds of each failure time can be obtained. For the above example, if we set ''CL''=0.9, from the calculated ''Q'' we can get the upper bound of the time for each failure. The calculated ''Q'' is given in the next figure:		Using the above equation, for a given ''Q'', we can get the corresponding time ''t''. The above calculation gives the '''median''' of each failure time for ''CL'' = 0.5. If we set ''CL'' at different values, the confidence bounds of each failure time can be obtained. For the above example, if we set ''CL''=0.9, from the calculated ''Q'' we can get the upper bound of the time for each failure. The calculated ''Q'' is given in the next figure:

	[[Image: Expected Failure Plot 0.9 Rank.png\|center\|~~600px~~]]		[[Image: Expected Failure Plot 0.9 Rank.png\|center\|]]

	If we set ''CL''=0.1, from the calculated ''Q'' we can get the lower bound of the time for each failure. The calculated ''Q'' is given in the figure below:		If we set ''CL''=0.1, from the calculated ''Q'' we can get the lower bound of the time for each failure. The calculated ''Q'' is given in the figure below:

	[[Image: Expected Failure Plot 0.1 Rank.png\|center\|~~600px~~]]		[[Image: Expected Failure Plot 0.1 Rank.png\|center\|]]

	==Example==		==Example==

Lisa Hacker: /* Expected Failure Times Plots */

2015-12-10T21:21:23Z

Expected Failure Times Plots

← Older revision		Revision as of 21:21, 10 December 2015
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	For example, given ''n'' = 4, ''r'' = 2 and ''CL'' = 0.5, the calculated ''Q'' is 0.385728. This means, at the time when the second failure occurs, the estimated system probability of failure is 0.385728. The median rank can be calculated in Weibull++ using the '''Quick Statistical Reference''', as shown below:		For example, given ''n'' = 4, ''r'' = 2 and ''CL'' = 0.5, the calculated ''Q'' is 0.385728. This means, at the time when the second failure occurs, the estimated system probability of failure is 0.385728. The median rank can be calculated in Weibull++ using the '''Quick Statistical Reference''', as shown below:

	[[Image: Expected Failure Plot Median Rank.png\|center\|~~600px~~]]		[[Image: Expected Failure Plot Median Rank.png\|center\|]]

	Similarly, if we set ''r'' = 3 for the above example, we can get the probability of failure at the time when the third failure occurs. Using the estimated median rank for each failure and the assumed underlying failure distribution, we can calculate the expected time for each failure. Assume the failure distribution is Weibull, then we know:		Similarly, if we set ''r'' = 3 for the above example, we can get the probability of failure at the time when the third failure occurs. Using the estimated median rank for each failure and the assumed underlying failure distribution, we can calculate the expected time for each failure. Assume the failure distribution is Weibull, then we know:

Kate Racaza: /* Expected Failure Times Plots */

2015-12-09T20:37:52Z

Expected Failure Times Plots

← Older revision		Revision as of 20:37, 9 December 2015
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	For example, given ''n'' = 4, ''r'' = 2 and ''CL'' = 0.5, the calculated ''Q'' is 0.385728. This means, at the time when the second failure occurs, the estimated system probability of failure is 0.385728. The median rank can be calculated in Weibull++ using the '''Quick Statistical Reference''', as shown below:		For example, given ''n'' = 4, ''r'' = 2 and ''CL'' = 0.5, the calculated ''Q'' is 0.385728. This means, at the time when the second failure occurs, the estimated system probability of failure is 0.385728. The median rank can be calculated in Weibull++ using the '''Quick Statistical Reference''', as shown below:

	[[Image: Expected Failure Plot Median Rank.png\|center]]		[[Image: Expected Failure Plot Median Rank.png\|center\|600px]]

	Similarly, if we set ''r'' = 3 for the above example, we can get the probability of failure at the time when the third failure occurs. Using the estimated median rank for each failure and the assumed underlying failure distribution, we can calculate the expected time for each failure. Assume the failure distribution is Weibull, then we know:		Similarly, if we set ''r'' = 3 for the above example, we can get the probability of failure at the time when the third failure occurs. Using the estimated median rank for each failure and the assumed underlying failure distribution, we can calculate the expected time for each failure. Assume the failure distribution is Weibull, then we know:
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	Using the above equation, for a given ''Q'', we can get the corresponding time ''t''. The above calculation gives the '''median''' of each failure time for ''CL'' = 0.5. If we set ''CL'' at different values, the confidence bounds of each failure time can be obtained. For the above example, if we set ''CL''=0.9, from the calculated ''Q'' we can get the upper bound of the time for each failure. The calculated ''Q'' is given in the next figure:		Using the above equation, for a given ''Q'', we can get the corresponding time ''t''. The above calculation gives the '''median''' of each failure time for ''CL'' = 0.5. If we set ''CL'' at different values, the confidence bounds of each failure time can be obtained. For the above example, if we set ''CL''=0.9, from the calculated ''Q'' we can get the upper bound of the time for each failure. The calculated ''Q'' is given in the next figure:

	[[Image: Expected Failure Plot 0.9 Rank.png\|center]]		[[Image: Expected Failure Plot 0.9 Rank.png\|center\|600px]]

	If we set ''CL''=0.1, from the calculated ''Q'' we can get the lower bound of the time for each failure. The calculated ''Q'' is given in the figure below:		If we set ''CL''=0.1, from the calculated ''Q'' we can get the lower bound of the time for each failure. The calculated ''Q'' is given in the figure below:

	[[Image: Expected Failure Plot 0.1 Rank.png\|center]]		[[Image: Expected Failure Plot 0.1 Rank.png\|center\|600px]]

	==Example==		==Example==

Kate Racaza at 20:30, 9 December 2015

2015-12-09T20:30:31Z

← Older revision		Revision as of 20:30, 9 December 2015
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	For example, given ''n'' = 4, ''r'' = 2 and ''CL'' = 0.5, the calculated ''Q'' is 0.385728. This means, at the time when the second failure occurs, the estimated system probability of failure is 0.385728. The median rank can be calculated in Weibull++ using the '''Quick Statistical Reference''', as shown below:		For example, given ''n'' = 4, ''r'' = 2 and ''CL'' = 0.5, the calculated ''Q'' is 0.385728. This means, at the time when the second failure occurs, the estimated system probability of failure is 0.385728. The median rank can be calculated in Weibull++ using the '''Quick Statistical Reference''', as shown below:

	[[Image: Expected Failure Plot Median Rank.png\|center~~\|500px\|~~ ]]		[[Image: Expected Failure Plot Median Rank.png\|center]]

	Similarly, if we set ''r'' = 3 for the above example, we can get the probability of failure at the time when the third failure occurs. Using the estimated median rank for each failure and the assumed underlying failure distribution, we can calculate the expected time for each failure. Assume the failure distribution is Weibull, then we know:		Similarly, if we set ''r'' = 3 for the above example, we can get the probability of failure at the time when the third failure occurs. Using the estimated median rank for each failure and the assumed underlying failure distribution, we can calculate the expected time for each failure. Assume the failure distribution is Weibull, then we know:
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	Using the above equation, for a given ''Q'', we can get the corresponding time ''t''. The above calculation gives the '''median''' of each failure time for ''CL'' = 0.5. If we set ''CL'' at different values, the confidence bounds of each failure time can be obtained. For the above example, if we set ''CL''=0.9, from the calculated ''Q'' we can get the upper bound of the time for each failure. The calculated ''Q'' is given in the next figure:		Using the above equation, for a given ''Q'', we can get the corresponding time ''t''. The above calculation gives the '''median''' of each failure time for ''CL'' = 0.5. If we set ''CL'' at different values, the confidence bounds of each failure time can be obtained. For the above example, if we set ''CL''=0.9, from the calculated ''Q'' we can get the upper bound of the time for each failure. The calculated ''Q'' is given in the next figure:

	[[Image: Expected Failure Plot 0.9 Rank.png\|center~~\|500px\|~~ ]]		[[Image: Expected Failure Plot 0.9 Rank.png\|center]]

	If we set ''CL''=0.1, from the calculated ''Q'' we can get the lower bound of the time for each failure. The calculated ''Q'' is given in the figure below:		If we set ''CL''=0.1, from the calculated ''Q'' we can get the lower bound of the time for each failure. The calculated ''Q'' is given in the figure below:

	[[Image: Expected Failure Plot 0.1 Rank.png\|center~~\|500px\|~~ ]]		[[Image: Expected Failure Plot 0.1 Rank.png\|center]]

	==Example==		==Example==

Richard House at 19:57, 26 September 2012

2012-09-26T19:57:53Z

Show changes

Richard House: /* Use Prior Information from Subsystem Tests */

2012-09-05T16:31:26Z

Use Prior Information from Subsystem Tests

← Older revision		Revision as of 16:31, 5 September 2012
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	::<math>\beta_{0}=\left(1-E\left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]</math>		::<math>\beta_{0}=\left(1-E\left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]</math>

	For each subsystem ''i'', from the beta distribution, we can calculate the expected value and the variance of the subsystem’s reliability <math>R_{i}\,\!</math> as [[Appendix: ~~Weibull References~~\|[38]]]:		For each subsystem ''i'', from the beta distribution, we can calculate the expected value and the variance of the subsystem’s reliability <math>R_{i}\,\!</math>, as discussed in Guo [[Appendix:_Life_Data_Analysis_References\|[38]]]:

	::<math>E\left(R_{i}\right)=\frac{s_{i}}{n_{i}+1}</math>		::<math>E\left(R_{i}\right)=\frac{s_{i}}{n_{i}+1}</math>
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	::<math>Var\left(R_{i}\right)=\frac{s_{i}\left(n_{i}+1-s_{i}\right)}{\left(n_{i}+1\right)^{2}\left(n_{i}+2\right)}</math>		::<math>Var\left(R_{i}\right)=\frac{s_{i}\left(n_{i}+1-s_{i}\right)}{\left(n_{i}+1\right)^{2}\left(n_{i}+2\right)}</math>

	Assuming that all the subsystems are in a series reliability-wise configuration, the expected value and variance of the system’s reliability <math>R\,\!</math> can then be calculated as [[Appendix: ~~Weibull References~~\|[38]]]:		Assuming that all the subsystems are in a series reliability-wise configuration, the expected value and variance of the system’s reliability <math>R\,\!</math> can then be calculated as per Guo [[Appendix:_Life_Data_Analysis_References\|[38]]]:

	::<math>E\left(R_{0}\right)=(i=1)^{k} E\left(R_{i}\right)=E\left(R_{1}\right)\times E\left(R_{2}\right)\ldots E\left(R_{k}\right)</math>		::<math>E\left(R_{0}\right)=(i=1)^{k} E\left(R_{i}\right)=E\left(R_{1}\right)\times E\left(R_{2}\right)\ldots E\left(R_{k}\right)</math>

Richard House: /* Expected Failure Times Plots */

2012-08-30T03:19:52Z

Expected Failure Times Plots

← Older revision		Revision as of 03:19, 30 August 2012
Line 192:		Line 192:

	where:		where:
	* <math>\beta </math> is the shape parameter		* <math>\beta \,\!</math> is the shape parameter
	* <math>\eta</math> is the scale parameter		* <math>\eta\,\!</math> is the scale parameter

	Using the above equation, for a given ''Q'', we can get the corresponding time ''t''. The above calculation gives the '''median''' of each failure time for ''CL'' = 0.5. If we set ''CL'' at different values, the confidence bounds of each failure time can be obtained. For the above example, if we set ''CL''=0.9, from the calculated ''Q'' we can get the upper bound of the time for each failure. The calculated ''Q'' is given in the next figure:		Using the above equation, for a given ''Q'', we can get the corresponding time ''t''. The above calculation gives the '''median''' of each failure time for ''CL'' = 0.5. If we set ''CL'' at different values, the confidence bounds of each failure time can be obtained. For the above example, if we set ''CL''=0.9, from the calculated ''Q'' we can get the upper bound of the time for each failure. The calculated ''Q'' is given in the next figure:

Richard House: /* Bayesian Non-Parametric Test Design */

2012-08-30T03:12:43Z

Bayesian Non-Parametric Test Design

Richard House: /* Test to Demonstrate MTTF */

2012-08-28T02:39:17Z

Test to Demonstrate MTTF

← Older revision		Revision as of 02:39, 28 August 2012
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	===Test to Demonstrate MTTF===		===Test to Demonstrate MTTF===

	Designing a test to demonstrate a certain value of the <math>MTTF</math> is identical to designing a reliability demonstration test, with the exception of how the value of the scale parameter <math>\phi </math> is determined. Given the value of the <math>MTTF</math> and the value of the shape parameter <math>\theta </math> , the value of the scale parameter <math>\phi </math> can be calculated. With this, the analysis can proceed as with the reliability demonstration methodology.		Designing a test to demonstrate a certain value of the <math>MTTF</math> is identical to designing a reliability demonstration test, with the exception of how the value of the scale parameter <math>\phi \,\!</math> is determined. Given the value of the <math>MTTF</math> and the value of the shape parameter <math>\theta \,\!</math> , the value of the scale parameter <math>\phi \,\!</math> can be calculated. With this, the analysis can proceed as with the reliability demonstration methodology.

	====Example====		====Example====

Richard House: /* Constant Failure Rate/Chi-Squared */

2012-08-26T22:58:49Z

Constant Failure Rate/Chi-Squared

← Older revision		Revision as of 03:12, 30 August 2012
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	::<math>1-CL=\text{Beta}\left(R,\alpha,\beta\right)=\text{Beta}\left(R,n-r+\alpha_{0},r+\beta_{0}\right)</math>		::<math>1-CL=\text{Beta}\left(R,\alpha,\beta\right)=\text{Beta}\left(R,n-r+\alpha_{0},r+\beta_{0}\right)</math>

	where <math>Beta</math> is the incomplete beta function. If <~~span class="texhtml" sab="459"~~>~~α<sub sab="460"~~>0</~~sub></span~~> and <~~span class="texhtml" sab="461"~~>~~β<sub sab="462"~~>0</~~sub></span~~> are known, then any quantity of interest can be calculated using the remaining three. The next two examples demonstrate how to calculate <~~span class="texhtml" sab="463"~~>~~α<sub sab="464"~~>0</~~sub></span~~> and <~~span class="texhtml" sab="465"~~>~~β<sub sab="466"~~>0</~~sub></span~~> depending on the type of prior information available.		where <math>Beta\,\!</math> is the incomplete beta function. If <math>{{\alpha}_{0}} >0\,\!</math> and <math>{{\beta}_{0}} >0\,\!</math> are known, then any quantity of interest can be calculated using the remaining three. The next two examples demonstrate how to calculate <math>{{\alpha}_{0}} >0\,\!</math> and <math>{{\beta}_{0}} >0\,\!</math> depending on the type of prior information available.

	===Use Prior Expert Opinion on System Reliability ===		===Use Prior Expert Opinion on System Reliability ===
	Prior information on system reliability can be exploited to determine <math>\alpha_{0}</math> and <math>\beta_{0}</math>. To do so, first approximate the expected value and variance of prior system reliability <~~span class="texhtml">''R''<sub~~>0</~~sub></span~~>. This requires knowledge of the lowest possible reliability, the most likely possible reliability and the highest possible reliability of the system. These quantities will be referred to as ''a'', ''b'' and ''c'', respectively. The expected value of the prior system reliability is approximately given as:		Prior information on system reliability can be exploited to determine <math>\alpha_{0}\,\!</math> and <math>\beta_{0}\,\!</math>. To do so, first approximate the expected value and variance of prior system reliability <math>R_{0}\,\!</math>. This requires knowledge of the lowest possible reliability, the most likely possible reliability and the highest possible reliability of the system. These quantities will be referred to as ''a'', ''b'' and ''c'', respectively. The expected value of the prior system reliability is approximately given as:

	::<math> E\left(R_{0}\right)=\frac{a+4b+c}{6} </math>		::<math> E\left(R_{0}\right)=\frac{a+4b+c}{6} </math>
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	::<math>Var({{R}_{0}})={{\left( \frac{c-a}{6} \right)}^{2}}</math>		::<math>Var({{R}_{0}})={{\left( \frac{c-a}{6} \right)}^{2}}</math>

	These approximate values of the expected value and variance of the prior system reliability can then be used to estimate the values of <math>\alpha_{0}</math> and <math>\beta_{0}</math>, assuming that the prior reliability is a beta-distributed random variable. The values of <math>\alpha_{0}</math> and <math>\beta_{0}</math> are calculated as:		These approximate values of the expected value and variance of the prior system reliability can then be used to estimate the values of <math>\alpha_{0}\,\!</math> and <math>\beta_{0}\,\!</math>, assuming that the prior reliability is a beta-distributed random variable. The values of <math>\alpha_{0}\,\!</math> and <math>\beta_{0}\,\!</math> are calculated as:
	<br><br>		<br><br>

	::<math> \alpha_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right] </math> <math> \beta_{0}=\left(1-E\left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right] </math>		::<math> \alpha_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right] </math> <math> \beta_{0}=\left(1-E\left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right] </math>

	With <math>\alpha_{0}</math> and <math>\beta_{0}</math> known, the above beta distribution equation can now be used to calculate a quantity of interest.		With <math>\alpha_{0}\,\!</math> and <math>\beta_{0}\,\!</math> known, the above beta distribution equation can now be used to calculate a quantity of interest.

	====Example====		====Example====
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	===Use Prior Information from Subsystem Tests ===		===Use Prior Information from Subsystem Tests ===
	Prior information from subsystem tests can also be used to determine values of alpha and beta. Information from subsystem tests can be used to calculate the expected value and variance of the reliability of individual components, which can then be used to calculate the expected value and variance of the reliability of the entire system. <math>\alpha_{0}</math> and <math>\beta_{0}</math> are then calculated as before:		Prior information from subsystem tests can also be used to determine values of alpha and beta. Information from subsystem tests can be used to calculate the expected value and variance of the reliability of individual components, which can then be used to calculate the expected value and variance of the reliability of the entire system. <math>\alpha_{0}\,\!</math> and <math>\beta_{0}\,\!</math> are then calculated as before:

	::<math>\alpha_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right] </math>		::<math>\alpha_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right] </math>
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	::<math>\beta_{0}=\left(1-E\left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]</math>		::<math>\beta_{0}=\left(1-E\left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]</math>

	For each subsystem ''i'', from the beta distribution, we can calculate the expected value and the variance of the subsystem’s reliability <math>R_{i}</math> as [[Appendix: Weibull References\|[38]]]:		For each subsystem ''i'', from the beta distribution, we can calculate the expected value and the variance of the subsystem’s reliability <math>R_{i}\,\!</math> as [[Appendix: Weibull References\|[38]]]:

	::<math>E\left(R_{i}\right)=\frac{s_{i}}{n_{i}+1}</math>		::<math>E\left(R_{i}\right)=\frac{s_{i}}{n_{i}+1}</math>
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	::<math>Var\left(R_{i}\right)=\frac{s_{i}\left(n_{i}+1-s_{i}\right)}{\left(n_{i}+1\right)^{2}\left(n_{i}+2\right)}</math>		::<math>Var\left(R_{i}\right)=\frac{s_{i}\left(n_{i}+1-s_{i}\right)}{\left(n_{i}+1\right)^{2}\left(n_{i}+2\right)}</math>

	Assuming that all the subsystems are in a series reliability-wise configuration, the expected value and variance of the system’s reliability <math>R</math> can then be calculated as [[Appendix: Weibull References\|[38]]]:		Assuming that all the subsystems are in a series reliability-wise configuration, the expected value and variance of the system’s reliability <math>R\,\!</math> can then be calculated as [[Appendix: Weibull References\|[38]]]:

	::<math>E\left(R_{0}\right)=(i=1)^{k} E\left(R_{i}\right)=E\left(R_{1}\right)\times E\left(R_{2}\right)\ldots E\left(R_{k}\right)</math>		::<math>E\left(R_{0}\right)=(i=1)^{k} E\left(R_{i}\right)=E\left(R_{1}\right)\times E\left(R_{2}\right)\ldots E\left(R_{k}\right)</math>

← Older revision		Revision as of 22:58, 26 August 2012
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	==Constant Failure Rate/Chi-Squared==<!-- THIS SECTION HEADER IS LINKED FROM OTHER PLACES IN THIS PAGE. IF YOU RENAME THE SECTION, YOU MUST UPDATE THE LINK(S). -->		==Constant Failure Rate/Chi-Squared==<!-- THIS SECTION HEADER IS LINKED FROM OTHER PLACES IN THIS PAGE. IF YOU RENAME THE SECTION, YOU MUST UPDATE THE LINK(S). -->
	Another method for designing tests for products that have an assumed constant failure rate, or exponential life distribution, draws on the chi-squared distribution. These represent the true exponential distribution confidence bounds referred to in [[The Exponential Distribution]]. This method only returns the necessary accumulated test time for a demonstrated reliability or <math>MTTF</math> , not a specific time/test unit combination that is obtained using the cumulative binomial method described above. The accumulated test time is equal to the total amount of time experienced by all of the units on test. Assuming that the units undergo the same amount of test time, this works out to be:		Another method for designing tests for products that have an assumed constant failure rate, or exponential life distribution, draws on the chi-squared distribution. These represent the true exponential distribution confidence bounds referred to in [[The Exponential Distribution]]. This method only returns the necessary accumulated test time for a demonstrated reliability or <math>MTTF\,\!</math> , not a specific time/test unit combination that is obtained using the cumulative binomial method described above. The accumulated test time is equal to the total amount of time experienced by all of the units on test. Assuming that the units undergo the same amount of test time, this works out to be:

	::<math>{{T}_{a}}=n\cdot {{t}_{TEST}}</math>		::<math>{{T}_{a}}=n\cdot {{t}_{TEST}}</math>

	where <math>n</math> is the number of units on test and <math>{{t}_{TEST}}</math> is the test time. The chi-squared equation for test time is:		where <math>n\,\!</math> is the number of units on test and <math>{{t}_{TEST}}\,\!</math> is the test time. The chi-squared equation for test time is:

	::<math>{{T}_{a}}=\frac{MTTF\cdot \chi _{1-CL;2f+2}^{2}}{2}</math>		::<math>{{T}_{a}}=\frac{MTTF\cdot \chi _{1-CL;2f+2}^{2}}{2}</math>

	where:		where:
	*<math>\chi _{1-CL;2f+2}^{2}</math> = the chi-squared distribution		*<math>\chi _{1-CL;2f+2}^{2}\,\!</math> = the chi-squared distribution
	*<math>{{T}_{a}}\,\!</math> = the necessary accumulated test time		*<math>{{T}_{a}}\,\!</math> = the necessary accumulated test time
	*<math>CL\,\!</math> = the confidence level		*<math>CL\,\!</math> = the confidence level
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	::<math>MTTF=\frac{t}{-ln(R)}</math>		::<math>MTTF=\frac{t}{-ln(R)}</math>

	and substituted into the chi-squared equation for developing a test that demonstrates reliability at a given time, rather than <math>MTTF</math> :		and substituted into the chi-squared equation for developing a test that demonstrates reliability at a given time, rather than <math>MTTF\,\!</math> :

	::<math>{{T}_{a}}=\frac{\tfrac{{{t}_{DEMO}}}{-ln(R)}\cdot \chi _{1-CL;2f+2}^{2}}{2}</math>		::<math>{{T}_{a}}=\frac{\tfrac{{{t}_{DEMO}}}{-ln(R)}\cdot \chi _{1-CL;2f+2}^{2}}{2}</math>