Template:Test Design Using Expected Failure Time Plots - Revision history

Kate Racaza: Redirected page to Reliability Test Design

2012-06-29T08:05:01Z

Redirected page to Reliability Test Design

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	~~{{template:LDABOOK\|29\|Expected Failure Time Plots}}~~		#REDIRECT [[Reliability Test Design]]

	~~{{expected failure time plots}}~~

Harry Guo: moved Template:Design Design Using Expected Failure Time Plots to Template:Test Design Using Expected Failure Time Plots

2012-03-01T18:41:56Z

moved Template:Design Design Using Expected Failure Time Plots to Template:Test Design Using Expected Failure Time Plots

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Harry Guo: moved Expected Failure Time Plots to Template:Design Design Using Expected Failure Time Plots: To include it in the Reliability Test Design Chapter

2012-02-08T16:16:08Z

moved Expected Failure Time Plots to Template:Design Design Using Expected Failure Time Plots: To include it in the Reliability Test Design Chapter

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Nicolette Young: Replaced content with '{{template:LDABOOK|29|Expected Failure Time Plots}} {{expected failure time plots}}'

2012-01-05T00:29:38Z

Replaced content with '{{template:LDABOOK|29|Expected Failure Time Plots}} {{expected failure time plots}}'

← Older revision		Revision as of 00:29, 5 January 2012
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	{{template:LDABOOK\|29\|Expected Failure Time Plots}}		{{template:LDABOOK\|29\|Expected Failure Time Plots}}

			{{expected failure time plots}}
	One of the new features in Weibull++ 8 is the Expected Failure Time Calculation (EFTC) tool. Given the population size and the failure time distribution <math>F(t) </math>, the EFTC tool estimates the expected value of the nth failure and its confidence interval at any confidence level.


	~~The EFTC tool works by first calculating the probability of failure ''F'' as a function of the sample size ''n'' and number of failures ''r''.~~
	~~This function is given by the equation <math> f\left(n,r,CL\right)=\frac{1}{1+\frac{n-r+1}{rF_{2r,2\left(n-r+1\right),1-CL}}} </math> where ''CL'' is the confidence level.~~

	For example, given ''n'' = 4, ''r'' = 2 and ''CL'' = 0.5, the median rank for ''F'' is <math> f\left(4,2,0.5\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.5}}}=0.385728 </math> The 80% 2-sided confidence interval on the probability of failure F is bounded by <math> f\left(4,2,0.1\right)=\frac{1}{1+\frac{4-2+1}{~~2F_~~{4,2\left(3\right),0.9}}}=0.142559 </math> and <math> f\left(4,2,0.9\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.1}}}=0.679539 </math> With the probabilities of failure known, and assuming the failure distribution is exponentially distributed with parameter \(\lambda\,\!\) = 1, the expected ~~times to~~ failure ~~can be calculated.~~

	~~For example, the median~~ time ~~to the second failure is calculated as <math> t=-\frac{ln\left(1-F\right)~~}{\lambda}=-\frac{1-0.385728}{1}=0.48732 </math> Similarly, the 10th percentile and 90th percentiles of time to failure are calculated as <math> t=-\frac{1-0.142559}{1}=0.153803 </math> and <math> t=-\frac{1-0.679539}{1}~~=1.137995 </math> The above results can also be seen in Weibull++ as shown below.~~

	~~[[Image:Eftc2.jpg\|thumb\|center\|500px\| ]]~~

Nicolette Young at 22:26, 19 August 2011

2011-08-19T22:26:31Z

← Older revision		Revision as of 22:26, 19 August 2011
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	For example, the median time to the second failure is calculated as <math> t=-\frac{ln\left(1-F\right)}{\lambda}=-\frac{1-0.385728}{1}=0.48732 </math> Similarly, the 10th percentile and 90th percentiles of time to failure are calculated as <math> t=-\frac{1-0.142559}{1}=0.153803 </math> and <math> t=-\frac{1-0.679539}{1}=1.137995 </math> The above results can also be seen in Weibull++ as shown below.		For example, the median time to the second failure is calculated as <math> t=-\frac{ln\left(1-F\right)}{\lambda}=-\frac{1-0.385728}{1}=0.48732 </math> Similarly, the 10th percentile and 90th percentiles of time to failure are calculated as <math> t=-\frac{1-0.142559}{1}=0.153803 </math> and <math> t=-\frac{1-0.679539}{1}=1.137995 </math> The above results can also be seen in Weibull++ as shown below.

	[[Image:Eftc2.jpg\|thumb\|center\|~~400px~~\| ]]		[[Image:Eftc2.jpg\|thumb\|center\|500px\| ]]

Nicolette Young at 22:25, 19 August 2011

2011-08-19T22:25:35Z

← Older revision		Revision as of 22:25, 19 August 2011
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	The EFTC tool works by first calculating the probability of failure ''F'' as a function of the sample size ''n'' and number of failures ''r''.		The EFTC tool works by first calculating the probability of failure ''F'' as a function of the sample size ''n'' and number of failures ''r''.
	This function is given by the equation <math> f\left(n,r,CL\right)=\frac{1}{1+\frac{n-r+1}{rF_{2r,2\left(n-r+1\right),1-CL}}} </math> where ''CL'' is the confidence level.		This function is given by the equation <math> f\left(n,r,CL\right)=\frac{1}{1+\frac{n-r+1}{rF_{2r,2\left(n-r+1\right),1-CL}}} </math> where ''CL'' is the confidence level.

	For example, given ''n'' = 4, ''r'' = 2 and ''CL'' = 0.5, the median rank for ''F'' is <math> f\left(4,2,0.5\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.5}}}=0.385728 </math> The 80% 2-sided confidence interval on the probability of failure F is bounded by <math> f\left(4,2,0.1\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.9}}}=0.142559 </math> and <math> f\left(4,2,0.9\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.1}}}=0.679539 </math> With the probabilities of failure known, and assuming the failure distribution is exponentially distributed with parameter \(\lambda\,\!\) = 1, the expected times to failure can be calculated.		For example, given ''n'' = 4, ''r'' = 2 and ''CL'' = 0.5, the median rank for ''F'' is <math> f\left(4,2,0.5\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.5}}}=0.385728 </math> The 80% 2-sided confidence interval on the probability of failure F is bounded by <math> f\left(4,2,0.1\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.9}}}=0.142559 </math> and <math> f\left(4,2,0.9\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.1}}}=0.679539 </math> With the probabilities of failure known, and assuming the failure distribution is exponentially distributed with parameter \(\lambda\,\!\) = 1, the expected times to failure can be calculated.

	For example, the median time to the second failure is calculated as <math> t=-\frac{ln\left(1-F\right)}{\lambda}=-\frac{1-0.385728}{1}=0.48732 </math> Similarly, the 10th percentile and 90th percentiles of time to failure are calculated as <math> t=-\frac{1-0.142559}{1}=0.153803 </math> and <math> t=-\frac{1-0.679539}{1}=1.137995 </math> The above results can also be seen in Weibull++ as shown below.		For example, the median time to the second failure is calculated as <math> t=-\frac{ln\left(1-F\right)}{\lambda}=-\frac{1-0.385728}{1}=0.48732 </math> Similarly, the 10th percentile and 90th percentiles of time to failure are calculated as <math> t=-\frac{1-0.142559}{1}=0.153803 </math> and <math> t=-\frac{1-0.679539}{1}=1.137995 </math> The above results can also be seen in Weibull++ as shown below.

	[[Image:Eftc2.jpg\|thumb\|center\|400px\| ]]		[[Image:Eftc2.jpg\|thumb\|center\|400px\| ]]

Nicolette Young at 22:24, 19 August 2011

2011-08-19T22:24:31Z

← Older revision		Revision as of 22:24, 19 August 2011
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	One of the new features in Weibull++ 8 is the Expected Failure Time Calculation (EFTC) tool. Given the population size and the failure time distribution <math>F(t) </math>, the EFTC tool estimates the expected value of the nth failure and its confidence interval at any confidence level.		One of the new features in Weibull++ 8 is the Expected Failure Time Calculation (EFTC) tool. Given the population size and the failure time distribution <math>F(t) </math>, the EFTC tool estimates the expected value of the nth failure and its confidence interval at any confidence level.

	The EFTC tool works by first calculating the probability of failure ''F'' as a function of the sample size ''n'' and number of failures ''r''. This function is given by the equation <math> f\left(n,r,CL\right)=\frac{1}{1+\frac{n-r+1}{rF_{2r,2\left(n-r+1\right),1-CL}}} </math> where ''CL'' is the confidence level. For example, given ''n'' = 4, ''r'' = 2 and ''CL'' = 0.5, the median rank for ''F'' is <math> f\left(4,2,0.5\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.5}}}=0.385728 </math> The 80% 2-sided confidence interval on the probability of failure F is bounded by <math> f\left(4,2,0.1\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.9}}}=0.142559 </math> and <math> f\left(4,2,0.9\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.1}}}=0.679539 </math> With the probabilities of failure known, and assuming the failure distribution is exponentially distributed with parameter \(\lambda\,\!\) = 1, the expected times to failure can be calculated. For example, the median time to the second failure is calculated as <math> t=-\frac{ln\left(1-F\right)}{\lambda}=-\frac{1-0.385728}{1}=0.48732 </math> Similarly, the 10th percentile and 90th percentiles of time to failure are calculated as <math> t=-\frac{1-0.142559}{1}=0.153803 </math> and <math> t=-\frac{1-0.679539}{1}=1.137995 </math> The above results can also be seen in Weibull++ as shown below.
			The EFTC tool works by first calculating the probability of failure ''F'' as a function of the sample size ''n'' and number of failures ''r''.
			This function is given by the equation <math> f\left(n,r,CL\right)=\frac{1}{1+\frac{n-r+1}{rF_{2r,2\left(n-r+1\right),1-CL}}} </math> where ''CL'' is the confidence level.
			For example, given ''n'' = 4, ''r'' = 2 and ''CL'' = 0.5, the median rank for ''F'' is <math> f\left(4,2,0.5\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.5}}}=0.385728 </math> The 80% 2-sided confidence interval on the probability of failure F is bounded by <math> f\left(4,2,0.1\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.9}}}=0.142559 </math> and <math> f\left(4,2,0.9\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.1}}}=0.679539 </math> With the probabilities of failure known, and assuming the failure distribution is exponentially distributed with parameter \(\lambda\,\!\) = 1, the expected times to failure can be calculated.
			For example, the median time to the second failure is calculated as <math> t=-\frac{ln\left(1-F\right)}{\lambda}=-\frac{1-0.385728}{1}=0.48732 </math> Similarly, the 10th percentile and 90th percentiles of time to failure are calculated as <math> t=-\frac{1-0.142559}{1}=0.153803 </math> and <math> t=-\frac{1-0.679539}{1}=1.137995 </math> The above results can also be seen in Weibull++ as shown below.

	[[Image:Eftc2.jpg\|thumb\|center\|400px\| ]]		[[Image:Eftc2.jpg\|thumb\|center\|400px\| ]]

Nicolette Young at 18:17, 15 August 2011

2011-08-15T18:17:32Z

← Older revision		Revision as of 18:17, 15 August 2011
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	One of the new features in Weibull++ 8 is the Expected Failure Time Calculation (EFTC) tool. Given the population size and the failure time distribution <math>\(F~~\left~~(t~~\right))\~~)</math>, the EFTC tool estimates the expected value of the nth failure and its confidence interval at any confidence level.		One of the new features in Weibull++ 8 is the Expected Failure Time Calculation (EFTC) tool. Given the population size and the failure time distribution <math>F(t) </math>, the EFTC tool estimates the expected value of the nth failure and its confidence interval at any confidence level.

	The EFTC tool works by first calculating the probability of failure ''F'' as a function of the sample size ''n'' and number of failures ''r''. This function is given by the equation <math> f\left(n,r,CL\right)=\frac{1}{1+\frac{n-r+1}{rF_{2r,2\left(n-r+1\right),1-CL}}} </math> where ''CL'' is the confidence level. For example, given ''n'' = 4, ''r'' = 2 and ''CL'' = 0.5, the median rank for ''F'' is <math> f\left(4,2,0.5\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.5}}}=0.385728 </math> The 80% 2-sided confidence interval on the probability of failure F is bounded by <math> f\left(4,2,0.1\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.9}}}=0.142559 </math> and <math> f\left(4,2,0.9\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.1}}}=0.679539 </math> With the probabilities of failure known, and assuming the failure distribution is exponentially distributed with parameter \(\lambda\,\!\) = 1, the expected times to failure can be calculated. For example, the median time to the second failure is calculated as <math> t=-\frac{ln\left(1-F\right)}{\lambda}=-\frac{1-0.385728}{1}=0.48732 </math> Similarly, the 10th percentile and 90th percentiles of time to failure are calculated as <math> t=-\frac{1-0.142559}{1}=0.153803 </math> and <math> t=-\frac{1-0.679539}{1}=1.137995 </math> The above results can also be seen in Weibull++ as shown below.		The EFTC tool works by first calculating the probability of failure ''F'' as a function of the sample size ''n'' and number of failures ''r''. This function is given by the equation <math> f\left(n,r,CL\right)=\frac{1}{1+\frac{n-r+1}{rF_{2r,2\left(n-r+1\right),1-CL}}} </math> where ''CL'' is the confidence level. For example, given ''n'' = 4, ''r'' = 2 and ''CL'' = 0.5, the median rank for ''F'' is <math> f\left(4,2,0.5\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.5}}}=0.385728 </math> The 80% 2-sided confidence interval on the probability of failure F is bounded by <math> f\left(4,2,0.1\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.9}}}=0.142559 </math> and <math> f\left(4,2,0.9\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.1}}}=0.679539 </math> With the probabilities of failure known, and assuming the failure distribution is exponentially distributed with parameter \(\lambda\,\!\) = 1, the expected times to failure can be calculated. For example, the median time to the second failure is calculated as <math> t=-\frac{ln\left(1-F\right)}{\lambda}=-\frac{1-0.385728}{1}=0.48732 </math> Similarly, the 10th percentile and 90th percentiles of time to failure are calculated as <math> t=-\frac{1-0.142559}{1}=0.153803 </math> and <math> t=-\frac{1-0.679539}{1}=1.137995 </math> The above results can also be seen in Weibull++ as shown below.

	[[Image:Eftc2.jpg\|thumb\|center\|400px\| ]]		[[Image:Eftc2.jpg\|thumb\|center\|400px\| ]]

Nicolette Young at 18:17, 15 August 2011

2011-08-15T18:17:08Z

← Older revision		Revision as of 18:17, 15 August 2011
Line 2:		Line 2:


	One of the new features in Weibull++ 8 is the Expected Failure Time Calculation (EFTC) tool. Given the population size and the failure time distribution \(F\left(t\right))\), the EFTC tool estimates the expected value of the nth failure and its confidence interval at any confidence level.		One of the new features in Weibull++ 8 is the Expected Failure Time Calculation (EFTC) tool. Given the population size and the failure time distribution <math>\(F\left(t\right))\)</math>, the EFTC tool estimates the expected value of the nth failure and its confidence interval at any confidence level.

	The EFTC tool works by first calculating the probability of failure ''F'' as a function of the sample size ''n'' and number of failures ''r''. This function is given by the equation <math> f\left(n,r,CL\right)=\frac{1}{1+\frac{n-r+1}{rF_{2r,2\left(n-r+1\right),1-CL}}} </math> where ''CL'' is the confidence level. For example, given ''n'' = 4, ''r'' = 2 and ''CL'' = 0.5, the median rank for ''F'' is <math> f\left(4,2,0.5\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.5}}}=0.385728 </math> The 80% 2-sided confidence interval on the probability of failure F is bounded by <math> f\left(4,2,0.1\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.9}}}=0.142559 </math> and <math> f\left(4,2,0.9\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.1}}}=0.679539 </math> With the probabilities of failure known, and assuming the failure distribution is exponentially distributed with parameter \(\lambda\,\!\) = 1, the expected times to failure can be calculated. For example, the median time to the second failure is calculated as <math> t=-\frac{ln\left(1-F\right)}{\lambda}=-\frac{1-0.385728}{1}=0.48732 </math> Similarly, the 10th percentile and 90th percentiles of time to failure are calculated as <math> t=-\frac{1-0.142559}{1}=0.153803 </math> and <math> t=-\frac{1-0.679539}{1}=1.137995 </math> The above results can also be seen in Weibull++ as shown below.		The EFTC tool works by first calculating the probability of failure ''F'' as a function of the sample size ''n'' and number of failures ''r''. This function is given by the equation <math> f\left(n,r,CL\right)=\frac{1}{1+\frac{n-r+1}{rF_{2r,2\left(n-r+1\right),1-CL}}} </math> where ''CL'' is the confidence level. For example, given ''n'' = 4, ''r'' = 2 and ''CL'' = 0.5, the median rank for ''F'' is <math> f\left(4,2,0.5\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.5}}}=0.385728 </math> The 80% 2-sided confidence interval on the probability of failure F is bounded by <math> f\left(4,2,0.1\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.9}}}=0.142559 </math> and <math> f\left(4,2,0.9\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.1}}}=0.679539 </math> With the probabilities of failure known, and assuming the failure distribution is exponentially distributed with parameter \(\lambda\,\!\) = 1, the expected times to failure can be calculated. For example, the median time to the second failure is calculated as <math> t=-\frac{ln\left(1-F\right)}{\lambda}=-\frac{1-0.385728}{1}=0.48732 </math> Similarly, the 10th percentile and 90th percentiles of time to failure are calculated as <math> t=-\frac{1-0.142559}{1}=0.153803 </math> and <math> t=-\frac{1-0.679539}{1}=1.137995 </math> The above results can also be seen in Weibull++ as shown below.

	[[Image:Eftc2.jpg\|thumb\|center\|400px\| ]]		[[Image:Eftc2.jpg\|thumb\|center\|400px\| ]]

Nicolette Young at 18:16, 15 August 2011

2011-08-15T18:16:35Z

← Older revision		Revision as of 18:16, 15 August 2011
Line 4:		Line 4:
	One of the new features in Weibull++ 8 is the Expected Failure Time Calculation (EFTC) tool. Given the population size and the failure time distribution \(F\left(t\right))\), the EFTC tool estimates the expected value of the nth failure and its confidence interval at any confidence level.		One of the new features in Weibull++ 8 is the Expected Failure Time Calculation (EFTC) tool. Given the population size and the failure time distribution \(F\left(t\right))\), the EFTC tool estimates the expected value of the nth failure and its confidence interval at any confidence level.

	The EFTC tool works by first calculating the probability of failure ''F'' as a function of the sample size ''n'' and number of failures ''r''. This function is given by the equation <math> f\left(n,r,CL\right)=\frac{1}{1+\frac{n-r+1}{rF_{2r,2\left(n-r+1\right),1-CL}}} </math> where ''CL'' is the confidence level. For example, given ''n'' = 4, ''r'' = 2 and ''CL'' = 0.5, the median rank for ''F'' is <math> f\left(4,2,0.5\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.5}}}=0.385728 </math> The 80% 2-sided confidence interval on the probability of failure F is bounded by <math> f\left(4,2,0.1\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.9}}}=0.142559 </math> and <math> f\left(4,2,0.9\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.1}}}=0.679539 </math> With the probabilities of failure known, and assuming the failure distribution is exponentially distributed with parameter \(\lambda\,\!\) = 1, the expected times to failure can be calculated. For example, the median time to the second failure is calculated as <math> t=-\frac{ln\left(1-F\right)}{\lambda}=-\frac{1-0.385728}{1}=0.48732 </math> Similarly, the 10th percentile and 90th percentiles of time to failure are calculated as <math> t=-\frac{1-0.142559}{1}=0.153803 ~~\end{equation}~~ and ~~\begin{equation}~~ t=-\frac{1-0.679539}{1}=1.137995 </math> The above results can also be seen in Weibull++ as shown below.		The EFTC tool works by first calculating the probability of failure ''F'' as a function of the sample size ''n'' and number of failures ''r''. This function is given by the equation <math> f\left(n,r,CL\right)=\frac{1}{1+\frac{n-r+1}{rF_{2r,2\left(n-r+1\right),1-CL}}} </math> where ''CL'' is the confidence level. For example, given ''n'' = 4, ''r'' = 2 and ''CL'' = 0.5, the median rank for ''F'' is <math> f\left(4,2,0.5\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.5}}}=0.385728 </math> The 80% 2-sided confidence interval on the probability of failure F is bounded by <math> f\left(4,2,0.1\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.9}}}=0.142559 </math> and <math> f\left(4,2,0.9\right)=\frac{1}{1+\frac{4-2+1}{2F_{4,2\left(3\right),0.1}}}=0.679539 </math> With the probabilities of failure known, and assuming the failure distribution is exponentially distributed with parameter \(\lambda\,\!\) = 1, the expected times to failure can be calculated. For example, the median time to the second failure is calculated as <math> t=-\frac{ln\left(1-F\right)}{\lambda}=-\frac{1-0.385728}{1}=0.48732 </math> Similarly, the 10th percentile and 90th percentiles of time to failure are calculated as <math> t=-\frac{1-0.142559}{1}=0.153803 </math> and <math> t=-\frac{1-0.679539}{1}=1.137995 </math> The above results can also be seen in Weibull++ as shown below.

	[[Image:Eftc2.jpg\|thumb\|center\|400px\| ]]		[[Image:Eftc2.jpg\|thumb\|center\|400px\| ]]