2 Paramter Exponential Distribution RRY - Revision history

Lisa Hacker: Redirected page to The Exponential Distribution

2012-08-08T07:32:58Z

Redirected page to The Exponential Distribution

← Older revision		Revision as of 07:32, 8 August 2012
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	[[~~Category: For Deletion~~]]		#REDIRECT [[The_Exponential_Distribution]]

Lisa Hacker: Replaced content with 'Category: For Deletion'

2012-08-08T06:26:59Z

Replaced content with 'Category: For Deletion'

← Older revision		Revision as of 06:26, 8 August 2012
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	~~<br><br>'''2 Parameter Exponential Distribution RRY'''~~		[[Category: For Deletion]]

	~~Fourteen units were being reliability tested and the following life test data were obtained:~~


	~~{\|align="center" border=1 cellspacing=0~~
	\|-
	~~\|colspan="2" style="text-align:center"\| Table - Life Test Data~~
	\|-
	~~\|- align="center"~~
	~~!Data point index~~
	~~!Time-to-failure~~
	~~\|- align="center"~~
	~~\|1 \|\|5~~
	~~\|- align="center"~~
	~~\|2 \|\|10~~
	~~\|- align="center"~~
	~~\|3 \|\|15~~
	~~\|- align="center"~~
	~~\|4 \|\|20~~
	~~\|- align="center"~~
	~~\|5 \|\|25~~
	~~\|- align="center"~~
	~~\|6 \|\|30~~
	~~\|- align="center"~~
	~~\|7 \|\|35~~
	~~\|- align="center"~~
	~~\|8 \|\|40~~
	~~\|- align="center"~~
	~~\|9 \|\|50~~
	~~\|- align="center"~~
	~~\|10 \|\|60~~
	~~\|- align="center"~~
	~~\|11 \|\|70~~
	~~\|-align="center"~~
	~~\|12 \|\|80~~
	~~\|- align="center"~~
	~~\|13 \|\|90~~
	~~\|- align="center"~~
	~~\|14 \|\|100~~
	\|}

	~~Assuming that the data follow a two-parameter exponential distribution, estimate the parameters and determine the correlation coefficient, <math>\rho </math>, using rank regression on Y.~~

	~~'''Solution'''~~

	~~Construct the following Table, as shown next.~~

	~~<center><math>\overset{{}}{\mathop{\text{Table}\text{- Least Squares Analysis}}}\,</math></center>~~

	~~<center><math>\begin{matrix}~~
	~~N & t_{i} & F(t_{i}) & y_{i} & t_{i}^{2} & y_{i}^{2} & t_{i} y_{i} \\~~
	~~\text{1} & \text{5} & \text{0}\text{.0483} & \text{-0}\text{.0495} & \text{25} & \text{0}\text{.0025} & \text{-0}\text{.2475} \\~~
	~~\text{2} & \text{10} & \text{0}\text{.1170} & \text{-0}\text{.1244} & \text{100} & \text{0}\text{.0155} & \text{-1}\text{.2443} \\~~
	~~\text{3} & \text{15} & \text{0}\text{.1865} & \text{-0}\text{.2064} & \text{225} & \text{0}\text{.0426} & \text{-3}\text{.0961} \\~~
	~~\text{4} & \text{20} & \text{0}\text{.2561} & \text{-0}\text{.2958} & \text{400} & \text{0}\text{.0875} & \text{-5}\text{.9170} \\~~
	~~\text{5} & \text{25} & \text{0}\text{.3258} & \text{-0}\text{.3942} & \text{625} & \text{0}\text{.1554} & \text{-9}\text{.8557} \\~~
	~~\text{6} & \text{30} & \text{0}\text{.3954} & \text{-0}\text{.5032} & \text{900} & \text{0}\text{.2532} & \text{-15}\text{.0956} \\~~
	~~\text{7} & \text{35} & \text{0}\text{.4651} & \text{-0}\text{.6257} & \text{1225} & \text{0}\text{.3915} & \text{-21}\text{.8986} \\~~
	~~\text{8} & \text{40} & \text{0}\text{.5349} & \text{-0}\text{.7655} & \text{1600} & \text{0}\text{.5860} & \text{-30}\text{.6201} \\~~
	~~\text{9} & \text{50} & \text{0}\text{.6046} & \text{-0}\text{.9279} & \text{2500} & \text{0}\text{.8609} & \text{-46}\text{.3929} \\~~
	~~\text{10} & \text{60} & \text{0}\text{.6742} & \text{-1}\text{.1215} & \text{3600} & \text{1}\text{.2577} & \text{-67}\text{.2883} \\~~
	~~\text{11} & \text{70} & \text{0}\text{.7439} & \text{-1}\text{.3622} & \text{4900} & \text{1}\text{.8456} & \text{-95}\text{.3531} \\~~
	~~\text{12} & \text{80} & \text{0}\text{.8135} & \text{-1}\text{.6793} & \text{6400} & \text{2}\text{.8201} & \text{-134}\text{.3459} \\~~
	~~\text{13} & \text{90} & \text{0}\text{.8830} & \text{-2}\text{.1456} & \text{8100} & \text{4}\text{.6035} & \text{-193}\text{.1023} \\~~
	~~\text{14} & \text{100} & \text{0}\text{.9517} & \text{-3}\text{.0303} & \text{10000} & \text{9}\text{.1829} & \text{-303}\text{.0324} \\~~
	~~\sum_{}^{} & \text{630} & {} & \text{-13}\text{.2315} & \text{40600} & \text{22}\text{.1148} & \text{-927}\text{.4899} \\~~
	~~\end{matrix}</math></center>~~


	~~The median rank values ( <math>F({{t}_{i}})</math> ) can be found in rank tables or they can be estimated using the '''Quick Statistical Reference''' in Weibull++.~~
	~~Given the values in the table above, calculate <math>\hat{a}</math> and <math>\hat{b}</math>:~~


	~~::<math>\begin{align}~~
	\hat{b}= & \frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}}{{y}_{i}}-(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}})(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}})/14}{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,t_{i}^{2}-{{(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}})}^{2}}/14} \\
	\\
	~~\hat{b}= & \frac{-927.4899-(630)(-13.2315)/14}{40,600-{{(630)}^{2}}/14}~~
	~~\end{align}</math>~~

	~~or:~~

	~~::<math>\hat{b}=-0.02711</math>~~

	~~and:~~

	~~::<math>\hat{a}=\overline{y}-\hat{b}\overline{t}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{t}_{i}}}{N}</math>~~

	~~or:~~

	~~::<math>\hat{a}=\frac{-13.2315}{14}-(-0.02711)\frac{630}{14}=0.2748</math>~~


	~~Therefore:~~

	~~::<math>\hat{\lambda }=-\hat{b}=-(-0.02711)=0.02711\text{ failures/hour}</math>~~


	~~and:~~

	~~::<math>\hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2748}{0.02711}</math>~~

	~~or:~~

	~~::<math>\hat{\gamma }=10.1365\text{ hours}</math>~~

	~~Then:~~

	~~::<math>f(t)=(0.02711)\cdot {{e}^{-0.02711(T-10.136)}}</math>~~


	~~The correlation coefficient can be estimated using equation for calculating the correlation coefficient:~~

	~~::<math>\hat{\rho }=-0.9679</math>~~


	~~This example can be repeated using Weibull++, choosing two-parameter exponential and rank regression on Y (RRY), as shown in the figure on the following page.~~

	~~The estimated parameters and the correlation coefficient using Weibull++ were found to be:~~

	~~::<math>\hat{\lambda }=0.0271\text{ fr/hr },\hat{\gamma }=10.1348\text{ hr },\hat{\rho }=-0.9679</math>~~

	~~[[Image:Exponential Example 2 Data Folio.png\|thumb\|center\|400px\|]]~~

	~~Please note that the user must deselect the '''Reset if Loc. Param > T1 on Exp RR''' option on the user setup page.~~

	~~The probability plot can be obtained simply by clicking the '''Plot''' icon.~~

	[[~~Image~~:~~Exponential Example 2 Plot.png\|thumb\|center\|400px\|~~]]

Nicolette Young at 18:49, 2 April 2012

2012-04-02T18:49:27Z

← Older revision		Revision as of 18:49, 2 April 2012
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	~~{\| class="FCK__ShowTableBorders" border="0" cellspacing="0" cellpadding="0" align="center"; style="width:100%;"~~		<br><br>'''2 Parameter Exponential Distribution RRY'''
	\|-
	~~\| valign="middle" align="left" bgcolor=EEEDF7\|[[Image:Weibull-Examples-banner.png\|400px\|center]]~~
	\|}<br><br>'''2 Parameter Exponential Distribution RRY'''

	Fourteen units were being reliability tested and the following life test data were obtained:		Fourteen units were being reliability tested and the following life test data were obtained:
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	[[Image:Exponential Example 2 Plot.png\|thumb\|center\|400px\|]]		[[Image:Exponential Example 2 Plot.png\|thumb\|center\|400px\|]]

	~~[[2P Exponential Example]]~~

Nicolette Young at 18:48, 2 April 2012

2012-04-02T18:48:48Z

← Older revision		Revision as of 18:48, 2 April 2012
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	[[Image:Exponential Example 2 Plot.png\|thumb\|center\|400px\|]]		[[Image:Exponential Example 2 Plot.png\|thumb\|center\|400px\|]]

			[[2P Exponential Example]]

Nicolette Young: Created page with '{| class="FCK__ShowTableBorders" border="0" cellspacing="0" cellpadding="0" align="center"; style="width:100%;" |- | valign="middle" align="left" bgcolor=EEEDF7|[[Image:Weibull-E…'

2012-03-21T23:31:06Z

Created page with '{| class="FCK__ShowTableBorders" border="0" cellspacing="0" cellpadding="0" align="center"; style="width:100%;" |- | valign="middle" align="left" bgcolor=EEEDF7|[[Image:Weibull-E…'

New page

{| class="FCK__ShowTableBorders" border="0" cellspacing="0" cellpadding="0" align="center"; style="width:100%;"
|-
| valign="middle" align="left" bgcolor=EEEDF7|[[Image:Weibull-Examples-banner.png|400px|center]]
|}<br><br>'''2 Parameter Exponential Distribution RRY'''

Fourteen units were being reliability tested and the following life test data were obtained:

{|align="center" border=1 cellspacing=0
|-
|colspan="2" style="text-align:center"| Table - Life Test Data
|-
|- align="center"
!Data point index
!Time-to-failure
|- align="center"
|1 ||5
|- align="center"
|2 ||10
|- align="center"
|3 ||15
|- align="center"
|4 ||20
|- align="center"
|5 ||25
|- align="center"
|6 ||30
|- align="center"
|7 ||35
|- align="center"
|8 ||40
|- align="center"
|9 ||50
|- align="center"
|10 ||60
|- align="center"
|11 ||70
|-align="center"
|12 ||80
|- align="center"
|13 ||90
|- align="center"
|14 ||100
|}

Assuming that the data follow a two-parameter exponential distribution, estimate the parameters and determine the correlation coefficient, <math>\rho </math>, using rank regression on Y.

'''Solution'''

Construct the following Table, as shown next.

<center><math>\overset{{}}{\mathop{\text{Table}\text{- Least Squares Analysis}}}\,</math></center>

<center><math>\begin{matrix}
N & t_{i} & F(t_{i}) & y_{i} & t_{i}^{2} & y_{i}^{2} & t_{i} y_{i} \\
\text{1} & \text{5} & \text{0}\text{.0483} & \text{-0}\text{.0495} & \text{25} & \text{0}\text{.0025} & \text{-0}\text{.2475} \\
\text{2} & \text{10} & \text{0}\text{.1170} & \text{-0}\text{.1244} & \text{100} & \text{0}\text{.0155} & \text{-1}\text{.2443} \\
\text{3} & \text{15} & \text{0}\text{.1865} & \text{-0}\text{.2064} & \text{225} & \text{0}\text{.0426} & \text{-3}\text{.0961} \\
\text{4} & \text{20} & \text{0}\text{.2561} & \text{-0}\text{.2958} & \text{400} & \text{0}\text{.0875} & \text{-5}\text{.9170} \\
\text{5} & \text{25} & \text{0}\text{.3258} & \text{-0}\text{.3942} & \text{625} & \text{0}\text{.1554} & \text{-9}\text{.8557} \\
\text{6} & \text{30} & \text{0}\text{.3954} & \text{-0}\text{.5032} & \text{900} & \text{0}\text{.2532} & \text{-15}\text{.0956} \\
\text{7} & \text{35} & \text{0}\text{.4651} & \text{-0}\text{.6257} & \text{1225} & \text{0}\text{.3915} & \text{-21}\text{.8986} \\
\text{8} & \text{40} & \text{0}\text{.5349} & \text{-0}\text{.7655} & \text{1600} & \text{0}\text{.5860} & \text{-30}\text{.6201} \\
\text{9} & \text{50} & \text{0}\text{.6046} & \text{-0}\text{.9279} & \text{2500} & \text{0}\text{.8609} & \text{-46}\text{.3929} \\
\text{10} & \text{60} & \text{0}\text{.6742} & \text{-1}\text{.1215} & \text{3600} & \text{1}\text{.2577} & \text{-67}\text{.2883} \\
\text{11} & \text{70} & \text{0}\text{.7439} & \text{-1}\text{.3622} & \text{4900} & \text{1}\text{.8456} & \text{-95}\text{.3531} \\
\text{12} & \text{80} & \text{0}\text{.8135} & \text{-1}\text{.6793} & \text{6400} & \text{2}\text{.8201} & \text{-134}\text{.3459} \\
\text{13} & \text{90} & \text{0}\text{.8830} & \text{-2}\text{.1456} & \text{8100} & \text{4}\text{.6035} & \text{-193}\text{.1023} \\
\text{14} & \text{100} & \text{0}\text{.9517} & \text{-3}\text{.0303} & \text{10000} & \text{9}\text{.1829} & \text{-303}\text{.0324} \\
\sum_{}^{} & \text{630} & {} & \text{-13}\text{.2315} & \text{40600} & \text{22}\text{.1148} & \text{-927}\text{.4899} \\
\end{matrix}</math></center>

The median rank values ( <math>F({{t}_{i}})</math> ) can be found in rank tables or they can be estimated using the '''Quick Statistical Reference''' in Weibull++.
Given the values in the table above, calculate <math>\hat{a}</math> and <math>\hat{b}</math>:

::<math>\begin{align}
\hat{b}= & \frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}}{{y}_{i}}-(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}})(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}})/14}{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,t_{i}^{2}-{{(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}})}^{2}}/14} \\
\\
\hat{b}= & \frac{-927.4899-(630)(-13.2315)/14}{40,600-{{(630)}^{2}}/14}
\end{align}</math>

or:

::<math>\hat{b}=-0.02711</math>

and:

::<math>\hat{a}=\overline{y}-\hat{b}\overline{t}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{t}_{i}}}{N}</math>

or:

::<math>\hat{a}=\frac{-13.2315}{14}-(-0.02711)\frac{630}{14}=0.2748</math>

Therefore:

::<math>\hat{\lambda }=-\hat{b}=-(-0.02711)=0.02711\text{ failures/hour}</math>

and:

::<math>\hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2748}{0.02711}</math>

or:

::<math>\hat{\gamma }=10.1365\text{ hours}</math>

Then:

::<math>f(t)=(0.02711)\cdot {{e}^{-0.02711(T-10.136)}}</math>

The correlation coefficient can be estimated using equation for calculating the correlation coefficient:

::<math>\hat{\rho }=-0.9679</math>

This example can be repeated using Weibull++, choosing two-parameter exponential and rank regression on Y (RRY), as shown in the figure on the following page.

The estimated parameters and the correlation coefficient using Weibull++ were found to be:

::<math>\hat{\lambda }=0.0271\text{ fr/hr },\hat{\gamma }=10.1348\text{ hr },\hat{\rho }=-0.9679</math>

[[Image:Exponential Example 2 Data Folio.png|thumb|center|400px|]]

Please note that the user must deselect the '''Reset if Loc. Param > T1 on Exp RR''' option on the user setup page.

The probability plot can be obtained simply by clicking the '''Plot''' icon.

[[Image:Exponential Example 2 Plot.png|thumb|center|400px|]]