Example Calculating System Reliability with Duty Cycles - Revision history

Lisa Hacker at 20:57, 18 September 2023

2023-09-18T20:57:33Z

← Older revision		Revision as of 20:57, 18 September 2023
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	<noinclude>{{Banner BlockSim Examples}}		<noinclude>{{Banner BlockSim Examples}}
	''This example appears in the [~~[Time-Dependent_System_Reliability_(Analytical)#Duty Cycle\|~~System ~~Analysis Reference book]~~]''.		''This example appears in the [https://help.reliasoft.com/reference/system_analysis System analysis reference]''.

	</noinclude>		</noinclude>

Richard House at 21:25, 5 January 2016

2016-01-05T21:25:06Z

← Older revision		Revision as of 21:25, 5 January 2016
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	[[Image:Fig 5.15.PNG\|center\|450px\|<div align="center"> Result for the computer system reliability.</div>]]		[[Image:Fig 5.15.PNG\|center\|450px\|<div align="center"> Result for the computer system reliability.</div>\|link=]]

Richard House at 23:08, 4 January 2016

2016-01-04T23:08:47Z

← Older revision		Revision as of 23:08, 4 January 2016
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	[[Image:BS5ex4_new.png\|center\|~~250px~~]]		[[Image:BS5ex4_new.png\|center\|400px\|link=]]

	'''Solution'''		'''Solution'''

Richard House at 16:47, 3 February 2014

2014-02-03T16:47:49Z

← Older revision		Revision as of 16:47, 3 February 2014
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	[[Image:~~BS5ex4~~.png\|center\|250px]]		[[Image:BS5ex4_new.png\|center\|250px]]

	'''Solution'''		'''Solution'''

Richard House at 21:24, 17 September 2012

2012-09-17T21:24:11Z

← Older revision		Revision as of 21:24, 17 September 2012
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	</noinclude>		</noinclude>
	Consider a computer system with three components: a processor, a hard drive and a CD drive in series as shown next. Assume that all three components follow a Weibull failure distribution. The processor has the following parameters, <math>{{\beta }_{1}}=1.5\,\!</math> and <math>{{\eta }_{1}}=5000\,\!</math>. For the hard drive, the parameters are <math>{{\beta }_{2}}=2.5\,\!</math> and <math>{{\eta }_{2}}=3000\,\!</math>, and for the CD drive they are <math>{{\beta }_{3}}=2\,\!</math> and <math>{{\eta }_{3}}=4000\,\!</math>. Determine the reliability of the computer system after one year (365 days) of operation, assuming that the CD drive is used only 30% of the time.		Consider a computer system with three components: a processor, a hard drive and a CD drive in series as shown next. Assume that all three components follow a Weibull failure distribution. The processor has the following parameters, <math>{{\beta }_{1}}=1.5\,\!</math> and <math>{{\eta }_{1}}=5000\,\!</math>. For the hard drive, the parameters are <math>{{\beta }_{2}}=2.5\,\!</math> and <math>{{\eta }_{2}}=3000\,\!</math>, and for the CD drive they are <math>{{\beta }_{3}}=2\,\!</math> and <math>{{\eta }_{3}}=4000\,\!</math>. Determine the reliability of the computer system after one year (365 days) of operation, assuming that the CD drive is used only 30% of the time.


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	= & {{e}^{-{{\left( \tfrac{365}{5000} \right)}^{1.5}}}} \\		= & {{e}^{-{{\left( \tfrac{365}{5000} \right)}^{1.5}}}} \\
	= & 0.9805\text{ or }98.05%		= & 0.9805\text{ or }98.05%
	\end{align}</math>		\end{align}\,\!</math>


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	= & {{e}^{-{{\left( \tfrac{365}{3000} \right)}^{2.5}}}} \\		= & {{e}^{-{{\left( \tfrac{365}{3000} \right)}^{2.5}}}} \\
	= & 0.9948\text{ or }99.48%		= & 0.9948\text{ or }99.48%
	\end{align}</math>		\end{align}\,\!</math>


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	= & {{e}^{-{{\left( \tfrac{0.3\times 365}{4000} \right)}^{2}}}} \\		= & {{e}^{-{{\left( \tfrac{0.3\times 365}{4000} \right)}^{2}}}} \\
	= & 0.9993\text{ or }99.93%		= & 0.9993\text{ or }99.93%
	\end{align}</math>		\end{align}\,\!</math>


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	= & 0.9805\cdot 0.9948\cdot 0.9993 \\		= & 0.9805\cdot 0.9948\cdot 0.9993 \\
	= & 0.9747\text{ or }97.47%		= & 0.9747\text{ or }97.47%
	\end{align}</math>		\end{align}\,\!</math>

Kate Racaza: Created page with '{{Banner BlockSim Examples}} ''This example appears in the System Analysis Reference book''.
2012-08-20T05:06:01Z

Created page with '<noinclude>{{Banner BlockSim Examples}} ''This example appears in the System Analysis Reference book''. </noinclude…'

New page
<noinclude>{{Banner BlockSim Examples}}
''This example appears in the [[Time-Dependent_System_Reliability_(Analytical)#Duty Cycle|System Analysis Reference book]]''.

</noinclude>
Consider a computer system with three components: a processor, a hard drive and a CD drive in series as shown next. Assume that all three components follow a Weibull failure distribution. The processor has the following parameters, <math>{{\beta }_{1}}=1.5\,\!</math> and <math>{{\eta }_{1}}=5000\,\!</math>. For the hard drive, the parameters are <math>{{\beta }_{2}}=2.5\,\!</math> and <math>{{\eta }_{2}}=3000\,\!</math>, and for the CD drive they are <math>{{\beta }_{3}}=2\,\!</math> and <math>{{\eta }_{3}}=4000\,\!</math>. Determine the reliability of the computer system after one year (365 days) of operation, assuming that the CD drive is used only 30% of the time.

[[Image:BS5ex4.png|center|250px]]

'''Solution'''

The reliability of the processor after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{processor}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{1}}} \right)}^{{{\beta }_{1}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{5000} \right)}^{1.5}}}} \\
= & 0.9805\text{ or }98.05%
\end{align}</math>

The reliability of the hard drive after 365 days of operation is given by:

::<math>\begin{align}
{{R}_{harddrive}}(365)= & {{e}^{-{{\left( \tfrac{365}{{{\eta }_{2}}} \right)}^{{{\beta }_{2}}}}}} \\
= & {{e}^{-{{\left( \tfrac{365}{3000} \right)}^{2.5}}}} \\
= & 0.9948\text{ or }99.48%
\end{align}</math>

The reliability of the CD drive after 365 days of operation (taking into account the 30% operation using a duty cycle of 0.3) is given by:

::<math>\begin{align}
{{R}_{CDdrive}}(365)= & {{e}^{-{{\left( \tfrac{{{d}_{c}}\times 365}{{{\eta }_{3}}} \right)}^{{{\beta }_{3}}}}}} \\
= & {{e}^{-{{\left( \tfrac{0.3\times 365}{4000} \right)}^{2}}}} \\
= & 0.9993\text{ or }99.93%
\end{align}</math>

Thus the reliability of the computer system after 365 days of operation is:

::<math>\begin{align}
{{R}_{s}}(365)= & {{R}_{processor}}(365)\cdot {{R}_{harddrive}}(365)\cdot {{R}_{CDdrive}}(365) \\
= & 0.9805\cdot 0.9948\cdot 0.9993 \\
= & 0.9747\text{ or }97.47%
\end{align}</math>

The result can also be obtained in BlockSim by creating an RBD of the system and using the Quick Calculation Pad (QCP) to calculate the reliability, as shown in the following figure.

[[Image:Fig 5.15.PNG|center|450px|<div align="center"> Result for the computer system reliability.</div>]]