Parallel Systems RBD - Revision history

Kate Racaza at 17:12, 31 July 2015

2015-07-31T17:12:07Z

← Older revision		Revision as of 17:12, 31 July 2015
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	{{Reference Example\|{{Banner BlockSim Reference Examples}}}}		{{Reference Example\|{{Banner BlockSim Reference Examples}}}}
	This example validates the results for ~~series~~ systems in BlockSim's analytical and simulation diagrams.		This example validates the results for parallel systems in BlockSim's analytical and simulation diagrams.



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	The data set is from example 4.6 on page 73 in the book ''Life Cycle Reliability Engineering'' by Dr. Guangbin Yang, John Wiley & Sons, 2007.		The data set is from example 4.6 on page 73 in the book ''Life Cycle Reliability Engineering'' by Dr. Guangbin Yang, John Wiley & Sons, 2007.


	{{Reference_Example_Heading2}}		{{Reference_Example_Heading2}}

	A lighting system uses three identical bulbs. The lifetimes of the bulbs are assumed to follow 2P-Weibull with parameters α = 35,800 hours (this is eta value in BlockSim ~~software~~) and β = 1.35. We calculate the reliability of the system after 8760 hours of use, and the number of bulbs needed to be connected in parallel to achieve 99.99% reliability at that time.		A lighting system uses three identical bulbs. The lifetimes of the bulbs are assumed to follow a 2-paremeter Weibull with parameters α = 35,800 hours (this is the eta value in BlockSim) and β = 1.35. We calculate the reliability of the system after 8760 hours of use, and the number of bulbs needed to be connected in parallel to achieve 99.99% reliability at that time.


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	Therefore, ~~it is concluded that~~ minimum of 5 bulbs in parallel configuration is needed to achieve 99.99% reliability.		Therefore, a minimum of 5 bulbs in parallel configuration are needed to achieve 99.99% reliability.


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	[[Image:parallel_rbd.png\|center]]		[[Image:parallel_rbd.png\|center]]

	Each bulb is modeled with 2P-Weibull distribution with ~~the given~~ parameters; eta = 35,800 hours and beta = 1.35.		Each bulb is modeled with a 2-parameter Weibull distribution with parameters eta = 35,800 hours and beta = 1.35.


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	The reliability after 8760 hours of use is estimated to be 99.73%, which is the same result as calculated in the reference book.		The reliability after 8760 hours of use is estimated to be 99.73%, which is the same result as calculated in the reference book.

	[[Image:parallel_qcp.png\|center]]		[[Image:parallel_qcp.png\|center\|500 px]]


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	The analysis shows that the number of bulbs needs to be increased to 1.55506. Therefore, the total number of bulbs needed in parallel to achieve 99.99% reliability after 8760 hours of use is 4.66518 bulbs, which is the sum of "Equivalent Parallel Units" column shown below. This number is almost exactly the same as the one calculated in the reference book.		The analysis shows that the number of bulbs needs to be increased to 1.55506. Therefore, the total number of bulbs needed in parallel to achieve 99.99% reliability after 8760 hours of use is 4.66518 bulbs, which is the sum of the "Equivalent Parallel Units" column shown below. This number is almost exactly the same as the one calculated in the reference book.

	[[Image:~~parallel_allocation~~.png\|center]]		[[Image:parallel_allocationresult.png\|center]]


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	We can also estimate the results by using the simulation tool in BlockSim. The simulation settings are shown below.		We can also estimate the results by using the simulation tool in BlockSim. The simulation settings are shown below.

	[[Image:parallel_sim.png\|center]]		[[Image:parallel_sim.png\|center\|500 px]]


	The point reliability after 8760 hours of use is estimated to be 99.77%.		The point reliability after 8760 hours of use is estimated to be 99.77%.

	[[Image:parallel_simqcp.png\|center]]		[[Image:parallel_simqcp.png\|center\|500 px]]

Kate Racaza at 16:58, 31 July 2015

2015-07-31T16:58:28Z

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 In BlockSim, the lighting system RBD is configured as shown below.
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 Each bulb is modeled with 2P-Weibull distribution with the given parameters; eta = 35,800 hours and beta = 1.35.
 '''Analytical Proof'''
--->
+[[Image:parallel_simqcp.png|center]]

Kate Racaza: Created page with '{{Reference Example|{{Banner BlockSim Reference Examples}}}} This example validates the results for series systems in BlockSim's analytical and simulation diagrams. {{Referenc…'

2015-07-31T16:42:55Z

Created page with '{{Reference Example|{{Banner BlockSim Reference Examples}}}} This example validates the results for series systems in BlockSim's analytical and simulation diagrams. {{Referenc…'

New page

{{Reference Example|{{Banner BlockSim Reference Examples}}}}
This example validates the results for series systems in BlockSim's analytical and simulation diagrams.

{{Reference_Example_Heading1}}

The data set is from example 4.6 on page 73 in the book ''Life Cycle Reliability Engineering'' by Dr. Guangbin Yang, John Wiley & Sons, 2007.

{{Reference_Example_Heading2}}

A lighting system uses three identical bulbs. The lifetimes of the bulbs are assumed to follow 2P-Weibull with parameters α = 35,800 hours (this is eta value in BlockSim software) and β = 1.35. We calculate the reliability of the system after 8760 hours of use, and the number of bulbs needed to be connected in parallel to achieve 99.99% reliability at that time.

{{Reference_Example_Heading3}}

Since the lives of the bulbs are modeled with the Weibull distribution, the reliability of a single bulb is calculated as:

::<math>R_{0} = exp \left [ -\left(\frac{t}{\lambda}\right)^{\beta} \right ] = exp\left [ -\left(\frac{8,760}{35,800}\right)^{1.35} \right ] = 0.8611\,\!</math>

Substituting the value of R<sub>0</sub> into Equation 4.11 on page 72:

::<math>R = 1 - (1-R_{0})^{n}\,\!</math>

where ''R'' is the system reliability and ''n'' is the number of identical components that form the system. The reliability of the system can be calculated as:

::<math>R = 1 - (1-0.8611)^{3} = 0.9973\,\!</math>

The minimum number of bulbs required to achieve 99.99% reliability after 8760 hours of use can be calculated via Equation 4.12 on page 72.

::<math>n = \frac{ln(1-R)}{ln(1-R_{0})} = \frac{ln(1-0.9999)}{ln(1-0.8611)} = 4.6658\,\!</math>

Therefore, it is concluded that minimum of 5 bulbs in parallel configuration is needed to achieve 99.99% reliability.

{{Reference_Example_Heading4|BlockSim}}