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This example validates the results for series systems in BlockSim's analytical and simulation diagrams.
Reference Case
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.
Data
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.
Result
Since the lives of the bulbs are modeled with the Weibull distribution, the reliability of a single bulb is calculated as:
- [math]\displaystyle{ 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 R0 into Equation 4.11 on page 72:
- [math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ 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.
Results in BlockSim