During design and development, Reliability Engineers often receive reliability parameters in many forms. The most common reliability parameter is the mean time to failure (MTTF), which can also be specified as the failure rate (this is expressed as a frequency or Conditional Probability Density Function (PDF)) or the number of failures during a given period.

These parameters are very useful for systems that are frequently operated, such as most vehicles, machinery, and electronic equipment. Reliability increases as the MTTFÂ increases. The MTTF is usually specified in hours, but can also be used with other units of measurement such as miles or cycles.

Once all the parameters are received, it then will become the Reliability Engineer’s responsibility to model the system using the appropriate reliability math models.Â Â Letâ€™s suppose that your components are modeled as a simple series system.Â Very easy right?Â Just multiply them. Sometimes is not that easy…

Consider that the component’s reliability isÂ provided in the following forms:

Component (1):Â WeibullÂ (Î²= 1.6 , Î· = 9,500)

Component (2):Â Exponential (Î»= 0.000087)

Component(3):Â LognormalÂ (Î¼Â = 7.5 , Ïƒ = 0.81)

The question you are asked is, “What is theÂ reliability of the system at 1,000 hours?”

You will now have to do a little math to calculate the reliability at t = 1,000 hours.

Component (1):Â WeibullÂ (Î²= 1.6 , Î· = 9,500)

$$ \displaystyle\large R\left( 1000 \right)={{e}^{-{{\left( {1000}/{9500}\; \right)}^{1.6}}}}=0.9731$$

Component (2):Â Exponential (Î»= 0.000087)

$$ \displaystyle\large R\left( 1000 \right)={{e}^{-\left( 0.000087 \right)\left( 1000 \right)}}=0.9166$$

Component(3):Â LognormalÂ (Î¼Â = 7.5 , Ïƒ = 0.81)

$$ \displaystyle\large \begin{array}{l}R\left( t \right)=1-\Phi \left( \frac{\ln t-\mu }{\sigma } \right)\\\text{where }\Phi \left( t \right)\text{ is the standard normal cumulative distribution}\\R\left( 1000 \right)=1-\Phi \left( \frac{\ln 1000-7.5}{0.81} \right)=0.7673\end{array}$$

Thus, R(1,000) = (0.9731)(0.9166)(0.7673) = 0.6844

Questions like these appear in the CRE exam, it will not just be a simple multiplication of 3 blocks in series.Â Be prepared,Â and study!!!

Related:

Weakest LinkÂ (article)

Exponential ReliabilityÂ (article)

Parallel SystemsÂ (article)

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