On occasion, we want to estimate the reliability of an item at a specific time.

Maybe we are considering extending the warranty period, for example, and want to know the probability of no failures over one year instead of over the current 3 months.

Or, letâ€™s say you talked to a bearing vendor and have the Weibull parameters and wish to know the reliability value over 2 years.

Whatever specific situation, you have the life distributions parameters. You just need to calculate reliability at a specific time. We can do that and letâ€™s try it with three distributions using their respective reliability functions: exponential, Weibull, and lognormal.

## The Exponential Reliability Function

Letâ€™s say the motor driver board has a data sheet value for Î¸ (commonly called MTBF) of 50,000 hours. Letâ€™s say we are interested in the reliability (probability of successful operation) over a year or 8,760 hours.

The reliability function for the exponential distribution is:

$$ \displaystyle\large R(t)={{e}^{-{}^{t}\!\!\diagup\!\!{}_{\theta }\;}}={{e}^{-\lambda t}}$$

Setting Î¸ to 50,000 hours and time, t, to 8,760 hours we find:

$$ \displaystyle\large R(t)={{e}^{-{}^{8,760}\!\!\diagup\!\!{}_{50,000}\;}}=0.839$$

Thus the reliability at one year is 83.9%.

Tip: check the units of the MTBF and time, t, values, they should match. If using failure rate, lambda, remember that Î¸ = 1 / Î», to also check the units. The time element of the failure rate should be in the same units as the time, t, value. If interested in hours, then the failure rate is in failures per hour.

We could use years, as weâ€™re interested in the reliability over a year, and would have to adjust the MTBF or failure rate value to the units of years. 50,000 hours becomes roughly 5.71 years MTBF, for example. Found by dividing 50,000 by the hours in a year, 8,760.

## The Weibull Reliability Function

A recent accelerated life test of bearings resulted in the Weibull parameters of Î² = 2.1 and Î· = 48,500 hours. What is the reliability at one year, or 8,760 hours?

The reliability function for the Weibull distribution is:

$$ \displaystyle\large R(t)={{e}^{-{{\left( {}^{t}\!\!\diagup\!\!{}_{\eta }\; \right)}^{\beta }}}}$$

Where Î² is the shape parameter and Î· is the scale parameter.

Given the Î² and Î· above what is the reliability at 8,760 hours?

$$ \displaystyle\large R(t)={{e}^{-{{\left( {}^{8,760}\!\!\diagup\!\!{}_{48,500}\; \right)}^{2.1}}}}=0.973$$

Thus given the Weibull parameters we expect a reliability at one year of 97.3%.

Remember that if time equals Î· it doesnâ€™t matter what the Î² value is, the reliability is 36.8%. Try it.

## The Lognormal Reliability Function

The lognormal life distribution is similar to the Weibull distribution, yet parameters are the mean and standard deviation of the transformed data (take the natural log of each data point).

Letâ€™s say we have the lognormal parameters of Î¼â€™ = 6.19Â and Ïƒâ€™ = 0.2642 (calculated using days as the unit of time within the example in Calculating Lognormal Distribution Parameters article).

What is the reliability at one year, or 365 days?

The reliability function of the lognormal distribution is:

$$ \displaystyle\large R(t)=1-\Phi \left( \frac{\ln \left( t \right)-{\mu }’}{{{\sigma }’}} \right)$$

Where the prime indicates the mean and standard deviation are of the transformed data. The Î¦(x) is the standard normal cumulative distribution function (think standard normal table here).

Using the given parameters and setting t to 365 we find:

$$ \displaystyle\large R(t)=1-\Phi \left( \frac{\ln \left( 365 \right)-6.19}{0.26} \right)=0.868$$

Pay attention to units once again.

## Summary

Given life distribution parameters all you need is reliability function for the distribution. A little skill with a calculator and you can quickly calculate the reliability value forÂ time, t.

## Leave a Reply