Weibull Reliability Calculator
Find the chance that a unit is still working at a given time, using Weibull shape and characteristic life.
What it does
Ask an engineer how long a motor will last and the honest answer is a curve, not a number. Some units fail during the first weeks, most run for years, and a few surprise everyone. The Weibull distribution is the standard way to describe that pattern, and it turns a pile of failure records into two numbers that you can use in a design or a spare parts plan.
The NIST and Sematech engineering statistics handbook calls the Weibull a very flexible life distribution model with two parameters. The scale parameter, the characteristic life, is the age at which 63.2 percent of the population has failed. The shape parameter sets the form of the failure rate. Because the two sit apart, the same distribution can describe a product that fails early, one that wears out, and one that fails at a steady rate.
The reliability function is a single line of algebra: the chance that a unit survives to time t is the exponential of minus the power of t over the characteristic life, raised to the shape parameter. The NIST handbook lists the same set of formulas for the density, the distribution function, the failure rate, the mean and the median, and the worked example it gives matches the numbers this page produces.
This page takes the two parameters and a time and answers the questions that come up in practice. How many units are still running at a given age, how many have already gone, what is the median life, what is the mean life, and how fast are failures arriving at that moment.
The limits are worth stating. The Weibull is a statistical model, so it only says something useful once the shape and scale have been fitted to real failure data. It assumes the population is the same kind of unit working under the same conditions, and it says nothing about why a particular unit failed.
How to use
- Enter the shape parameter. Use a value below 1 for a product with early failures, 1 for a constant failure rate, and above 1 for wear out.
- Enter the characteristic life, which is the age at which 63.2 percent of the units have failed. Keep the unit consistent with the time you will use next.
- Enter the time in service for the question you want answered.
- Read the reliability, which is the share of units still working at that age, and the failure share right below it.
- Check the failure rate row to see whether failures are speeding up or slowing down at that moment.
- Compare the mean life with the median life. A large gap tells you the population has a long tail of units that last far longer than the rest.
Formula
- R(t)
- Share of the units that are still working at time t, between 0 and 1.
- t
- Time in service, in the same unit as the characteristic life.
- eta
- Characteristic life, the age at which 63.2 percent of the units have failed.
- beta
- Shape parameter. Below 1 failures come early, 1 is a constant rate, above 1 failures come late.
The mean life is eta times the gamma function of one plus one over beta, and the median life is eta times the natural log of two raised to one over beta. The failure rate at time t is beta over eta times the power of t over eta, raised to beta minus one. When beta is exactly 1 the Weibull becomes the exponential distribution and the mean life equals the characteristic life.
Worked example
Example: pumps with a shape of 1.5 and a characteristic life of 5000 hours
Defaults: a shape parameter of 1.5, a characteristic life of 5000 hours and a time in service of 1000 hours.
- The ratio of the time to the characteristic life is 1000 / 5000 = 0.2.
- Raise that to the shape: 0.2 raised to 1.5 = 0.0894427.
- Reliability: e^(-0.0894427) = 0.9144, so 91.44 percent of the units are still working.
- The share that have failed is 1 - 0.9144 = 0.0856, which is 8.56 percent.
- The median life is 5000 x (ln 2)^(1 / 1.5) = 3916.1 hours, and the mean life is 5000 x gamma of 1 plus 1 over 1.5 = 4513.7 hours.
- The failure rate at 1000 hours is 1.5 / 5000 x 0.2 raised to 0.5, which is 0.1342 failures per 1000 hours.
- The tool shows 91.44 %, 8.56 %, 3916.1 h, 4513.7 h, 0.1342 per 1000 h and an increasing failure rate.
Reference table
| Time in service | Reliability | Share failed | Failure rate per 1000 h |
|---|---|---|---|
| 500 h | 96.89 % | 3.11 % | 0.0949 |
| 1000 h | 91.44 % | 8.56 % | 0.1342 |
| 2000 h | 77.65 % | 22.35 % | 0.1897 |
| 3000 h | 62.83 % | 37.17 % | 0.2324 |
| 4000 h | 48.89 % | 51.11 % | 0.2683 |
| 5000 h | 36.79 % | 63.21 % | 0.3000 |
Limits
- The model only means something once the shape and the characteristic life have been fitted to real failure data for that population.
- It assumes every unit in the population is the same design working under the same conditions, which is rarely true of a field fleet.
- It describes when units fail, not why they fail, and it does not model a mixture of several different failure causes.
- It gives the share of units that are still working, not a date for the failure of one particular unit.
Common mistakes
- Mixing units. A characteristic life in hours with a time in days gives an answer that is off by a factor of 24.
- Reading the characteristic life as the average life. It is the age at which 63.2 percent have failed, and the mean life is a different number whenever the shape is not 1.
- Using the exponential model, which is the Weibull with a shape of exactly 1, for a product that clearly has early failures.
- Forgetting that the same shape and scale only describe the population they were fitted to, so a new batch of parts may need new numbers.
- Treating a reliability of 91 percent as a promise about one unit rather than a statement about a population.
Try it yourself
Quick checks. Answers sit tucked under each question.
How many of 1000 units with a shape of 1.5 and a characteristic life of 5000 h are still working at 1000 h?
About 914 units, because the reliability is 91.44 percent. The same population at 5000 h has only 36.79 percent left, which is 368 units, since the characteristic life is the point where 63.2 percent have already gone.
A population has a shape of 2.5 and a characteristic life of 1000 h. What is the reliability at 400 h?
90.38 percent, so 9.62 percent have failed. The calculation is e^(-(0.4 raised to 2.5)), and the higher shape puts more of the failures in the early part of life than the shape of 1.5 does.
Why does the mean life sit above the median life when the shape is 1.5?
With a shape above 1 the distribution has a long tail of units that last far longer than the rest, and that tail pulls the average up. The median is 3916.1 h while the mean is 4513.7 h, a gap of nearly 600 hours.
Questions
What is the Weibull distribution used for?
It describes the life of a population of units such as bearings, motors, capacitors or lamps. It is the standard model in reliability work because two parameters cover both early failures and wear out.
What does the characteristic life mean?
It is the age at which 63.2 percent of the population has failed. It comes from the shape of the exponential term in the formula, and the NIST handbook notes that a Weibull with a shape of 1 reduces to the exponential distribution with the same mean life.
How do I read the shape parameter?
Below 1 the failure rate falls with age, which points to problems with manufacture or installation. Exactly 1 gives a constant rate. Above 1 the rate rises with age, which is the normal pattern for wear out.
Can I use any time unit I like?
Yes, as long as the characteristic life and the time in service use the same one. The formula is a ratio of the two, so hours with hours and days with days both work.
Where do the two parameters come from?
From failure records. The shape can be read off a Weibull probability plot, where a good fit shows as a straight line, and the characteristic life comes from the scale of that fit.
What is the difference between the mean life and the median life?
The median life is the age at which half the units have failed, and the mean life is the average age of failure across the population. They match only when the shape is 1, and the gap grows as the shape moves away from 1.
Sources
- NIST and Sematech engineering statistics handbook, 8.1.6.2: Weibull formulas
- NIST and Sematech engineering statistics handbook, 8.1.2.2: reliability or survival function
- NIST: chapter 8 of the engineering statistics handbook on product reliability