It may be possible to pass the CRE exam knowing one formula.

The math elements of the exam may take a bit of time to solve, and knowing reliability statistics well is a good plan heading into the exam. Knowing the exponential distribution reliability function is one that you should memorize.

For some reason, it is a favorite formula for questions. Maybe because it is simple, is a key concept or because a lot of questions have been written using the formula. I donâ€™t know.

The reliability function provides the probability of success or surviving till a time of interest. So, we want to know what is the chance our new car will survive 5 years if we have the failure rate (or MTBF) we can calculate the probability. The results may be since the carâ€™s reliability over 5 years. I hope my car is 99% or better over five years.

The exponential distribution has only one parameter, lambda or itâ€™s inverse, MTBF (we use theta commonly). Like all distributions, the exponential has probability density, cumulative density, reliability and hazard functions.

The Reliability Function for the Exponential Distribution

$$ \large\displaystyleÂ R(t)={{e}^{-\lambda t}}$$

Given a failure rate, lambda, we can calculate the probability of success over time, t. Cool.

## An Example

Letâ€™s say we want to know if a new product will survive 850 hours.

We have data on 1,650 units that have operated for an average of 400 hours. Overall there have been 145 failures. We are assuming an exponential distribution – thus we do not need to know the time to failure for each failure, just the total time and number of failures.

Assuming an exponential distribution and interested in the reliability over a specific time, we use the reliability function for the exponential distribution, shown above.

The problem does not provide a failure rate, just the information to calculate a failure rate. For lambda we divided the number of failures by the total time the units operate. Weâ€™re given 1,650 its ran on average 400 hours, thus 400 time 1,650 provides the total time.

$$ \large\displaystyleÂ \lambda =\frac{\text{ }\!\!\#\!\!\text{ failures}}{\text{Total Time}}=\frac{145}{1,650\times 400}=0.0002197/\text{hour}$$

With the failure rate we can calculate the reliability at 850 hours

$$ \large\displaystyleÂ R(850)={{e}^{-0.0002197\times 850}}=0.829=83%$$

## Conclusion

Based on some testing we find a failure rate and can calculate the probability of success (reliability) over a time period of interest.

The basic assumption is the failures have an equal chance to occur each hour of operation, which if true (and in real life, you should check) then this is a pretty simple calculation.

Related:

Common formulasÂ (article)

Exponential ReliabilityÂ (article)

Reliability FunctionÂ (article)

Tom Roltsch says

I have a B.S. in physics and Master’s in Engineering. I took the CRE exam and I needed every minute of the 4 hours that was allotted for the exam. Memorizing one formula is not going to empower someone to pass. The test covers all aspects of reliability, probability, statistics, parallel and series systems, reliability block diagrams and more.

Fred Schenkelberg says

Hi Tom,

Thanks for the comment and note on the CRE exam – it is a tough exam for most. I agree that knowing only one formula isn’t sufficient, yet in my experience knowing this one formula helps you solve many of the math related questions (certainly not all). Better is to have a command of reliability statistics. I have found e to the negative lambda t to find it’s way into a range of different questions mostly as it is easy to write a question based on the formula’s underlying concepts.

Further, the CRE is not the end of one’s work to improve and master the range of topics we should know as a reliability professional. hence the Accendo Reliability site and the many topics discussed here.

Cheers,

Fred

A. Nony Mous Jr says

I’m Studying for an exam in Digital System Design and Reliability Engineering, a level 2 module in my BEng Electronic and Computer Engineering .

This article has really cleared up some of the topics that we have gone over in lectures. Really appreciate you writing this, Thanks!

Fred Schenkelberg says

good luck and in practice please be sure to use the appropriate distribution, as the exponential is rarely appropriate or useful. cheers, Fred

rishi says

hi.

I understand up to : R (850) = e -.002197 *850

I am not sure sure how you got the .829 and the 83.

Currently sitting the CQE exam

Fred Schenkelberg says

Enjoy the process of prep for the CRE exam and learn and use that information.

I just double checked the math, and e raised to (-0.0002197 times 850) equals 0.829, which I then rounded to 0.83

hope that helps

Cheers,

Fred

Atrace says

rishi

On a Ti-30 calculator, -0.00002197 x850=-0.166745

On Ti-30

-0.166745

Hit 2nd and button LN

This will give your reliability of 0.829655 round to 83%

Fred Schenkelberg says

thanks for the step by step calculation instructions, well done. cheers, Fred

teacher says

A pump operates continuously with a mean time to fail of 200 hours that follows the exponential distribution. A second, identical pump is placed in standby redundancy, and the mean time to fail while the pump is inactive is 1,000 hours. The standby time to fail is also exponentially distributed. What is the mean time to fail for the system, and what is the system reliability at time = 300 hours?

Fred Schenkelberg says

Hi

Take a look at the article https://accendoreliability.com/standby-redundancy-equal-failure-rates-imperfect-switching/ it may provide a clue on how to solve this one.

cheers,

Fred

PS: please do not use MTBF

Debo says

Is that right that preventive maintenance should not be performed on a component when the failure rate is constant (Exponential Distribution)?

Fred Schenkelberg says

Hi and yes, preventative maintenance is not useful when the underlying failures actaully occur with a constant failure rate…. which, btw, failure rarely actually occur with a constnat hazard rate or along the exponential distribuiton.

The element repaired or replaced simply has the same hazard rate before and after the maintenance action – thus, what’s the point? Also, the maintenance action introduces a finite yet real chance for damage and additional correction maintenance.

cheers,

Fred

Alexander says

Is testing 1000 parts for 1-hour equivalent to testing 1 part for 1000 hours if an exponential distribution is assumed?

Fred Schenkelberg says

You are correct – consider that the item if actually (almost nothing does) follow an exponential distribution and has the constant hazard rate – then each hour for the unit has the same chance of failure. Doing the math for setting a test of items that enjoy a constant hazard rate does have the feature that 1 unit for 1000 hours creates the same result as testing 1000 items for 1 hour each… odd, yet that is what math says is true. in practice, keep in mind that very, very few items actually have a constant hazard rate, which doesn’t change just because we assume it is true.

cheers,

Fred

Bob says

Hi

I am just wondering why would your solution be correct.

You state that

Î»= # failures/Total Time=145/(1,650Ã—400)=0.0002197/hour

So, I use your result for Î»

R(400)=e^âˆ’0.0002197Ã—400=0.9159

This corresponds ca. 1512 survivors (0.9159×1650) which means that at time 400h there failed 138 pieces.

This is however incorrect as in your assignment you state there are 145 failures and rounding is not an issue.

A have no idea where your formula originates but seems incorrect.

A correct solution should be based on:

R(t)=e^-Î»t

R(400)=(1650-145)/1650=e^âˆ’Î»Ã—400 where Î»=0.000022996

and then R(850)=e^âˆ’Î»Ã—850=0.8225

Could you clarify, please?

Kind regards

Fred Schenkelberg says

I see your point, we have evidence that with the sample set of 1650 units we enjoyed an average run time of 400 hours – some of those items had failed at some point which may have been at any point such that the overall tally of time operating for all units is 400. Some may have run much longer as some may have failed very early.

The unbias estimator for the failure rate, lambda, for the exponential distribution is the total count of failures divided total time of units operating, failed or not. It is just an estimator for the distribution that best fits the data in the sample.

In the calculation, you have the success rate set equal to the exponential distribution reliability function. The R in the reliability function is the probability of a unit surviving till the time, t. it is not the same as the count of successes to a time t. The probability of success and failure or success rate are not equitable.

At least that is how I understand it.

cheers,

Fred

Bob says

Hi, I will use your formula, as this pops up in CQE trial tests as an instantaneous failure rate estimator. The derivations are clear now, still not sure why one estimator is better than another. This kind of higher math is not my cup of a tea

Thank you for your feedback!

Bob

Derek Brown says

Fred,

Can you give an example of items that do have a constant hazard rate?

Cheers

Fred Schenkelberg says

Hi Derek,

The only thing I have found to have a constant failure rate, well described by the exponential distribution is a ceramic mug – with the primary risk being drop breakage during the course of normal home or office use.

Nothing else and no dataset I’ve taken a look at or failure mechanism considered justifies the use of a exponential distribution model.

cheers,

Fred