Reliability Predictions

[ramseng]

I have an easy question which I should know the answer for but can't find the right terms to search for:

I'm conducting a reliability analysis on some equipment. I have some data and have used a chi-squared method to calculate a failure rate, lambda, and then reliability with R(t)=e^-(lambda*t).

This is no problem, but this reliability is for the equipment as a whole. How do I then use this to calculate the probability of a particular piece of equipment failing?

So for example, say you have a fleet of trucks. You've have a years data over your fleet for say, failure of the starter motor. With this data you've calculated R(t)=0.6, and hence unreliability is 0.4. This is then the reliability and unreliability for your fleet's starter motors as a whole. This is the probability of a starter motor failing somewhere in my whole fleet over whatever period I have used. But the likelihood of an individual truck's starter motor failing in this period will be different to this. How do I calculate this?

Hope this makes sense.

 

[IRstuff]

There is a reliability forum should you post a new question in the future.

Your single motor failure rate is simply the total failure rate divided by the number of trucks used to calculate the aggregate failure rate.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529

 

[ramseng]

I had a look for a reliability forum but couldn't find it..

I'm not sure I have an aggregrate failure rate - what I have is an estimated failure rate from some sample data which applies to starter motors in general. Or, I could be in the same situation by using a manufacturer's quoted MTBF. This statistic tells me how starter motors fail across the whole population of starter motors in existence. What I want to know is given a specific truck, what is the risk of its starter motor failing? I can't divide by the number of trucks. If I have ten trucks and my competitor has one, the likelihood of one my specific trucks failing is the same as his. Dividing the failure rate by the number of trucks leads to conclusion I can make a specific truck more reliable by buying more trucks!

This is analogous to the same birthday problem - the probability of two people in a group of ten sharing any birthday is x. But the probability of two people sharing the birthday of 10 May is much less. This is a similar problem - finding the probability of a specific occurrence.

I hope I'm making sense - I'm not sure I am!

 

[cloa]

http://www.eng-tips.com/threadminder.cfm?pid=961

Looking for a forum use the search function that is at top in the white area.

http://www.eng-tips.com/threadminder.cfm?pid=1529

Use translation assistance for Engineers forum

Note the rules include No Student posting

 

[IRstuff]

You're over complicating things. You have your failure rate for the fleet, since you've managed to calculate the reliability of the fleet, since
R(t) = exp(-λt)

Then λ/n is the fleet average failure rate for a single truck. You can calculate from your historical data the specific λ for a specific truck, but there's no guarantee that a single truck's reliability data is exactly representative of that truck

You are confused about: "If I have ten trucks and my competitor has one, the likelihood of one my specific trucks failing is the same as his."

The likelihood of one of your trucks failing is the same as one of his, assuming their they're the same equipment. Your FLEET failure rate is 10 times the failure rate of a single truck. This is how failure rates of mass-produced products are calculated: (failures of aggregate) / (total aggregate operational hours) = failure rate of single product. Implicit in the calculation is the number of products, so if you have 10 failures in 100 units over an aggregate of 80,000 hr you get 1.1 fails/yr for the aggregate, and 0.011 fails/yr for a single unit.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529

 

[EmmanuelTop]

Please note that expression R(t) = exp(-λt) applies only for constant failure rate. Given that trucks have multiple failure modes and increasing failure rate over time, this will be just an approximation (overestimated availability).

Dejan IVANOVIC
Process Engineer, MSChE

 

[IRstuff]

"R(t) = exp(-λt) applies only for constant failure rate"

And it only applies in the flat part of the bathtub curve, which is after infant mortality and before wearout, The bottom of the bathtub may have a slight positive slope, but is still treated as a constant failure rate region.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529

 

[ramseng]

Please note that expression R(t) = exp(-λt) applies only for constant failure rate. Given that trucks have multiple failure modes and increasing failure rate over time, this will be just an approximation (overestimated availability).

Cheers EmmanuelTop I am aware of this - I'm not actually analysing trucks, they are just an analogy. I'm pretty happy to assume a constant failure rate for the equipment I'm looking at.

IRStuff, thanks very much for taking the time to respond.

You're over complicating things. You have your failure rate for the fleet, since you've managed to calculate the reliability of the fleet, since
R(t) = exp(-λt)

Ah, I didn't make this clear - I've estimated λ for a single starter motor.

I think I have confused myself. I've already started with the figure I wanted - the likelihood of a single truck failing. With this I can then calculate probabilities for the fleet as a whole if I wanted.

So if I want to calculate the probability of one specific starter motor failing in a year, it is R(t) = exp(-λ*8760) assuming λ was in hours^-1. I can then use, using normal probability formula to calculate anything else I wished, e.g. the likelihood at least two starters failing, or half of them failing, etc.

 

[MikeHalloran]

Starter motors are a bad example, because their life should relate more strongly to number of starts than to total run time, so the math only works for fleets with the exact same duty cycles.



Mike Halloran
Pembroke Pines, FL, USA