Calculating Reliability for Subsystems in Series

[JeantheBigone]

I'm hoping some Covers can help me.

I'm studying for one of the ASQ exams and am stumped by a sample test question on reliability. This is not my area so maybe I'm missing the obvious.....

the question is:
A series system has three subsystems with the following distributions:

1) exponential life, MTBF=10,000 h

2) Weibull, theta = 12,000 hrs, beta = 0.42

3) exponential life, MTBF = 25,000 hrs

What is the reliability of the system at t=5000 hrs?

I get R(1) = 0.606, R(2) = 0.839, R(3) = 0.819,

so R(1)x R(2)x R(3) = 0.416, which is one of the answers.

The answer key however says that the correct answer is 0.249.

Can anyone help me see what I'm doing wrong?

 

[Steve Prevette]

My answer was incorrect, deleted.

 

[Miner]

I'm hoping some Covers can help me.

I'm studying for one of the ASQ exams and am stumped by a sample test question on reliability. This is not my area so maybe I'm missing the obvious.....

the question is:
A series system has three subsystems with the following distributions:

1) exponential life, MTBF=10,000 h

2) Weibull, theta = 12,000 hrs, beta = 0.42

3) exponential life, MTBF = 25,000 hrs

What is the reliability of the system at t=5000 hrs?

I get R(1) = 0.606, R(2) = 0.839, R(3) = 0.819,

so R(1)x R(2)x R(3) = 0.416, which is one of the answers.

The answer key however says that the correct answer is 0.249.

Can anyone help me see what I'm doing wrong?

You made your mistake in the reliability of R(2) at 5000 hrs. Your other numbers and calculations are correct.

 

[Miner]

If R(1) is probability of failure, you have calculated the probability that all three subsystems fail. I suspect you need to go with (1-R1)x(1-R2)x(1-R3) is the survival probability, and then . . .

Nope. That is the formula for a parallel system.

 

[JeantheBigone]

Thank you.

Any chance you could elaborate? Or point me to a reliable source of information? I'm using the ASQ CQE book along with assorted miscellaneous references and this one has me stumped.

Edited to add: I got it.....thanks again.

 

[normzone]

[JeantheBigone] - when you have the time, could you possibly unpack how this works using words having three four or less syllables for those of us that never passed math classes? I'm aware that this stuff can be calculated but never learned it.

Thanks -

 

[Miner]

Thank you.

Any chance you could elaborate? Or point me to a reliable source of information? I'm using the ASQ CQE book along with assorted miscellaneous references and this one has me stumped.

Edited to add: I got it.....thanks again.

See Weibull distribution for details about the method. Attached is a file showing the actual calculation.

 

[JeantheBigone]

normzone, I'm honored that you're asking me for my help. Well actually I'm a little startled. :mg: Of all the areas that are covered in the ASQ CQE, system reliability is probably my weakest. So anything you'd get from me would be "book larnin'."

Still, I can try to pass on the little bit that I've learned, which is probably just enough to be dangerous. What is it you're looking for? Prose explanations? Plug and chuck equations? I'll do my best.

I guess I'm not allowed to use the word "probability" though, huh?

 

[normzone]

Thanks [JeantheBigone]. Yeah, a prose summary of the logic and method would be great. The shorter you can make it, the more valuable it would be.

Why? Because I'll be passing it on to folks who know even less than I do about it. Part of what I get paid in the day job is being a knowledge meme carrier.

 

[JeantheBigone]

normzone, this is for you. I hope it's what you were looking for.

Instead of talking about system reliability and probability, I'll use an example that will hopefully be easy to digest and then do a "bait and switch" at the end.

Let's say you and your partner like to play tennis doubles Saturdays at 10 AM. You know two people, Bob and Jane, and Bob and Jane are not a couple, but they both play tennis and can play together as a double.

The chances that Bob will show up to the club are 50-50 and for Jane the same, 50-50.

What are the chances that you'll get to play against Bob and Jane? Should be pretty easy to see that it's 1 in 4, or 25 %, right? Because they could both be no-shows, Bob could show up but not Jane, or Jane but not Bob, or both of them. Those are the only four possibilities, and only one of those four means you get to play.

Now let's say that one of them is 100 % reliable, and make it Bob. Now it should be easy to see that the chances of having them to play against is 50 %, because it's only a matter of whether Jane shows up, which is 50-50.

If the chances of Bob or Jane showing up depend on the time of day, say zero at 6 AM, 50-50 at 10 AM, 75 % at 2 PM etc, you can have time-dependent reliability information which you use to make the calculation. If you can find an equation which gives R as a function of t, you can use that R(t).

So far so good?

Here comes the switch. Instead of asking whether two people are reliable enough to show up at the same time to play tennis, make it a more realistic system. Say you have a machine that makes donuts. There are three steps, extruding the dough into a donut shape into the hot oil, keeping the oil at the right temperature, and lifting the donuts out of the hot oil at the right time. If any of these three steps isn't done right, no donuts. This is an example of a system with three subsystems in series. Just like the tennis players, each of the three subsystems has its own probability, and if you play with simple probabilities (like 100 % or 50-50 for each of the three subsystems and write out the possible outcomes, you should be able to convince yourself, or someone who isn't good at math, that you multiply the individual probabilities to get the system reliability.

Hope that hit the mark.