Quizzes & Puzzles13 mins ago
How Long Until The First Of 60 Leds Fail?
13 Answers
A mirror has 60 LEDs, each lasts 5000 hours on average. When, on average, will the first LED fail. Obviously it depends on the spread of failure rates. 60 candles would all fail at similar times. So assume that 5000 hours is a half-life, as for radioactivity.
Answers
The radioactive analogy only really works for large numbers of LEDs, since just checking it the implication of a radioactive decay formula is that if you had one LED with a half-life of -- well, any half-life really, then you can be sat there until the end of the universe waiting for it to decay. On the other hand, if you had say a million LEDs then the first one...
22:30 Mon 25th Jan 2016
No way of telling without more information. You can't assume that 5000 hours is equivalent to a half life by the way as it is doubtful that any of the LEDs woulld last even 10,000 hours but we really don't know. If they have been manufactured to very stringent standards of uniformity they could all be defunct at 5001 hours. (not likely though)
Does this help?http://www.thelightbulb.co.uk/resources/light_bulb_average_rated_life_time_hours
"The two important factors when understanding the life expectancy of your chosen Light bulb are the Hours and the L rating stated." If your 5000 hours is at "L50" then "half life" calculations may be useful.
"The two important factors when understanding the life expectancy of your chosen Light bulb are the Hours and the L rating stated." If your 5000 hours is at "L50" then "half life" calculations may be useful.
Thanks to everyone. I've heeded Old_Geezer and decided to buy a different mirror. But I was really interested in the statistical answer to the question: If x items have a half-life of t, when will the first fail, on average. This is a well-defined question fiction-factory, but you are correct that we should need the variance for other distributions e.g. Gaussian. Yes, Jonathan-Joe, I was assuming that 5000 hours was L50, but I've no evidence that half-life is appropriate for LEDs. Can we please forget LEDs and answer the question as an exercise in probability?
This is chap 1 of a textbook but I cant remember which one
it is the same one where you find standard cable theory
The prob was first enumerated I think by Edison in 1875
and is all to do with ... if one fails then the whole thing fails
(telephones and valve computers )
and you know I think it is 300 h
I agree with the others that you need a standard deviation ( or perhaps ooer a standard error of the mean ) but I am also sure that this question first surfaced at the end of the nineteenth century - I remember reading about it in 1970 - but I cant remember the answer ! eek
it is the same one where you find standard cable theory
The prob was first enumerated I think by Edison in 1875
and is all to do with ... if one fails then the whole thing fails
(telephones and valve computers )
and you know I think it is 300 h
I agree with the others that you need a standard deviation ( or perhaps ooer a standard error of the mean ) but I am also sure that this question first surfaced at the end of the nineteenth century - I remember reading about it in 1970 - but I cant remember the answer ! eek
google
reliability of thermionic valves
and it is pretty obvious you need a failure rate
and the average can be the same 5000 with a spread that is different
BUT
where is a mathematician ? pleez
if you say there is a constant failure rate - then there is only one that will fit an average of 5000
yes yes if one fails an hour then of 60 then 30 will have failed within 30 hours - actually on the linear model - half have failed at 5000 h
and zoop say it is 50 because I really cant be ossed to divide 5/6 it is late
then twenty five will have failed by 5000 hours and so the first one fails at 200 h
you have assumed radio activity as a model
I have assumed linear failure rate as I found it as a reasonable assumption in a book on thermionic valve failures....
ola chico
reliability of thermionic valves
and it is pretty obvious you need a failure rate
and the average can be the same 5000 with a spread that is different
BUT
where is a mathematician ? pleez
if you say there is a constant failure rate - then there is only one that will fit an average of 5000
yes yes if one fails an hour then of 60 then 30 will have failed within 30 hours - actually on the linear model - half have failed at 5000 h
and zoop say it is 50 because I really cant be ossed to divide 5/6 it is late
then twenty five will have failed by 5000 hours and so the first one fails at 200 h
you have assumed radio activity as a model
I have assumed linear failure rate as I found it as a reasonable assumption in a book on thermionic valve failures....
ola chico
I am also sure we need to know what the distribution is. If RevG wants to assume it's exponential we could calculate it, but as PP says it may be much more like a linear distribution. Or it could be a normal distribution.
This book here covers the topic. I've only glanced at it but I think it refers to an exponential rate
https:/ /books. google. co.uk/b ooks?id =A5-NAg AAQBAJ& amp;pg= PA140&a mp;lpg= PA140&a mp;dq=f ailure+ rates+o f+light +bulbs+ distrib ution&a mp;sour ce=bl&a mp;ots= eFzJW4p tkH& ;sig=dq q3OziQ3 5d0hhG5 xY36rFn fkPs&am p;hl=en &sa =X& ved=0ah UKEwjjq sDA2L_K AhUG1xQ KHcMSDR 8Q6AEII zAA#v=o nepage& amp;q=f ailure% 20rates %20of%2 0light% 20bulbs %20dist ributio n&f =false
This book here covers the topic. I've only glanced at it but I think it refers to an exponential rate
https:/
Hi FF
nice to see you contributing
I thought a failure rate was more realistic if it were poisson
we have an average life of 5000
and for poisson mu = sigma - the variance = the average
but I am not sure if mu is 5000
my brain has kinda burnt out on that
and I wondered if ( since you are good at math and I am pretty crap to be honest ) if you could construct a model on a poisson failure rate rather than linear. I couldnt
nice to see you contributing
I thought a failure rate was more realistic if it were poisson
we have an average life of 5000
and for poisson mu = sigma - the variance = the average
but I am not sure if mu is 5000
my brain has kinda burnt out on that
and I wondered if ( since you are good at math and I am pretty crap to be honest ) if you could construct a model on a poisson failure rate rather than linear. I couldnt
Hi PP- Poisson: sounds a bit fishy to me. Sorry, it's actually a good call. But could be chi-squared. It was always easier in exams when they told you which distribution to assume.
I think you are right about poisson-
If X ∼ Po(λ) then
µ = λ and σ²= λ
There is this another formula I recall for variance
Var(X) = E(X²) – [E(x)]²
But we'd need some more info on the distribution I think before we can find E(x²)
But I've got the excuse that I've got to set off to work now so i'll leave you to it. There are a few around on here like jim360 who, unlike me, have done this stuff within the 40 years or so. I still think we need more info on the distribution though
I think you are right about poisson-
If X ∼ Po(λ) then
µ = λ and σ²= λ
There is this another formula I recall for variance
Var(X) = E(X²) – [E(x)]²
But we'd need some more info on the distribution I think before we can find E(x²)
But I've got the excuse that I've got to set off to work now so i'll leave you to it. There are a few around on here like jim360 who, unlike me, have done this stuff within the 40 years or so. I still think we need more info on the distribution though
Using what you've all said, I estimate 2 days, but please can someone who has studied statistics correct me. Here's my "solution": Assume the LEDs flash randomly with 30 flashes in 5000 days. So the average flash rate is 1 per week. For a Poisson distribution, the time to the first flash will be 1^1 * e^(-1) / 1! i.e. 0.368 weeks or about 2.5 days. If instead the LEDs fail at their first flash, there would be fewer than 30 failures in 5000 hours (because no LED could be counted more than once). If each LED has a half-life of 5000 hours we need a higher failure rate. As a pure guess, let's increase the failure rate by a factor of 1.414. So the time to the first failure is 2.5/1.4 or about 1.8 days. (I see that asking people to forget LEDs and consider radioactivity was as successful as asking someone not to think of an elephant.)
The radioactive analogy only really works for large numbers of LEDs, since just checking it the implication of a radioactive decay formula is that if you had one LED with a half-life of -- well, any half-life really, then you can be sat there until the end of the universe waiting for it to decay. On the other hand, if you had say a million LEDs then the first one would probably fail in a little under 30 seconds. So the expected failure time is heavily correlated with the number of LEDs you had to start with.
Anyway, taking 5000 hours to be a half-life, then (I think!) the formula to calculate the expected time of failure of the first LED if you started would be:
t1 = 5000/ (ln (2)) * ln (60/59)
Which I reckon comes to a little over 121 hours.
As I've mentioned, this calculation probably doesn't really make sense for such a small number of LEDs -- at that level, adding a single extra LED reduces the expected first failure time (EFFT) by over two hours, whereas with, say, 60 million LEDs the EFFT (less than half a second) doesn't give a toss about the difference between 60 million and 60 million and one.
I think it's reasonable enough to assume, though, that all LEDs are made to essentially the same standard, so that a radioactive decay model would do pretty well (for large LED numbers, at least).
Anyway, taking 5000 hours to be a half-life, then (I think!) the formula to calculate the expected time of failure of the first LED if you started would be:
t1 = 5000/ (ln (2)) * ln (60/59)
Which I reckon comes to a little over 121 hours.
As I've mentioned, this calculation probably doesn't really make sense for such a small number of LEDs -- at that level, adding a single extra LED reduces the expected first failure time (EFFT) by over two hours, whereas with, say, 60 million LEDs the EFFT (less than half a second) doesn't give a toss about the difference between 60 million and 60 million and one.
I think it's reasonable enough to assume, though, that all LEDs are made to essentially the same standard, so that a radioactive decay model would do pretty well (for large LED numbers, at least).
I did decide to think about this question a little more, since some of the calculations I had yesterday troubled me, in terms of LEDs at least. I suppose the thing about radioactive decay is that the effective variance around the half-life of an individual atom is so large that you can't really ask the question "when will this atom decay?" and expect a sensible answer. That single atom will decay now, or next week, or in a million years, and you would learn very little about the half-life -- it's only when you have a huge collection, around billions of billions or more, that you can start saying "how long until about a third of the atoms have decayed?" or some such. The resulting exponential decay curve is robust experimentally, to be sure. But I was wondering if it really was appropriate for LEDs after all. They aren't, after all, quite as unstable as atoms (and if you had seen the calculations above you would see that an LEDs could fail in about half a second if 5000 hours was a radioactive half-life).
Instead, the decay curve I had in mind was something that was largely flat and linear until around 5000 hours or so, before falling off fairly sharply afterwards until maybe twice or three times that, by which time virtually all, if not all, LEDs would have failed. A curve that, incidentally, looks not unlike this one (the blue curve)...
http:// www.aax atech.c om/arti cles/UH Pversus LED_Cur ve.jpg
A not unreasonable fit to that curve is something like the following:
N = N0 ArcTan[-(t-t_max)/la] / (ArcTan[t_max/la])
Where the last bit is just to ensure that it normalises correctly, t_max is the absolute maximum time any one LED can be expected to last, and then "la" is some factor that scales the decay rate appropriately -- probably it can be chosen to fix the value of time at which about 10% or so of the LEDs can be expected to have decayed.
I've made up a couple of parameters to give an idea of behaviour -- roughly speaking, though, if no LED lasts more than 15,000 hours and la is about 4,300, then around 10% of LEDs fail after 5,000 hours, 33% after 10,000 hours and -- to answer the question in the OP -- if you have 60 LEDs to start with then the first one would probably fail after 1,100 hours in this model.
I think it's preferable to a radioactive decay model, and certainly seems to fit LED reality better -- although I suspect there's an even more accurate model out there. I was pleased to find the ArcTan fit though.
Instead, the decay curve I had in mind was something that was largely flat and linear until around 5000 hours or so, before falling off fairly sharply afterwards until maybe twice or three times that, by which time virtually all, if not all, LEDs would have failed. A curve that, incidentally, looks not unlike this one (the blue curve)...
http://
A not unreasonable fit to that curve is something like the following:
N = N0 ArcTan[-(t-t_max)/la] / (ArcTan[t_max/la])
Where the last bit is just to ensure that it normalises correctly, t_max is the absolute maximum time any one LED can be expected to last, and then "la" is some factor that scales the decay rate appropriately -- probably it can be chosen to fix the value of time at which about 10% or so of the LEDs can be expected to have decayed.
I've made up a couple of parameters to give an idea of behaviour -- roughly speaking, though, if no LED lasts more than 15,000 hours and la is about 4,300, then around 10% of LEDs fail after 5,000 hours, 33% after 10,000 hours and -- to answer the question in the OP -- if you have 60 LEDs to start with then the first one would probably fail after 1,100 hours in this model.
I think it's preferable to a radioactive decay model, and certainly seems to fit LED reality better -- although I suspect there's an even more accurate model out there. I was pleased to find the ArcTan fit though.
Related Questions
Sorry, we can't find any related questions. Try using the search bar at the top of the page to search for some keywords, or choose a topic and submit your own question.