Dear Alok,

You could proceed in two differents ways:

1 - You can use a frequentist approach: number of events of failure divided by the number of observations;

2 - You can use the Komogorov-Smirnov test to define a probability distribution for those failures.

Best regards,

Glauco

Check out the Weibull Distribution and survival analysis. That is what we use in Reliability Analysis, which tries to measure MTTF (Mean time to failure), etc. 

Component reliability R(t) and failure rate h(t) are related to each other via

the relationship R(t)=exp[-\Integral_{0}^{t} h(u)du].

Ilya Gertsbakh

Everything read so far is true, but not an answer to the question.

First, assume, that the failure rate is the same all the time, i.e. it does not become larger or smaller if you switch on the instrument. Further, assume that the repair time is negligible in comparison to the six hours interval between uses.

Then, there are two possibilities:

a) The six h are truly random. In this case, the probability to find the component failed approaches lamda times 6 hours.

b) The six h are approximately constant. In this case, the value is lamda times 6 hours divided by 2.

Best regards

Günter Becker

Thanks @ Günter Becker

I was looking for something like this and my prediction is also somewhat like u described in (a) i.e. lambda times 3 minutes (operation time).

1. I guess it should be 3 min which is time of operation (and not six hours).

2. How to accommodate number of operation in one day which is 6 times. (or it is not required to be accounted) ?

If the failure rate is constant, as I assumed, then the probability that it fails during 6 h is about 120 times larger than then the probability it fails during 3 minutes. (6*60/3=120). This is why I suggested to neglect the 3 minutes of operation.

What do you know about the failure rate? Is it per hour of calendar time, or is it per hour of operation?

Sorry, you said six times per day. Then, it is not 6 h, but 4 h.

@Becker, thank for your answer. for further clarification;

1. The failure rate is constant

2. Six hours(demand) are random

3.Failure rate is per hour of operation (since operation is on demand and not continuous)

In above scenario failure probability = lambda(per hr) *  6(times)*3 (minutes each time)/ 60

Is this correct approach?

And this failure probability will be for per operation.

And how to go about if this lambda is per hour of calendar year.

Most technical devices can fail even if they are not in operation. Causes may be corrosion, vibration, sometimes, e.g. for highly integrated semi-conductors even (cosmic) radiation.

But let us consider the case, when your device can fail only during operation. In this case, if repair time is negligable in comparison with the four hours time between successive demands, the device will be intact when you activate it. Thus, it must fail during three minutes. This probability of failure is Q(t)=1-exp(-lamda*t) and by taylor expansion approximately lamda*t/2, where t is 3 minutes

If repairtime is not negligible, things become a bit more complicated. But obviously, you then need the pdf of repair time, because your component may be defect, because ist has failed last time and has not been repaired yet, it has failed the time before last time and has not been repaired yet etc.

Your approach is to assume, that repair is not possible during the day considered, but it deterministically succeeds over-night. But even then, the argument using the Taylor expansion holds, and you could divide your result by 2 as long as 6*lamda*t is much smaller than 1.

@Becker, thanks for your response.

I guess in such cases the component may be considered as Operating, Non-repairable component, since in such a short time of operation it is not possible to repair. therefore approximation of lambda*t/2 is best suited in such a case.

Hello, the failure rate is probably calculated with the normal operational schedule of the instrument so if it's switched on 6 times a day and switched back off, that's how it was working before and the gathered data is based on that.

So as Jean C. Salazar has responded Jean C. Salazar

if this is not the case and the failure rate is for the ON state only, I believe you can use a markov chain where you have 3 nodes, the first of which represents the state of OFF, the second one represents the state of ON and the final node represents the state of failure. you calculate the rates for each arrow and then using the differential equations you can calculate the probability of the failure node happening. also you can calculate pi which represents the percentage of time that each node is active during your mission time to check if it's correct.

Hope this can help.

;-)