$R(t)$ = reliability = 1 - unreliability (probability that item is still operational for a given time t)

unreliability = Cumulative Distribution Function (CDF) = (probability that item has failed for a given time $t$)

So if item's time-to-failure data is assumed to be distributed exponentially; then

$R(t) = e^{-\lambda t}$ where $\lambda$: Failure Rate of the item

And the example requirement is

item reliability must be no lower than 95% with a 1-Sided Lower Confidence Bound of 99% for a mission of 18 hours.

and the statement of the example requirement in formal statistical way: $$ P(R > 95\% ) =P(e^{-\lambda \times 18}>95\% ) = 99\% $$

Knowing the $t$, duration value; I need to find my target $\lambda$, that is the maximum allowable value of $\lambda$

I have 2 questions that are very related.

Question 1

Will $\lambda_{target}$ be affected by the 1-Sided Confidence Bound existence

Question 2

If the answer to my question 1 is "YES"; then how will I find the value of $\lambda_{target}$

For this question, an answer with the methodology explained is of course the best for me however it's also pretty good for me that you only give me the technical keywords (name of the methodology required as specific and detailed as possible ) for me to search on web.

by the way, this is not a homework question, this question is related to my job, I had 2 semesters of statistics course however it's been nearly 15 years.

best regards


My aim is to calculate, find the target failure rate at the beginning of a design project. At the beginning, i have no data to estimate something. Situation is i'll calculate the target lambda (depending on my requirement) and give the target failure rate value to the design team as a derived requirement.

  • $\begingroup$ Not quite sure what you're asking. There seem to be two questions here: (1) How do you find the maximum allowable value of $\lambda$ - well you just solve $R(t) = e^{-\lambda t}$ for given values of $R$ & $t$. (2) How do you calculate a lower confidence bound for the true value $\lambda$ from a sample? That depends on the nature of your sample - e.g. whether you have censored observations - & we'd need more information to answer. (Note also your formal statement of the requirement seems to treat $\lambda$ as a random variable so & doesn't tally with the "confidence bound" wording above.) $\endgroup$ Jun 23, 2016 at 13:43
  • $\begingroup$ Hi, i added my EDIT at the end of my q. I hope i am clear now, i really want to learn what i asked. $\endgroup$ Jun 23, 2016 at 17:02

1 Answer 1


If you need a reliability of at least $r$ over time $t$, then when failure times are exponentially distributed the upper specification limit (I wouldn't call it a "target") for the rate parameter is given by

$$\lambda_\mathrm{u} = \frac{-\log r}{t}$$

You might also want to specify what'll count as enough evidence that the true rate parameter $\lambda < \lambda_\mathrm{u}$; observations will give you an estimate $\hat\lambda$, but it's subject to a non-negligible uncertainty unless the no. observations is very large. One way of doing this is to require that a statistical test of the null hypothesis that $\lambda\geq\lambda_\mathrm{u}$ is performed & that it rejects the null hypothesis at a significance level $\alpha$ (or equivalently that $\lambda_u$ is not contained within a (one-sided) $1-\alpha$ confidence interval for $\lambda$).

In your example $\lambda_\mathrm{u} =\frac{-\log 0.95}{18 \, \mathrm{h}}=0.00285\, \mathrm{h}^{-1}$ while $\alpha=1-0.99=0.01$. In your example $\lambda_\mathrm{u} =\frac{-\log 0.95}{18 \, \mathrm{h}}=0.00285\, \mathrm{h}^{-1}$ while $\alpha=1-0.99=0.01$. Suppose the plan's to observe 10 independent failure times $t_1\ldots t_{10}$. You'd estimate the rate as $\frac{10}{\sum_{i=1}^{10} t_i}$, so observed values of $\sum_{i=1}^{10} t_i <3509\,\mathrm{h}$ are entirely compatible with the null hypothesis; but for $\sum_{i=1}^{10} t_i >3509\,\mathrm{h}$, the longer the total time until 10 failures the more the evidence mounts against it. Let $\lambda=\lambda_\mathrm{u}$ (defining the out-of-spec. distribution that's closest to the in-spec. one, & hardest to distinguish from it): the random variable $\sum_{i=1}^{10} T_i$ follows an Erlang distribution with rate $\lambda_\mathrm{u}$ & shape 10, from which you can find the critical value that $\sum_{i=1}^{10} T_i$ exceeds with probability 0.01, $(\sum_{i=1}^{10} T_i)_\mathrm{c}=5511\,\mathrm{h}$.


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