1. ## Life duration problem!

If I have got 10000 PCs,each with a life duration of 9000 hours,how often will I have a loss if each PC works 8 hours a day,5 days a week?
Thanks.

2. I assume thats an expected life duration or a value given so that most will last longer. Such things usually follow a normal distribution but much more data is needed to do any calculations andthe ones required will vary depending on whether you give us sample data or population data (mean and standard deviation).

Your question would still be impossible to work out as it is a random process. For weeks you would expect no failures then nearer to 9000 hours you would expect much more. So what time period are you talking about?

3. Originally Posted by Glaysher
I assume thats an expected life duration or a value given so that most will last longer. Such things usually follow a normal distribution but much more data is needed to do any calculations andthe ones required will vary depending on whether you give us sample data or population data (mean and standard deviation).

Your question would still be impossible to work out as it is a random process. For weeks you would expect no failures then nearer to 9000 hours you would expect much more. So what time period are you talking about?
If these PC are in equilibrium, that is we are many lifetimes from the start
of the experiment, with PCs being replaced after failure you will have a
failure rate of 1.111 per hour of operation or a mean time between failures
of 0.9 hours, and the interval between failures will have a negative exponential distribution.

RonL

4. Originally Posted by CaptainBlack
If these PC are in equilibrium, that is we are many lifetimes from the start
of the experiment, with PCs being replaced after failure you will have a
failure rate of 1 every ~1.111 hours of operation, and the interval between
failures will have a negative exponential distribution.

RonL
I've not heard of a negative exponential distribution though it makes me think of the Poisson distribution. What is it?

5. Originally Posted by Glaysher
I've not heard of a negative exponential distribution though it makes me think of the Poisson distribution. What is it?
Same thing as the exponential distribution with density:

$\displaystyle p(t)=\left\{ {\lambda e^{-\lambda t},\ \ \ t\ge 0 \atop 0\ \ ,\ \ \ \ t<0} \right{}$,

yes it is related to the poison distribution. The number of occurrences in an
interval of a negative exponential distributed rv has a poison distribution.

RonL

6. Originally Posted by CaptainBlack
Same thing as the exponential distribution with density:

$\displaystyle p(t)=\left\{ {\lambda e^{-\lambda t},\ \ \ t\ge 0 \atop 0\ \ ,\ \ \ \ t<0} \right{}$,

yes it is related to the poison distribution. The number of occurrences in an
interval of a negative exponential distributed rv has a poison distribution.

RonL
But the negative exponential distribution is continuous instead? Under what conditions do you use it?

7. Originally Posted by Glaysher
But the negative exponential distribution is continuous instead? Under what conditions do you use it?
The negative exponential distribution is usually used to model things like
the interval between the arrival times of busses, or other phenomena
assumed to occur randomly in time. Which is a continuous RV.

Now the number of busses that arrive in an 1 hour interval is discrete
(and if the inter-arrival times is neg-exp is poison).

Note: One of the methods of generating poison RV's is to sum neg-exp
RV's until the sum exceeds the time interval, and then the poison RV is
one less than the number of neg-exps used. Of course there are some
fiddles used in practise to make this more efficient, but this is the basic idea).

RonL

8. Thanks

9. Originally Posted by Glaysher
Thanks
Hi, thanks for your response. Let's say the life duration follows an exponential distribution with loss rate 1/50000 and we are in equalibrium.

10. Originally Posted by ioannisp
Hi, thanks for your response. Let's say the life duration follows an exponential distribution with loss rate 1/50000 and we are in equalibrium.
Is this a new problem? The failure rate is now different having been 1/9000 per hour, what else is different?

RonL

11. Originally Posted by CaptainBlack
Is this a new problem? The failure rate is now different having been 1/9000 per hour, what else is different?

RonL
Hi again. I apologise for the typo.

It is 1/9000.

12. Originally Posted by ioannisp
Hi again. I apologise for the typo.

It is 1/9000.
I gave the answer in an earlier post.

RonL

13. Originally Posted by CaptainBlack
I gave the answer in an earlier post.

RonL
Yes I remember. How exactly is that derived?

Thanks.

14. Originally Posted by ioannisp
Yes I remember. How exactly is that derived?

Thanks.
You have a failure rate for a single computer of 1/t per unit time, for n
you have n/t failing per unit time. That's all there is to it.

RonL

15. Originally Posted by CaptainBlack
You have a failure rate for a single computer of 1/t per unit time, for n
you have n/t failing per unit time. That's all there is to it.

RonL
OK, so it is 1,111 failures per unit time,thus the failure rate is 1 per 0,9 hours of operation.

Not 1 failure every 1,111 hours, as you mentioned in the first answer.Right?

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