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.
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 startOriginally Posted by Glaysher
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
Same thing as the exponential distribution with density:Originally Posted by Glaysher
$\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
The negative exponential distribution is usually used to model things likeOriginally Posted by Glaysher
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