# the Poisson result and Exponential interpretation for spare part requirement analysis

I am confused with with the Poisson result and Exponential interpretation for spare part requirement analysis. I try to calculate the required number of spare parts for a disposable remove-replace item.

From a real life case; if:

1. warranty period for a part is $t=400 hr$
2. constant failure rate for the part is $\lambda=0,003534$ 1/hour so mean time to failure $MTTF=1/\lambda=283{\rm hours}$

then if I use Poisson distribution to calculate what is the required number of spares NOT to be in shortage for 400 hours for this part with at least a probability of 95%, solution is 4 pieces of spares. (the reason is:

1st x value which satisfies below inequality is 4.

$${P(x=0)\lor P(x=1)\lor P(x=2)\lor P(x=3)\lor P(x=4) }>0,95$$

)

However logicly $MTTF$ says that average time between failure events will be 283 hours so even the part fails at 1st hour then I will need 1 piece of spare.

So in this point I'm very confused since 1 and 4 are very different and has very serious consequences. (cost, customer satisfaction etc.)

I also had the graph for Reliability for this part. (${\rm Reliability}=e^{-\lambda(t)}$ )

Observing the graph also didn't help me!

Can you please explain "where and why my logic fails"? I think I've some trouble with understanding the true nature of exponential distribution and meaning of failure rate or MTTF.

Notice that with a poisson distribution for the number of failures, the expected number of failures in 400 hours is 1.4. It's important to realize the distinction between the following two phrases "what you expect to happen" and how that differs from "what will happen most of the time".

If you simulated this 400 hour period many times, the average number of failures across the simulations would be 1.4. It would probably be helpful to think of expected value in this way. Notice that the expected number of failures coincides with your faulty logic regarding the MTTF. However, the proportion of simulations with failures less than or equal to 4 would be .985. In assessing how to use this information, it seems more sensible to base your decision on what will happen most of the time as opposed to what happens on average.

• really appreciated your answer. easy to understand because of your ability to teach I think, thank you. – Andre Chenier May 2 '14 at 6:25
• You're welcome. I'm glad I was able to help. – jsk May 2 '14 at 7:56