Central Limit Theorem and Mean time between failures

Question

I was reading up about RAID, and the text said:

Suppose that the mean time to failure of a single disk is $100000$ hours. Then the mean time to failure of some disk in an array of 100 disks will be $\frac{100000}{100} = 1000$ hours, or $41.66$ days

I don't understand this. Why should the average lifetime of a disk reduce if the number of disks increases? Let's say $X_i$ is a random variable that equals the life of $i^{\text{th}}$ disk. All $X_i$'s are identically distributed and independent. Then, by Central limit theorem:

$$\hat{X} = \dfrac{X_1 + X_2 + \ldots + X_N} { N }$$

follows normal distribution as $N \to \infty $, with mean $E(X_i)$. What's wrong with this way of thinking?

Answer 1

I think that's a misunderstanding. The passage you quote is a bit informally phrased. By "the mean time to failure of some disk", they don't mean the mean time to failure of each individual disk but the mean time it takes until any one of the $100$ disks fails. If the failures form a Poisson process, then having $100$ disks instead of $1$ disk will increase the rate of the process by a factor $100$ and thus reduce the mean time between two events in the process by a factor $100$. Of course this doesn't reduce the lifetimes of the individual disks.

Answer 2

The

mean time to failure of some disk

is not the mean time to failure of those disks: it is the mean time before at least one of those 100 disks fails. In other words, the time to failure of some disk $T$ is the minimum of the time to failure of all disks, i.e.

$$T = \min_{i \leq i \leq N} \{X_i\}.$$

A standard of model is that aging has no memory (the failure rate of a new disk is the same as a 10-year olds disk), which implies that the $X_i$ have an exponential distribution of some parameter $\lambda$ (here, $\lambda$ is 100000 /hours). We may also assume that the $X_i$ are independent (we neglect the possibility that a meteor falls on the data center, that there is a manufacturer's defect on a large proportion of these disks, etc.)

Then $T$ follows an exponential distribution of parameter $N \lambda$, and thus has expectation $(\lambda N)^{-1}$. Hence,on average, $T$ is $1/N$ times the average lifetime of a disk; with $N = 100$ disks, you get on average $1000$ hours before some disk fails.

Answer 3

Compare it to lottery. If you own one ticket, the probability to win is very low. If you own 100 tickets it's much more likely that you win (something). Likewise, the probability that one disk fails is quite low but the probability that any of the 100 disks fails is higher, and therefore the mean time to failure is lower.

Answer 4

Well, both are wrong.

Take 100 brand new disks, each with an expectancy of 100,000 hours to the first failure. How long does it take until the first one fails? Certainly before 100,000 hours, but you'd need to know at least the variance, and better the distribution. You might have disks that all without exception fail after 99,000 to 101,000 hours, then you expect the first failure shortly after 99,000 hours. Or the fail randomly between 0 and 200,000 hours. Then I would expect the first failure around 2000 hours. Or you have disks where 20% fail within 100 hours, linear distribution, while the 80% are expected to last 125,000 hours. I expect the first failure within 5 hours, twenty or maybe some more within 150 hours, and then 100,000 hours until the next one fails.