I am reading about disk redudancy.
I read the following:

Suppose that the mean time to failure of a disk is 100,000 hours. Then the mean time to failure of some disk in an array of 100 disks will be 100,000/100 = 1000 hours

I don't understand this. Why isn't it 100,000^100 instead?
Then the same textbook says in the next paragraph concerning 2 mirrored disks:

If the mean time to failure of a single disk is 100,000 hours and the mean time to repair is 10 hours then the mean time to data loss is (100,000^2)/2*100

Now here it multiplies for the 2 disks but before the MTTF was divided.

Can anyone please help me figure out how we calculate this.
In case it matters the text book is Database Concepts from Silberschatz 4th edition paragraph 11.3.1 (Improvement of reliability via redundancy) page 403


2 Answers 2


Think of it this way. A single disk fails on average after $100,000$ hours. Now you have $100$ disks. How long before one of them fails? It will almost certainly take much less than $100,000$ hours for the first to fail, and much more than $100,000$ hours for the last to fail. (This of course depends on the distribution of failures, which is assumed to be exponential here.)

Here is another example: if you keep throwing one die, it will show $6$ after an average of $6$ throws. Now suppose that you throw two dice. How long do you expect to wait until some $6$ shows up? Certainly less than $6$ throws. (In fact, $36/11 \approx 3.27$ throws on average.)

In the second example, you need both disks to fail in order for data loss to occur. Not only this, but you also want both failure "events" to be within $10$ hours of each, since otherwise the system will recover after repairing the faulty disk. Hopefully the book explains the calculation somewhere, if not, the book is useless and you should tell your instructor to use a different one (perhaps with less pictures but more content, and not $n$th edition, prepared to finance the publisher's new yacht).

  • $\begingroup$ I guess you mean that because both the disks must fail then the average time is multiplied.No, this calculations are not explained in the book.If you could recommend another book to help me out I would be very happy. $\endgroup$
    – Jim
    Nov 12, 2013 at 23:15
  • 1
    $\begingroup$ I would recommend changing a textbook. Your textbook is useless if it only throws some calculations at you without explaining them. $\endgroup$ Nov 12, 2013 at 23:15
  • $\begingroup$ Do you have some other book(s) in mind to propose/recommend me to help me out? $\endgroup$
    – Jim
    Nov 12, 2013 at 23:21
  • $\begingroup$ Unfortunately I've never studies this subject, so in particular I can't recommend any books. $\endgroup$ Nov 12, 2013 at 23:37
  • $\begingroup$ @YuvalFilmus I think the textbook he needs is "Introduction to Probability" $\endgroup$ Aug 17, 2015 at 20:49

I think this problem is interesting. So I want to provide a explanation, and I think it will help me understand it better.

The die example Yuval provides is interesting. Because the distribution is known and very simple. Let P(k) be the probability of 6 shown in the kth try but not before the kth try.

$P(k=1) = \frac{1}{6}$;

$P(k=2) = (1-\frac{1}{6}) * \frac{1}{6} $;

$P(k) = (1-\frac{1}{6})^{(k-1)} * \frac{1}{6} $;

So the expected number of try that we can see 6 is

$\Sigma_{k=1}^{k=\infty} kP(k)=k(1-\frac{1}{6})^{(k-1)}*\frac{1}{6} = 6.$

If we throw two dies at the same time and let P(k) be the probability of 6 shown in the kth try but not before the kth try. Then:

$P(k=1) = 1 - (1-\frac{1}{6})^2 = 1 - \frac{25}{36} = \frac{11}{36}$

$P(k=2) = (1 - \frac{1}{6})^2 * (1 - (1 - \frac{1}{6})^2) = \frac{25}{36}*\frac{11}{36}$

$P(k) = (\frac{25}{36})^{k-1}*\frac{11}{36}$

so expected number of try that we can see 6 is:

$\Sigma_{k=1}^{k=\infty}kP(k) = \frac{36}{11}$

The sum equation could derived from this formula.

$\Sigma_{k=1}^{k=\infty} kx^{k-1} = \frac{1}{(1-x)^2}\ for\ x>0\ and\ x < 1.$

If we doesn't know the distribution of the events, we could roughly estimate the value. For the die problem, we know that every 6 tries will produce one 6. So if we throw 2 at the same time. Every 6 tries will produce two 6. So the rate of 6 happening is 2 / 6 = 1/ 3. 1/rate = 3. So roughly every 3 tries produce a 6. 3 is very close to $\frac{36}{11}$.

In the case of 100 drives. The mean time to failure of a disk is 100,000 hours, if we assume that the drive will definitely fail before 100,000 hours. So within 100,000 hours, we would have 100 failed drives. In average every $\frac{100,000}{100}$ hours we will have a failed drive, so we can assume the first failed drive will appear before $\frac{100,000}{100}$ hours.

For the two disk mirrored case, we assume A disk and B disk. In order to lose data, A and B need to be failed at the same time. if A is already failed and within 100,000 hours B disk will fail, then data will be lost. The other case is B is already failed and within 100,000 hours A will fail and then data will be lost.

For the first case, A disk is failed for 100 hours every 100,000 hours. so in order to make B to fail, it will need 100,000^2 / 100 hours. Because the other case, the time is reduced to 100,000^2/(2*100)

  • $\begingroup$ Welcome to the site! I don't follow your reasoning when you start to talk about disk drives. If we assume that a drive will definitely fail before 100,000hrs, its MTBF simply cannot be 100,000hrs. And whether or not we would expect the first failed drive before 1,000hrs depends very much on the distribution of the failures. It is completely invalid to assume that there will be a failure before 1,000hrs. Consider, e.g., the case where 50% of drives fail after 99,999hrs and the rest after 100,001hrs. The MTBF is 100,000hrs but no drives will fail before 99,999hrs. $\endgroup$ Jan 29, 2017 at 13:00
  • $\begingroup$ You also need to keep "mean time between failures" and "life expectancy" separate. 100,000 hours is over ten years and your hard drive won't last that long. MTBF is to estimate failure rates during the lifetime of a device. So during any hour, 1 in 100,000 drives will fail on average, until they all get too old. $\endgroup$
    – gnasher729
    Jan 29, 2017 at 18:39

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