# What is the system's mean time to failure?

I have the following homework problem:

A 10 TB disk drive has an MTTF of 6,000,000 hours. How much data can we store in a system comprised of these disks, if we want the system MTTF to be at least 1.2M hours?

If we are allowed to make it redundant (i.e., two of them operating in a parallel mode), and if the MTTR is 120 hours, what is the system MTTF?

I think it is

$$\qquad \operatorname{MTTF}(\text{System}) = \frac{1}{\sum_i \frac{1}{\operatorname{MTTF}_i}}$$.

• What are MTTF, MTTR? Where is the problem here, isn't this just computation? Please include your own thoughts and attempts.
– Raphael
Feb 6, 2013 at 7:46
• @Raphael MTTF usually stands for Mean Time To Failure. No sure about MTTR, but the R might stand for Replace. Feb 6, 2013 at 12:46
• @user6722 What did you try? Is there some specific concept you don't understand? Feb 6, 2013 at 12:48
• There is formula I updated my post above that I don't see how I plug in the numbers form the problem. I'm not even certain if I'm going about it in a right way. Feb 6, 2013 at 15:51
• MTTR is meant time to repair. Feb 6, 2013 at 15:54

For part 1, its simple. You need to find the $x$ (as defined below) for which MTTF(system) = 1.20M.

So $1.20 * 10^6 = \frac{1}{[(\frac{1}{6*10^6})*x]} \implies x = 5$.

We can have 5 disks (50 TB) if we want an MTTF(System) = 1.2M.

Part 2 is homework for you.

• In part 2 the MTTR is 120 hours. Feb 6, 2013 at 16:48
• @RND, I get a really large number, is that normal? The formula is MTTF (system)=MTTF (component)^2 / 2 x MTTR (component) So 6, 000, 000^2 / 2x120 = 150, 000, 000, 000 That's a lot of freeking hours! Feb 7, 2013 at 2:57

Suppose $X,Y$ are exponential random variables with means $\lambda^{-1},\mu^{-1}$. Then $$\Pr[\max(X,Y) < t] = (1-e^{-\lambda t})(1-e^{-\mu t}) = 1 - e^{-\lambda t} - e^{-\mu t} + (\lambda + \mu)e^{-(\lambda+\mu)t},$$ and from this one can compute \begin{align*} \mathbb{E}[\max(X,Y)] &= \int_0^{\infty} t \frac{d}{dt} [1 - e^{-\lambda t} - e^{-\mu t} + (\lambda + \mu)e^{-(\lambda+\mu)t}] \, dt \\ &= \int_0^{\infty} t(\lambda e^{-\lambda t}+\mu e^{-\mu t}-(\lambda+\mu)e^{-(\lambda+\mu)t}) \, dt \\ &= \frac{1}{\lambda} + \frac{1}{\mu} - \frac{1}{\lambda+\mu}. \end{align*} When $\lambda = \mu$, we increase the MTTF from $\lambda^{-1}$ to $(3/2)\lambda^{-1}$. A similar computation shows that by taking three drives, we increase the MTTF to $(3-3/2+1/3)\lambda^{-1} = (11/6)\lambda^{-1}$. Using four drives, we get $(25/12)\lambda^{-1}$.

• I don't think the idea here is correct. The time to failure of $n$ disks with lifetimes $X_1, X_2,\ldots, X_n$ is $\min\{X_1, X_2,\ldots, X_n\}$ because the overall system crashes as soon as one disk crashes. Then, $$P\{\min_i X_i > t\} = \prod_i P\{X_i > t\} = e^{-\sum_i \lambda_i t}$$ showing that the minimum is also an exponential random variable with mean $$\left(\sum_i\lambda_i\right)^{-1} = \frac{1}{\sum_i\lambda_i} = \frac{1}{\sum_i \frac{1}{E[X_i]}}.$$ For the special case of all the $\lambda_i$ the same, the MTTF is smaller by a factor of $n$, giving $n=5$ as RDN points out. Feb 7, 2013 at 4:13
• So what's the point of adding redundancy if it shortens the MTTF? Feb 7, 2013 at 5:08
• In this problem, it is not redundancy that is being added but storage capacity. A 10 TB disk has MTTF 6,000,000 hours. A bank of 5 such disks can store 50 TB of data, but the MTTF reduces to 1,200,000 = 1.2M hours. The question asked was "How much data can we store in a bank of such disks if the MTTF of the bank is 1.2M hours?", and the answer is 50 TB. You are answering the question "If we store identical 10 TB data on $n$ disks and read from a non-failed disk as long as there is at least one such disk, what is the MTTF?" which increases the MTTF to more than 6M hours, not what was asked. Feb 7, 2013 at 14:36