# Tag Info

## Hot answers tagged reliability

6

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 ...

2

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.

2

Computers rely on physics. Transistors have some known characteristics, given some assumptions on making process. Parts that does not meet the exepected characteristics are just trashed (or sold with lower expectations, usually lower max frequency). This ensure that computations are correct while complying operation conditions (stable power supply, EMC, ...)...

1

In Case 1, for the service to not fail we need three things to not fail: computer A, component a and component b. In contrast, in Case 2 we also need computer B to not fail. Plainly the probability in the latter case is smaller, by a factor of 0.99. You can calculate the non-failure probability by just multiplying the non-failure probabilities for the ...

1

Ignore the software reliability for now, and only focus on the machines. In case 1 there is only one machine that can fail, and it fails 1% of the time. In case 2 there are two machines that can fail. However, even if just one fails the service goes down. The chance that at least one machine fails is $1 - 0.99^2 = 1.99\%$. So instead of spreading the risk ...

1

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. $P(k=1) = \frac{1}{6}$; $P(k=2) = (1-\frac{1}{6}) * \... 1 For your particular problem, the null hypothesis is$H_0: \mu_{manifacturer} = \mu_{competitors}$and the alternative hypothesis is$H_1: \mu_{manifacturer} > \mu_{competitors}$. Basically, the null hypothesis states that there is no difference between the widget under test and the widgets made by other competitors. The alternative hypothesis states ... 1 Suppose$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 ...

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