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This is a discrete problem. You cannot have 2.9 replicas instead of 3. The fact that 24% of these drives failed without smart indicators means no test exists wh
by honkhonkpants 10y ago
This is a discrete problem. You cannot have 2.9 replicas instead of 3. The fact that 24% of these drives failed without smart indicators means no test exists which can lower your replication from 3 to 2.
- nitrogen 10y agoCan having ~2.9 replicas be approximated by using an error-correcting code across a large enough number of drives? Isn't that how RAID 5/6/Z work?
- honkhonkpants 10y agoSort of but again no. He problem is if you have really wide coded stripes your I/O costs to reconstruct after a failure will be astronomical, and the probability of one failure increases with the number of participating devices. Besides you could not reshape the stripe in response to anticipated failures without reading and writing the whole thing, in which case you'd be better off in terms of I/O costs just evacuating the device in question.
- matt4077 10y agoWell, by that measure it doesn't even matter if a drive fails with 10% or 5% probability beyond the replacement costs. Because a test with 50% sensitivity effectively halves the failure rate you have to use. Since it's all probabilities, triple redundancy does not guarantee complete absence of data loss. On the other side of the spectrum, single replication might offer a better cost/safety ratio for some applications. A failure that is known in advance is equal to no failure in these term and even when discrete, it will make the difference once in a while.
- kryptiskt 10y agoYou can reduce your risk by requiring that no more than one of the three replicas is on an iffy drive, creating new replicas on healthy drives if that's not true.