4 ms·

While the probability of failure is nearly a function of the number of drives, the MTBF/MTTF calculations do not work that way. For example, if there were a pr

While the probability of failure is nearly a function of the number of drives, the MTBF/MTTF calculations do not work that way. For example, if there were a probability of 5% that the disk would fail within three years, in a three disk RAID0 array, that probability of failure would be: P=(1-(1-.05)^3)=.14263 In other words, 14.3% probability of failure within three years. That doesn't mean it will fail in that time frame. It means if you have a large population of that configuration, that is the rate you would be dealing with for drive replacement planning. The MTBF and MTTF calculations apply to populations of drives (e.g. a given model) not to a given drive. The values provide no predictability for the failure of any specific drive. Using the values for that purpose is a common misapplication. A drive with a MTTF of 1,000,000 power-on hours can fail in 15 minutes or never during its useful life. As a result, a three drive array will have a higher probability of failure over a given interval, but the MTTF/MTBF of the drives is essentially unchanged. Think of it this way... The probability of winning the lottery is one in 20,000,000. The probability that someone (anyone) will win the lottery in a given week may be one out of ten - 10%. In other words, some person wins the lottery, on average, one time in ten weeks. That doesn't mean that your probability of winning the lottery is 10%. It also doesn't mean that the average probability of winning the lottery is 10%. It also doesn't change the probability of winning the lottery; it's still one in 20,000,000, even if three people win in a 10 week interval.

moe 16y ago
Hm. Thanks for repeating what I just said, I guess. But what was your point again?
- macemoneta 16y ago
  tl;dr: For RAID0 arrays there is a non-linear increase in the probability of failure, but the MTTF/MTBF doesn't change much.
  - moe 16y ago
    Could it be you're just arguing for arguments sake? My original point was: A RAID0 over 3 disks is about 3 times more likely to fail than a single disk running standalone. Fail means "total data loss". You confirm that point with your own math, yet still seem to be trying to argue that there was no difference. Sorry, that makes no sense to me.
    - macemoneta 16y ago
      Your statement was: "A RAID0 over three disks has about 1/3 the MTBF of a single disk." This is incorrect, the MTTF and MTBF are not significantly changed. Assuming you meant failure probability, my issue with the probability variance is the linear relationship you imply. If the variation were linear, a RAID array composed of drives with a 5% failure probability would reach certainty of failure (1.00 probability) within the interval at 20 drives. In actuality, it takes 225 drives to reach that probability. The difference is a real world consideration for capacity management. What it means is that RAID0 arrays are not as failure prone as people think they are.
      - moe 16y ago
        > "A RAID0 over three disks has about 1/3 the MTBF of a single disk." This is incorrect, the MTTF and MTBF are not significantly changed. Wikipedia disagrees; http://en.wikipedia.org/wiki/Standard_RAID_levels#RAID_0_failure_rate http://en.wikipedia.org/wiki/Standard_RAID_levels#RAID_0_fai... array_MTTF = avg(drive_MTTF) / number_of_drives
        macemoneta 16y ago
        Which is at odds with the (correct) definition of MTTF as a rate-based calculation: http://en.wikipedia.org/wiki/Failure_rate http://en.wikipedia.org/wiki/Failure_rate The person that wrote the Wikipedia article you referenced read the same mythology you did; repeating it doesn't make it true. The plural of anecdote is not fact. Think about it yourself for a moment. If two cars are traveling 50mph, does that make their average speed 25mph (50/2)? Applying a divisor to a failure rate based on the number of devices is nonsensical.

moe 16y ago

Hm. Thanks for repeating what I just said, I guess. But what was your point again?

macemoneta 16y ago
tl;dr: For RAID0 arrays there is a non-linear increase in the probability of failure, but the MTTF/MTBF doesn't change much.
- moe 16y ago
  Could it be you're just arguing for arguments sake? My original point was: A RAID0 over 3 disks is about 3 times more likely to fail than a single disk running standalone. Fail means "total data loss". You confirm that point with your own math, yet still seem to be trying to argue that there was no difference. Sorry, that makes no sense to me.
  - macemoneta 16y ago
    Your statement was: "A RAID0 over three disks has about 1/3 the MTBF of a single disk." This is incorrect, the MTTF and MTBF are not significantly changed. Assuming you meant failure probability, my issue with the probability variance is the linear relationship you imply. If the variation were linear, a RAID array composed of drives with a 5% failure probability would reach certainty of failure (1.00 probability) within the interval at 20 drives. In actuality, it takes 225 drives to reach that probability. The difference is a real world consideration for capacity management. What it means is that RAID0 arrays are not as failure prone as people think they are.
    - moe 16y ago
      > "A RAID0 over three disks has about 1/3 the MTBF of a single disk." This is incorrect, the MTTF and MTBF are not significantly changed. Wikipedia disagrees; http://en.wikipedia.org/wiki/Standard_RAID_levels#RAID_0_failure_rate http://en.wikipedia.org/wiki/Standard_RAID_levels#RAID_0_fai... array_MTTF = avg(drive_MTTF) / number_of_drives
      - macemoneta 16y ago
        Which is at odds with the (correct) definition of MTTF as a rate-based calculation: http://en.wikipedia.org/wiki/Failure_rate http://en.wikipedia.org/wiki/Failure_rate The person that wrote the Wikipedia article you referenced read the same mythology you did; repeating it doesn't make it true. The plural of anecdote is not fact. Think about it yourself for a moment. If two cars are traveling 50mph, does that make their average speed 25mph (50/2)? Applying a divisor to a failure rate based on the number of devices is nonsensical.