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Are Solid State Drives Worth the Money?

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Re: Are Solid State Drives Worth the Money?

#11
post #10

Earlier quoted context omitted.

You'll probably find this interesting reading on the internal architecture of SSDs: http://www.denali.com/wordpress/index.php/dmr/2010/02/02/ssd-interfaces-and-performance-effects Also, while RAID0 reduces the MTBF, it's not linear. Drive life is not magically shortened as a result of the drive being in a RAID array (if you take care to isolate synchronous vibration). The life of the array is equal to the shortest dr…

Also, while RAID0 reduces the MTBF, it's not linear. Well, it is inverse proportional. The life of the array is equal to the shortest drive life. Erm. To be clear: Your risk of having a RAID0-set (over 3 disks) fail during a given timespan is 3 times higher than having a single-disk-"set" fail in the same timespan. In other words, if a drive would have failed after 25,000 hours in standalone operation, it will still…

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.

Re: Are Solid State Drives Worth the Money?

#12
Totally worth it with one major caveat.

SSD drives don't actually delete stuff when you 'delete' them, it just tells your OS that that it can write to that space. Normal hard drives do this too, the problem comes when you actually go to write stuff to the space that is 'marked for deletion'.

Unlike normal hard drives that just overwrite the disk space, SSD drives can only write to disk in groups called 'blocks'. So when the drive is filled up with stuff (or you've been using it for some time), it has to corral the block, make sure its safe to write to, if it isn't, it wipes it, before finally writing to it ... which is really slow.

This manifests itself in your system basically freezing on you from time to time after you've been using it. Its fully explained here http://www.anandtech.com/show/2738/8

It can be REALLY frustrating, but even with that ... I would NEVER NEVER go back to a normal hard drive. Why? My photoshop opens in 7 seconds, Netbeans (bloated Java IDE) in 10, Windows XP boots in bootcamp in 2 minutes, and OS X is fully loaded in 30 seconds. So yeah ... you can pry my ssd from my cold, dead, mutilated fingers.

There is a TRIM OS command that allows your OS to basically clean out the space on your hard drive that is marked for deletion while you're not using the computer, so that this problem does not occur, but OS X does not support it ... Windows 7 and Linux are the only OS's I know that do. http://www.anandtech.com/show/2738/10

The other thing to consider is that SSDs are very fast on sustained writes ... copying a 1GB file from one location to another. However, modern OS'es employ frequent but small writes in their operation ... and certain drives that look good on paper stink up the joint in this department (things have changed recently though, since anandtech called them out on it).

What you want to do when you're looking at specs is find out what the throughput is (MB/s) for writes on 1KB, 4KB pieces of data, and compare it to the sustained write speeds, to see how big the difference is. If you want a shortcut ... buy an intel SSD ... and thank me later.

Re: Are Solid State Drives Worth the Money?

#13
post #10

Earlier quoted context omitted.

Also, while RAID0 reduces the MTBF, it's not linear. Well, it is inverse proportional. The life of the array is equal to the shortest drive life. Erm. To be clear: Your risk of having a RAID0-set (over 3 disks) fail during a given timespan is 3 times higher than having a single-disk-"set" fail in the same timespan. In other words, if a drive would have failed after 25,000 hours in standalone operation, it will still…

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 f…

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

Re: Are Solid State Drives Worth the Money?

#14
post #13

Earlier quoted context omitted.

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 f…

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

tl;dr: For RAID0 arrays there is a non-linear increase in the probability of failure, but the MTTF/MTBF doesn't change much.

Re: Are Solid State Drives Worth the Money?

#15
post #13

Earlier quoted context omitted.

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

tl;dr: For RAID0 arrays there is a non-linear increase in the probability of failure, but the MTTF/MTBF doesn't change much.

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.

Re: Are Solid State Drives Worth the Money?

#16
post #15

Earlier quoted context omitted.

tl;dr: For RAID0 arrays there is a non-linear increase in the probability of failure, but the MTTF/MTBF doesn't change much.

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.

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.

Re: Are Solid State Drives Worth the Money?

#17
post #15

Earlier quoted context omitted.

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.

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) wit…

> "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_fai...

array_MTTF = avg(drive_MTTF) / number_of_drives

Re: Are Solid State Drives Worth the Money?

#18
post #17

Earlier quoted context omitted.

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) wit…

> "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_fai... array_MTTF = avg(drive_MTTF) / number_of_drives

Which is at odds with the (correct) definition of MTTF as a rate-based calculation:

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.

Re: Are Solid State Drives Worth the Money?

#19
post #17

Earlier quoted context omitted.

> "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_fai... array_MTTF = avg(drive_MTTF) / number_of_drives

Which is at odds with the (correct) definition of MTTF as a rate-based calculation: 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…

If you are so convinced then why don't you correct the wikipedia article?

Perhaps also call up LSI and Adaptec, who use the same formula in their documentation.

http://storageadvisors.adaptec.com/2005/11/01/raid-reliabili...

But what do they know, they only build raid controllers...

Re: Are Solid State Drives Worth the Money?

#20
post #19

Earlier quoted context omitted.

Which is at odds with the (correct) definition of MTTF as a rate-based calculation: 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…

If you are so convinced then why don't you correct the wikipedia article? Perhaps also call up LSI and Adaptec, who use the same formula in their documentation. http://storageadvisors.adaptec.com/2005/11/01/raid-reliabili... But what do they know, they only build raid controllers...

You're right, there's no reason to try to correct the 20% of the population that believes the Sun revolves around the Earth. It's a lost cause; you win.
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