Are Solid State Drives Worth the Money?
lifehacker.com
lifehacker.com
On a desktop, a three-drive RAID0 provides about the same performance, and gives you nearly 40x the storage for a given price-point.
On a mobile platform the physical space, vibration/motion, and power constraints coupled with the increased performance may make SSDs worthwhile.
Quite a terrible idea unless you enjoy backup recovery sessions.
If the noise of 3 drives doesn't bother you then you can just as well get 4 and setup a RAID10, which is nearly as fast but much less likely to fail.
Erm. Actually they get their performance because they write to NAND instead of spinning platters. SSD controllers are also quite a bit smarter than plain RAID0.
RAID0 works very well with good quality drives.
A RAID0 over three disks has about 1/3 the MTBF of a single disk.
That can still be a worthwhile trade-off if you need the extra capacity, but if you're mostly after performance and reliability then a pair of SSDs, or even a single SSD, is the better choice.
http://www.denali.com/wordpress/index.php/dmr/2010/02/02/ssd-interfaces-and-performance-effects
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 fail in 25,000 hours in an array.
That calculation makes no sense. If you have a single drive then that will fail, on average, after 25k hours. If you stripe over three of these drives then your array will, on average, fail after 8333 hours.
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.
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.
"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.
Wikipedia disagrees; http://en.wikipedia.org/wiki/Standard_RAID_levels#RAID_0_fai...
array_MTTF = avg(drive_MTTF) / number_of_drives
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.
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...
Erm wait, didn't I just suggest the exact opposite?
If you really think everyone has been wrong about this all the time then please, by all means, correct wikipedia or write a blog post about the matter.
This "false" formula has been out there for quite some time and you find it in pretty much every write-up on the topic, including those from RAID-vendors who (I'd hope) have spent some thought on these things.
On the flip-side I haven't found a single source to support your thesis. Thus I'd say the burden of proof is on you.
I know a couple of friends with them, and they say that applications load vastly faster, and their machines also boot faster.
It doesn't help with sustained throughput, but that's not normally a problem.
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.