May be far-map must be

Disk 1 Disk 2 Disk 3 Disk 4
—— —— —— ——
A1 A2 A3 A4
A5 A6 A7 A8
.. .. .. ..
.. A1 A2 A3
A4 A5 A6 A7
A8 .. .. ..

Then lost any folowing 2 drive – lost some info
Ex: Disk2+Disk3 -> lost A2 and A6
Disk1+Disk3 -> all data OK

PS Last map is 4/5 of 1 disk cap

Regards
Donserg

8 drives, 2 controllers, 2 sockets, 4 cores each socket – (or 8 controllers?) would allow 8 cpu execution threads on windows ? assuming you stripe the two ssd raids physical raid-0 into 4 logical drives per physical raid then strip them using windows software raid all back into one?

like a short stroke without the stroke – if an ssd can handle multiple reads at once faster than single – then creating logical volumes of smaller blocks would create artificial higher queue depth than a hard drive which mostly cannot read from two places at once (untrue there is a drive with more than one actuator). nested raid could use variable strip size and a smart i/o drive to create logical volumes of 4,16,64,128 and attempt to send “like” requests and use a simple QOS algorithm like a router.

http://www.dslreports.com/front/shutdown.html

https://docs.google.com/document/d/1kll86bDn_MgWoo6Ja7oHo_yvI0SCqggEvNWwPWIcrHY/edit

Capacity = 3/5 * capacity of single disk

should read:
Capacity = 3n/5 * capacity of single disk

no?

How it _should_ work is: if a user tries to create a RAID 0+1 volume, the RAID engine should make a RAID 1+0 volume instead.

RAID 10 cost more in disk space since, space efficiency is 1/n
RAID 5 is cheaper in disk space, space efficiency is 1-1/n

Performance wise, the RAID 10 is the better choice.
But very often money is the first criteria of the management …

Wikipedia has an excellent explanation on subtility between RAID types.
See http://en.wikipedia.org/wiki/RAID

For simple RAID-10, the article is correct. The capacity of RAID-01 (which no one would ever use) and RAID-10 are both n/2, and can tolerate the loss of any one disk. If the array is made from more than 1 set of RAID-1 pairs, you can loose 1 disk from each pair – but not two disks from the same pair.

But for mdadm RAID10, the article is mostly wrong. It starts off by missing out the “offset” option as an alternative to “near” and “far” – though much less used than “near” or “far”, it is possibly slightly faster for some loads.

The efficiency of single-mirror RAID10 is always 50%, or n/2 – regardless of the number of disks. Think about it – each block of data is written to two disks. So if you have three disks, the capacity is 3/2 – not 2/3 (or 2n/3). Regardless of the number of disks, the RAID10 will tolerate the loss of any one disk. If you have enough disks (4 or more), it /may/ tolerate the loss of other disks, depending on the spread of the data.

With four-way mirroring (near-2 and far-2), the capacity is n/4 (not n/2), and it will tolerate the loss of any /3/ disks (assuming you have at least four disks). This is the point of having multiple-copy mirroring – for each mirror, it costs you space for a duplication of the data, but you get extra redundancy.

It is also possible to use three-way mirroring (such as far-3), for n/3 capacity and two disk redundancy.

The whole last section of the article (from “Linux is special”) needs a re-write, using correct information about sizes, redundancy, and speeds. The maths is so simple for RAID-10, regardless of the number of disks and the types of mirrors, that it is hard to understand how the author got this so muddled. It is good practice to read the relevant wikipedia article before writing your own article!