nVraid question

D

Duck

[H]ard|Gawd
Joined
Apr 17, 2000
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K, here's my dilemma

I have an ASUS A8N32-SLI deluxe. I Built the computer using 2 raptor 74's in raid0 for the primary OS drive attached to ports 1&2 on the nVraid controller. Additionally I have 2 storage drives that I would like to connect to ports 3&4 as single non-raid drives.

When i disable RAID on those ports in BIOS and connect my drives the computer hangs right after the RAID bios loads. If i enable raid on those ports it wont allow me to use them as single non-raid drives.

Any ideas on how to fix this condition?
 
With this current setup, I do not believe you will be able to. Most likely, you will need to purchase a promise card in order to get data drives to be accessed.
 
I'm pretty sure that I've run a mix of RAID and non-RAID at the port level with nForce; I don't know why you're having troubles. Try updating the BIOS / chipset drivers? Contact your MB support?

Failing all that, as a last resort, you can run nVIDIA drives in single drive RAID 0 as a workaround. Not one I enjoy recommending, but something which would solve the problem

Before the "experts" post again that RAID 0 takes more than one drive, yes of course it does in theory. But with nVIDIA RAID, this is a sometimes-important feature of their implementation (which you even need for some in-place migrations).
 
Madwand said:
Before the "experts" post again that RAID 0 takes more than one drive, yes of course it does in theory. But with nVIDIA RAID, this is a sometimes-important feature of their implementation (which you even need for some in-place migrations).
Click to expand...
Raid 0 takes N drives. Nothing in theory says that you need more than one, except implementation-specific code. Heck, a 2-disk raid 5 array is perfectly legal, too. It looks almost like a raid 1 array, with two big differences - can anyone derive them from the definition of raid 5? I shall award two Bonus Points to anyone that gets both right, and an additional Point to anyone who comes up with an additional theoretical (not implementation-based) difference :D
 
found the problem, turns out it was something silly. When i connected the other two drives the motherboard decided they should be "disk 1" and "disk 2" while relegating my array to "disk 3" so the system was trying to boot off the data drives and failing (and hanging). All i had to do was go into bios after attaching them and correct the drive priority order

so long story short PEBKAC.............
 
unhappy_mage said:
Raid 0 takes N drives. Nothing in theory says that you need more than one, except implementation-specific code. Heck, a 2-disk raid 5 array is perfectly legal, too. It looks almost like a raid 1 array, with two big differences - can anyone derive them from the definition of raid 5? I shall award two Bonus Points to anyone that gets both right, and an additional Point to anyone who comes up with an additional theoretical (not implementation-based) difference :D
Click to expand...

What are the answers?
 
unhappy_mage said:
Raid 0 takes N drives. Nothing in theory says that you need more than one, except implementation-specific code. Heck, a 2-disk raid 5 array is perfectly legal, too. It looks almost like a raid 1 array, with two big differences - can anyone derive them from the definition of raid 5? I shall award two Bonus Points to anyone that gets both right, and an additional Point to anyone who comes up with an additional theoretical (not implementation-based) difference :D
Click to expand...

First difference between 2-disk RAID 1 and 2-disk RAID 5 is that the data is striped in RAID 5, not mirrored, so the data on both drives will not be identical (as it would be with RAID 1).

Second difference is that with RAID 5, parity data will be interleaved between the stripes - and this will also not be identical from one drive to the next.

Net result is that reading the striped data on the RAID 5 is fast (theoreticalyl similar to RAID 0), but writing requires writing both the data and parity to both drives. Either drive could be used to reconstruct the array so no data is lost if one drive fails.

I'll take a stab at the theoretical too.

With a 2-drive RAID 1, the capacity will be N, where N is the size of the smaller of the two drives used. With a 2-disk RAID 5, I believe that the parity data is smaller than the corresponding striped data, so the capacity of the RAID 5 should be slightly larger than N. Wait...no, I don't think that makes sense with only two drives. Oh well.
 
Sounds about right. The first thing I was thinking of was rotation - the difference between raid 4 and 5 - and the second was parity as opposed to data. So instead of "1 2 3 4" and "1 2 3 4" on raid 1, you get "1 ~2 3 ~4" and "~1 2 ~3 4" on 2-disk raid 5, where ~ is the bitwise not operator.

The capacity is the same as raid 1, though. They work essentially the same way, and use blocksize*floor(disksize/blocksize) bytes of the disk.

Aaaaand this thread's derailed :eek: Glad you got your question figured out, OP.

 
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