My orignal question is listed below, I choose the best response I got and
posted it here. Generally speaking no one thought my design was very well
thought out. Mostly I opened myself up to power supply failures by
placing the mirrored drives on the same shelf.
Alan's idea seemed the best of the lot to me, although there are more ways
to do this (depending on which unit you have), I think his block diagram
below is the best.
Thanks to everyone for their help!
-Ed Silver
==========
With 8 drives in the array and only six backend SCSI
busses, you don't have the leisure to put each disk
on a separate bus. So, you have to decide:
o Avoid single points of failure
o Avoid some potential performance problems
You might be able to avoid more of the first than the
second. But you'll have to end up with four disks of
the whole array on the same two busses.
The SW300 backplane runs power horizontally and the
SCSI busses vertically. You don't want any given
pair of mirrored devices on the same power supply(s).
That way if a power supply fails (and you don't have
redundant power), only one disk in the mirror fails.
This is not what you have now. Taking just M1 as the
example, both disks are on the same row. If the power
supplies for that row fail, then M1 fails and the whole
stripe set fails. An alternate scheme would be to mirror
diagonally:
M1 = disk300 + disk410
M2 = disk310 + disk400
M3 = disk320 + disk430
M4 = disk330 + disk420
If a particular row fails, you'll still lose two disks
but they won't belong to the same mirror.
You also don't want mirrored pairs to share the same
bus in case that fails.
From a striping point of view, you want each disk to
be a on separate bus. You have 6 to work with, but
you're only using two. With a sufficiently small I/O
size, the stripe set may not be able to saturate the
bus with data movement and you can still get some
seek overlapping, but you won't get multiple I/Os
(especially writes) transferring at the same time.
The alternate choice is to spread the four disks of each
group across four busses. Then spread the other four
across the remaining two busses and two of the busses
in use by the first group.
A block diagram might look like:
1 2 3 4 5 6
+-----+-----+-----+-----+-----+-----+
| | | | 430 | | 630 | 3
+-----+-----+-----+-----+-----+-----+
| | | 320 | | 520 | | 2
+-----+-----+-----+-----+-----+-----+
| | 210 | | 410 | | | 1
+-----+-----+-----+-----+-----+-----+
| 100 | | 300 | | | | 0
+-----+-----+-----+-----+-----+-----+
M1 = disk100 + disk630
M2 = disk210 + disk520
M3 = disk320 + disk410
M4 = disk420 + disk300
S1 = M1 + M2 + M3 + M4
Each pair of mirrorer disks is both on a separate bus and
power supply. Only four disks overlap for the stripe sets.
Other arrangements should be possible. If my numbering
doesn't precisely doesn't match the SW300 or BA370, it is
because I don't have one in front of me.
-------------------------------
Somewhat new to RAID I wanted to get some advice on setting up a SW300
RAID box.
1. I want to have RAID 0+1 (striped mirroset). So I setup four Mirrors
and than created one Stripeset.
M1=(disk300)(disk400)
M2=(disk310)(disk410)
M3=(disk320)(disk420)
M4=(disk330)(disk430)
S1= M1+M2+M3+M4.
This seems to work fine, but is it the best way to lay the disks out in
the SW300? The manual is a little confusing about the layout of the disks
for optimal performance. Right now I've got each mirrorset next to each
other in the same bus, and than the stripeset is across multiple buses
made up of 4 mirrors. Seems like the best way to me, but I wanted to see
what others thought.
Any advice is appreciated.
-Ed Silver
Received on Sat Jan 27 2001 - 15:20:36 NZDT