Has anyone successfully mixed FC8s and FC9s on the same FCAL loop?
We currently have four FC8s connected to our F740, one FC8 not connected (no drives in it), and two FC9s just arrived (we ordered FC8s, but tech said that they'll co-exist just fine).
The instructions indicate that when you connect one FC9 to an existing loop containing FC8s, you have to make it ID-0. What about the second one? (I'm assuming that we just make it ID-1, but the documentation isn't clear.)
I'll probably just price what NetApp would charge to just upgrade your existing FC8s to FC9s. I'd rather have a homogeneous set of shelves, honestly.
Also, I heard a nasty rumor that the F740 actually maxes out at 7 trays worth of (18Gig) disks. Is this true or can we order our 8th tray without worries?
-- Rob ---------------------------------------- Rob Windsor E-Mail - mailto:rob_windsor@adc.com Senior Unix Systems Administrator Voice - phone:972-680-6919 Computer Services Fax - phone:972-680-0370 Broadband Access and Transport Group __o ADC Telecommunications _`<,_ Richardson, TX 75082 (_)/ (_)
On Wed, 10 May 2000, Rob Windsor wrote:
Has anyone successfully mixed FC8s and FC9s on the same FCAL loop?
We currently have four FC8s connected to our F740, one FC8 not connected (no drives in it), and two FC9s just arrived (we ordered FC8s, but tech said that they'll co-exist just fine).
The instructions indicate that when you connect one FC9 to an existing loop containing FC8s, you have to make it ID-0. What about the second one? (I'm assuming that we just make it ID-1, but the documentation isn't clear.)
On one of my F740's I have 3 FC8 shelves and a single FC9 shelf. The 3 FC8's are the original shelves and the FC9 was added later. They are all on a single FCAL loop. Since the 3 FC8's were there first, they are number 0,1,2 and the FC9 was added as id 3. I do not recall seeing anything in the instructions about making the FC9 be id 0. When I first hooked up the shelf I had a problem that turned out to be a loose cable but as a result I was on the phone to tech support for almost an hour. During that phone call I described in exact detail what I was doing and they never told me to make the FC9 be id 0. This configuration has been up and running in a production environment for over 4 months with no problems. If it makes any difference, all of the drives that are currently in the filer are 18GB drives.
I'll probably just price what NetApp would charge to just upgrade your existing FC8s to FC9s. I'd rather have a homogeneous set of shelves, honestly.
I am also considering upgrading the FC8's to FC9's, not so much to be homogeneous but to be able to upgrade all the disks to 36GB.
Also, I heard a nasty rumor that the F740 actually maxes out at 7 trays worth of (18Gig) disks. Is this true or can we order our 8th tray without worries?
Are you saying "maxes out" in that it will not allow you to hook up the last shelf or that it runs out of horsepower?
On Thu, 11 May 2000, David H. Brierley wrote:
On Wed, 10 May 2000, Rob Windsor wrote:
Has anyone successfully mixed FC8s and FC9s on the same FCAL loop?
On one of my F740's I have 3 FC8 shelves and a single FC9 shelf. The 3 FC8's are the original shelves and the FC9 was added later. They are all on a single FCAL loop. Since the 3 FC8's were there first, they are number 0,1,2 and the FC9 was added as id 3. I do not recall seeing
Our VAR came in just this morning and added the new shelf to two older ones (after the LRC upgrade in the field alert) on one of our F760's. The new shelf is numbered three, no more, no less. Three shalt be the number thou shalt count, and the number of the counting shall be three. Four shalt thou not count, nor either count thou two, excepting that thou then proceed to three. Five is right out!
Oh, sorry, where was I? Oh, yes...here is a snippet of `sysconfig -v`:
slot 4: Fibre Channel Host Adapter 4 (Qlogic ISP2100 rev. 3, 32-bit) Firmware rev: 1.15.72 Host Loop Id: 7 FC Node Name: 2:000:00e08b:01096f Cacheline size: 8 FC Packet size: 2048 0: SEAGATE ST118202FC FDF6 Size=17.0GB (35566480 blocks) [...] 16: SEAGATE ST318203FC F304 Size=17.0GB (35566480 blocks) Shelf 0: EDM Kernel Version : 0.4 App. Version : 0.0 Shelf 1: EDM Kernel Version : 0.4 App. Version : 0.0 Shelf 2: VEM Kernel Version : 2.5 App. Version : 3.0
Until next time...
The Mathworks, Inc. 508-647-7000 x7792 3 Apple Hill Drive, Natick, MA 01760-2098 508-647-7001 FAX tmerrill@mathworks.com http://www.mathworks.com ---
On Thu, 11 May 2000, Todd C. Merrill wrote:
F760's. The new shelf is numbered three, no more, no less. Three shalt
Damn. So much for the humor. It is the third shelf, numbered two. Forgot the first is "zero." What'd'ya expect, I had to be at work at 6:45 AM this morning to install it. Time to go home.
Until next time...
Todd C. Merrill The Mathworks, Inc. 508-647-7000 x7792 3 Apple Hill Drive, Natick, MA 01760-2098 508-647-7001 FAX tmerrill@mathworks.com http://www.mathworks.com ---
On Thu, 11 May 2000 12:02:35 EDT, "David H. Brierley" wrote:
I do not recall seeing anything in the instructions about making the FC9 be id 0.
Yeah, I read that wrong.
I am also considering upgrading the FC8's to FC9's, not so much to be homogeneous but to be able to upgrade all the disks to 36GB.
Also, I heard a nasty rumor that the F740 actually maxes out at 7 trays worth of (18Gig) disks. Is this true or can we order our 8th tray
without
worries?
Are you saying "maxes out" in that it will not allow you to hook up the last shelf or that it runs out of horsepower?
You hit a maximum in one of the specifications. As per Dennis Haag from NetApp:
The F740 maxxes out at 16 shelves (8 per loop), 112 disks (56 per loop), or 928GB, whichever you hit first. So with 18GB disks 928/18 ~= 51.5 disks. Parity disks and hot spares would not be counted against the 928GB limit.
(This is in the technical specifications for the F740 running Data ONTAP v5.x)
-- Rob ---------------------------------------- Rob Windsor E-Mail - mailto:rob_windsor@adc.com Senior Unix Systems Administrator Voice - phone:972-680-6919 Computer Services Fax - phone:972-680-0370 Broadband Access and Transport Group __o ADC Telecommunications _`<,_ Richardson, TX 75082 (_)/ (_)
On Thu, 11 May 2000, Rob Windsor wrote:
On Thu, 11 May 2000 12:02:35 EDT, "David H. Brierley" wrote:
Are you saying "maxes out" in that it will not allow you to hook up the last shelf or that it runs out of horsepower?
You hit a maximum in one of the specifications. As per Dennis Haag from NetApp:
The F740 maxxes out at 16 shelves (8 per loop), 112 disks (56 per loop), or 928GB, whichever you hit first. So with 18GB disks 928/18 ~= 51.5 disks. Parity disks and hot spares would not be counted against the 928GB limit.
But.... According to my F740, an 18GB disk is not 18GB. The output of 'sysconfig -v' identifies the drives as follows:
2: SEAGATE ST118202FC FDF6 Size=17.0GB (35566480 blocks)
Based on my understanding of the way mathematics works, 928 divided by 17 is 54.5882. Since a raid group is limited to 28 drives, filling the filer this full requires at least 2 parity disks. Eight full shelves, at 7 drives per shelf, is 56 drives, minus the 2 parity drives, yields 54 possible data drives. The last time I looked, 54 was less than 54.5882. And this even ignores the question of spare drives, which I'm assuming you have at least one of and possibly more. So, it should be possible to completely fill eight shelves with "18GB" drives and not exceed any of the defined limits.
On Thu, 11 May 2000 18:37:11 EDT, "David H. Brierley" wrote:
Are you saying "maxes out" in that it will not allow you to hook up
the last
shelf or that it runs out of horsepower?
You hit a maximum in one of the specifications. As per Dennis Haag
from
NetApp:
The F740 maxxes out at 16 shelves (8 per loop), 112 disks (56 per
loop),
or 928GB, whichever you hit first. So with 18GB disks 928/18 ~= 51.5
disks.
Parity disks and hot spares would not be counted against the 928GB
limit.
But.... According to my F740, an 18GB disk is not 18GB. The output of 'sysconfig -v' identifies the drives as follows:
2: SEAGATE ST118202FC FDF6 Size=17.0GB (35566480 blocks)
Based on my understanding of the way mathematics works, 928 divided by 17 is 54.5882. Since a raid group is limited to 28 drives, filling the filer this full requires at least 2 parity disks. Eight full shelves, at 7 drives per shelf, is 56 drives, minus the 2 parity drives, yields 54 possible data drives. The last time I looked, 54 was less than 54.5882. And this even ignores the question of spare drives, which I'm assuming you have at least one of and possibly more. So, it should be possible to completely fill eight shelves with "18GB" drives and not exceed any of the defined limits.
Yes, I had already conceeded that, perhaps I wasn't clear.
We run with 13 disk raid groups and we have a hot spare for each raid group. That makes it an even "6 drives per shelf" when calculating the capacity of the filer. Even assuming a full 18Gig on the drives, we can fully populate 8 trays.
-- Rob ---------------------------------------- Rob Windsor E-Mail - mailto:rob_windsor@adc.com Senior Unix Systems Administrator Voice - phone:972-680-6919 Computer Services Fax - phone:972-680-0370 Broadband Access and Transport Group __o ADC Telecommunications _`<,_ Richardson, TX 75082 (_)/ (_)
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David H. Brierley -
Rob Windsor -
Rob Windsor -
Todd C. Merrill