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August 31st, 2009 04:00

How many host per FA port?

There should be a formula for calculating -how many host per FA port is possible to attach.
Or some rule of thumb? I need connection with my calculated IOPS.
Somebody?
Thanks!

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August 31st, 2009 06:00

Hi llello,

take a look at interop matrix and search for "fc-fanout"

http://www.emc.com/collateral/elab/emc-support-matrices.pdf

9 Posts

September 1st, 2009 00:00

Thanks, but there is no written connection between IOPS (port speed) and number of hosts.

11 Legend

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September 1st, 2009 05:00

9 Posts

September 1st, 2009 05:00

Thanks again.
Lets ask another way: There is a question in one EMC exam....."if I have 6000 iops max storage port speed and 1200 iops per host. How many hosts can I attach to this port?! I know the answer, but I don't understand it.

46 Posts

September 1st, 2009 12:00

Thanks again.
Lets ask another way: There is a question in one EMC
exam....."if I have 6000 iops max storage port speed
and 1200 iops per host. How many hosts can I attach
to this port?! I know the answer, but I don't
understand it.


I'm guessing the answer is 5. The storage array does a max of 6k iops for a single port, if every host does 1.2k iops your aggregate iops peaks at 5 hosts. i.e. 5 hosts * 1200 iops/host == 6000 max iops the port supports. If you had a sixth host you have gone past the ports limit on the number of iops (7200)

Not sure it really means anything in real life, as workloads tend to be more random and vary per host.

9 Posts

September 1st, 2009 22:00

Unfortunately the answer should be "3", but I don't know why, and I am looking for this answer 3 weeks already....:(

6 Operator

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September 2nd, 2009 01:00

I guess it's simply a queuing problem.
It's a quite complex and interesting field. :D

http://en.wikipedia.org/wiki/Queueing_theory

To make a long story short, if a system can "handle" N operations per second, the response time won't grow proportionally to iops (since the requests from hosts follow a random pattern). And usually it tends to "diverge" (grow very fast) if you reach 60/70% of utilization. That's what usually we call "the knee".

Now guess what's 60/70% of 5 :D

If you google a little you can find a lot more details on this .. Try googling for "queue theorie response time".

Here's an couple of examples:

http://www.cmg.org/measureit/issues/mit60/m_60_9.html

http://ramya-moorthy.blogspot.com/2008/03/response-time-vs-queue-length-vs-server.html

9 Posts

September 2nd, 2009 01:00

Thanks a lot.
Now I remember,that in some EMC Clariion documents it is written, that everything in the system should be utilized no more than 70%.
This is the best possible answer.
Thanks again :)

6 Operator

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September 2nd, 2009 01:00

I'm pretty glad it solved your clue :D
However considere (as you can see in the links I posted) that it isn't related only to clariion or storage. It's a general rule that applies everywhere you have a "server" (a generic term referring to something/someone serving requests) and its clients :-)

September 8th, 2009 12:00

Has your question been answered?

Message was edited by:
KCSCP

9 Posts

September 9th, 2009 00:00

Yes thank you.

6 Operator

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September 9th, 2009 03:00

I guess KCSCP simply wanted to remind you that you should mark the question as "answered" and eventually award points (green/yellow stars) to better/good answers you received.. ;-)
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