Re: What is a disk arm?

From the page...

"The following formula shows a worst case scenario.."

Also
"The amount of time required to shut down a system typically is a small
percentage of this value."

Lastly it's important to realize you aren't doing a "suspend" as is the
case with a Windows PC. With a suspend in the PC world, all you are doing
is writing out a copy of the memory to a reserved section of the disk.
With the i, you are updating all the objects on disk that have dirty pages
in memory and doing additional house keeping as required. Not to mention
you are ending all jobs...

Charles


On Fri, Jun 14, 2013 at 9:54 AM, Buck Calabro <kc2hiz@xxxxxxxxx> wrote:

On 6/13/2013 5:39 PM, DrFranken wrote:

Thee be having a math problem in here somewhere.!.

Aye, 'tis indeed a mathematical conundrum I be wanting to solve! The
magick formula I'm pondering is this: ((0.0554*(MS size in MB))/(# of
disk arms)) + 1.6 = # of minutes

First off those ARE real disk arms and you have 20 of 'em.

Doing the math then, I have 32GB of RAM. Is this (32*1024) or (32*1000)
for the purposes of 'size in MB'? Let's go worst case and make it 1024.
Gives me 1815.35. Divide that by my 20 'arms' and I get 90.77 minutes.
Plus 1.6 minutes is 92.4 minutes to write a completely dirty main store
to disk.

You can see why I question what IBM means when they say 'disk arms' - an
hour and a half to do a 'suspend' operation seems insane. Especially
when a typical PWRDWNSYS takes less than 10 minutes.

Those disk arms have a 3GB interface and one single arm could handle
about 1Gb sustained write performance. So in theory one single drive
could accept all 32GB in about 4.3 minutes or all 20 drives writing flat
out could handle it in under 15 seconds, IN THEORY!

Of course it doesn't work ANYTHING Like that in reality with RAID reads
and updates it's twice that at minimum and even then that's in a perfect
world.

Understood. I'm not looking for perfection - the documentation
specifically says that this won't be an exact number. If 'disk unit'
means 'disk arm' then there's my answer.

We can do theory math all day but what I would do is wait until a time
when you can shut the system down. Then do a PWRDWNSYS *IMMED and time
it. Of course you'll not have any users on then and not have any
significant number of files open etc so this will be a 'Best Case' power
down time. Figure double that and you're likely safe.

If my disaster plan calls for a UPS to keep the machine up for a half
hour, what I really need - in order to guarantee that half hour - is a
UPS that will give me over 2 hours of run time. I'd set QUPSDLYTIM to
1800 and whan a power fail occurs, the UPS would carry normal operations
for a half hour, then give me 90 minutes to shutdown.

Having said that, I personally agree that the chances of a completely
dirty main store are infinitesimal, and 15 minutes from normal
operations to cold shutdown are far more likely. This means that my
half our UPS needs to be sized to carry me for 45 minutes to an hour,
which is much more reasonable than a two hour run time.

I'm going to send reader feedback to have IBM clarify what a disk arm is
in these calculations so that the next person reading that manual page
can benefit from this discussion.
--buck
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