× The internal search function is temporarily non-functional. The current search engine is no longer viable and we are researching alternatives.
As a stop gap measure, we are using Google's custom search engine service.
If you know of an easy to use, open source, search engine ... please contact support@midrange.com.



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

As an Amazon Associate we earn from qualifying purchases.

This thread ...

Follow-Ups:
Replies:

Follow On AppleNews
Return to Archive home page | Return to MIDRANGE.COM home page

This mailing list archive is Copyright 1997-2024 by midrange.com and David Gibbs as a compilation work. Use of the archive is restricted to research of a business or technical nature. Any other uses are prohibited. Full details are available on our policy page. If you have questions about this, please contact [javascript protected email address].

Operating expenses for this site are earned using the Amazon Associate program and Google Adsense.