That is the reason that IBM allows installations to set the MEMLIMIT.  While
it is true that even a reasonable limit can be abused by multiple jobs, you
would have the same exposure in a 31 bit world if you allow jobs to use
REGION=0M.
Wayne Driscoll
Product Developer
JME Software LLC
NOTE: All opinions are strictly my own.
  

-----Original Message-----
From: IBM Mainframe Discussion List [mailto:[EMAIL PROTECTED] On Behalf
Of Veilleux, Jon L
Sent: Tuesday, July 18, 2006 6:26 AM
To: [email protected]
Subject: Re: 64-bits is a really big number! - was z/OS level for SETFRR for
AMODE(64)

I always get blank looks when I ask what would happen if someone actually
tried to exploit 64bit addressing to the fullest. How do we provide page
space to back these requests? 
As someone responsible for keeping our mainframes up and running, it worries
me that application type people have the ability to crash the system just by
a stupid mistake. One STORAGE loop in a 64bit address space and the paging
subsystem is toast. I know that there are other exposures that application
folks can use to crash the system, but, to me, this looks like an accident
waiting to happen.


Jon L. Veilleux
[EMAIL PROTECTED]
(860) 636-2683 


-----Original Message-----
From: IBM Mainframe Discussion List [mailto:[EMAIL PROTECTED] On Behalf
Of Craddock, Chris
Sent: Tuesday, July 18, 2006 1:05 AM
To: [email protected]
Subject: 64-bits is a really big number! - was z/OS level for SETFRR for
AMODE(64)

> >> AFAIK, the total capacity of all DASD ever manufactured is still 
> >> insufficient to fully back even ONE 64-bit address space....
> 
> At the rate 80 - 200 gigabyte drives and higher have been produced and

> sold, is the statement being made for only mainframe DASD?  I realize 
> that no z box could handle that much DASD space and wonder if either i

> or p boxes could in theory.

Nothing could.

Let's make a few simplifying assumptions to make the math easy. First let's
use 256GiB as the drive size. Yeah, that is way bigger than any MF drive
ever shipped, or ever thought of, but it's about the size of the drive in my
PC and you can buy bigger drives now for a few hundred bucks. <snicker>

Anyway, one of those puppies is (2**8)*(2**30) = 2**38 bytes And a single
64-bit address space is, well 2**64 bytes. 

So assuming you could use the whole drive, the number of these current
generation drives that you would need to back a single address space is

 (2**64)/(2**38) = 2**(64-38) = 2**26 = 67,108,864

Ok, so on that basis I was a little pessimistic about the time we would need
to build enough disk to back one address space. It is becoming plausible
that some time in the relatively near future there will be enough disk
storage capacity on Earth to map a single 64-bit address space. Of course,
housing, powering, wiring, configuring, initializing and addressing that
much disk storage would be somewhat of a problem and doing it all over again
to map that second address space will be a real b1tch!

In other words, 2**64 is a phenomenally large number. In terms of actual
usable storage, it is wildly beyond the limits of our current technology and
certainly beyond my feeble imagination about what one might actually fill it
up with. An RFID for every star and galaxy perhaps? Endless re-runs of "I
Love Lucy" in HD? Hell if I know.

And for a real giggle-fest, take a look through your own JCL libraries and
see how many jobs are still running with "REGION=something_really_small". I
am glad that the wizards have made it all work out, but in terms of
functional needs, 64-bit addressing is overkill on a cosmic scale.

CC

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