X-Mailer: ELM [version 2.3 PL0]
Newsgroups: bit.listserv.novell
Date:         Sun, 12 Apr 1992 19:28:09 CDT
Reply-To:     Novell LAN Interest Group <NOVELL@SUVM.BITNET>
From:         wittmann@ENGR.WISC.EDU
Subject:      Re: How Much memory per unit diskspace (long winded)
X-To:         novell@suvm.acs.syr.edu

>On Thu, 9 Apr 1992 20:51:43 GMT Ken Lam said:
>>I know what NOVELL recommends, but I don't think that most users are loading
>>68MB of memory in a server which is mounting 2GB of disk with namespaces.
>>
>>Anyone have suggestions(from experience preferred).
>>
>
>Ken, I think you used the wrong formula. Old manuals had a decimal
>point out of place.
>
>The formula for disk space memory requirements for a DOS-only name
>space, is
>
>megs = (.023*disk-space-in-megs)/(blocksize-in-kbytes).
>
>If the block size is 4k (the default), and you have a 2gig disk, you
>need 12 megs (plus about 2 megs for file server normal stuff plus
>whatever NLMS you need.
>
>
>For each name space, you need an additional
>
>
>   (.032*disk-space-in-megs)/(blocksize-in-kbytes)
>
>
>or 16 megs.
>
>Thus if you have Mac name space, for example, you will need
>
>   12+16+2 = 30 megs!
>
>And if you want OS/2 as well, you are up to 46 megs!
>
>
>I don't know how Netware 3.11 uses the name space stuff. For
>example, if you only have two files in Mac format, and 20,000 in DOS
>format, does the system still require all that memory? Or does it
>just need enough for the two Mac files?
>
>
>What I did here was set up a small volume (50 meg) for Mac users,
>and the rest for DOS users. But that isn't too good if people want
>to share WordPerfect files, for example.
>
>Does anyone out there (Art?) know how Novell uses memory for name
>space support?
>

Hmmm,

How old is this formula?  As I understand it, the only reason that
extra memory is required for each name space is so that the hash table
can be enlarged.  Each file on disk will have name space information
for each name space installed on that volume.  This is regardless of
whether the file was ever accessed under the particular name space
(this is in contrast to netware 2.1x & 2.2 where an additional directory
entry is allocated only if the file is accessed by a mac).  The hash table
must contain an entry for every name by which a file might be known, so
if there are three name spaces on a particular volume, there must be three
hash table entries.

Now, hash table entries are quite small (6 or 8 bytes I believe).  This
is in contrast to Netware 3.x where each hash table entry took considerably
more space.  Anyway, the size of the volume is not particularly important,
it's really the number of files that are likely to be held on the volume.
On a 2 Gbyte volume, it might be reasonable to expect a maximum of 250,000
files (our 2Gbyte volume presently has 186,759 files and is quite full).
At that rate, the hash table for single name space should take 8 X 250,000 =
2 Mbytes.  So for each name space added, I would expect that you'd need
about 2 Mbytes of server RAM.  The formula shown above allocated 32 bytes
per block (it apparently assumes that the average file size will be below
4k, that seems a bit small to me).  I think 32 bytes per block is too
much, probably 4 to 8 bytes is closer to the right number.

On the server I mentioned above, the OS & associated processes use about
18 Mbytes (the server has 32 Mbytes of memory, about 14 Mbytes are in
cache buffers).  The hash table is 4,498,520 bytes, so that sounds pretty
close the number I calculated.  (Size of the hash table is found in the
monitor screen under Cache Movable Memory -> SERVER.NLM Directory Hash Index)

The File allocation table may grow too, I'm not sure about that.  In any
event, if you want to use the above formula, remember, it's really an
estimate of the number of files that you're like to store on the disk.  The
disk space involved, is byproduct of how many files the disk could store.
So, if you wanted a 2 Gbyte volume that was only going to hold a couple
very large files, the above formula is completlely useless.  Anyway, the
.032, should probably be more like .008 in that formula.

Art

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Art Wittmann                                  Phone: (608) 263-1748
Network Manager                               Email: wittmann@engr.wisc.edu
Computer Aided Engineering Center                or: wittmann@cae.wisc.edu
University of Wisconsin, Madison
Comments: Gated by NETNEWS@AUVM.AMERICAN.EDU