[comp.sys.mac] More questions on SE/30

cmkuo@cs.utexas.edu (Chin-Ming Kuo) (02/14/89)

I am seriously considering a SE/30, and  have the following questions
for you Mac gurus.

1. Since there's a little chance I can get enough money for all 4 1M
SIMMs, I am planning to use the 256K SIMMS to bring the SE to 2M
total. There's a chance I might get a hold of some 256Kx1 chips and 
some 256Kx4 SIMMs ( what is this!), all are 120ns. can I make
something useful out of these? Or a simple question, what exactly is
the configuration of the 256K SIMM used in the SE30/ II? Are there 8
256Kx1 on a module and I need 4 of the modules to make 1M? 

2. Why AUX requires a min. 80M hard drive? Is it simply the
requirement of size, or the system drive must be at least 80M?
Eventually I will get AUX, so can I get a 40M hd for now and get a
second 40M hd later when I really need to run AUX?

Thanks,
--
Chin-Ming Kuo
cmkuo@cs.utexas.edu
cs.utexas.edu!cmkuo


-- 
Chin-Ming Kuo
cmkuo@cs.utexas.edu
cs.utexas.edu!cmkuo

mithomas@bsu-cs.UUCP (Michael Thomas Niehaus) (02/15/89)

In article <4776@cs.utexas.edu>, cmkuo@cs.utexas.edu (Chin-Ming Kuo) writes:
> 1. Since there's a little chance I can get enough money for all 4 1M
> SIMMs, I am planning to use the 256K SIMMS to bring the SE to 2M
> total. There's a chance I might get a hold of some 256Kx1 chips and 
> some 256Kx4 SIMMs ( what is this!), all are 120ns. can I make
> something useful out of these? Or a simple question, what exactly is
> the configuration of the 256K SIMM used in the SE30/ II? Are there 8
> 256Kx1 on a module and I need 4 of the modules to make 1M? 

I don't understand.  I think that you may have been confused by previous
postings.  There are currently two types of SIMMs that are usable in the
Mac: 256K SIMMs and 1MB SIMMs.  The '256Kx4' means that you are using four
of the 256K chips.  With the SE/30, you must put the SIMMs into the machine
in sets of four.  That means if you buy a 1MB SE/30, you can install four
of the 256k SIMMs (256kx4) to bring your machine to two meg.  But to go any
farther than that, you will have to remove some (4) of those and install
some (4) 1MB chips to bring your system to 5MB.

> 2. Why AUX requires a min. 80M hard drive? Is it simply the
> requirement of size, or the system drive must be at least 80M?
> Eventually I will get AUX, so can I get a 40M hd for now and get a
> second 40M hd later when I really need to run AUX?

A/UX is like any UNIX system: it needs lots of disk space for utilities.
I suppose you could get by with running A/UX on a system with a 40MB
hard drive, but you would probably have to strip it down (maybe get rid
of manual pages, utilities that are not used, etc.).  

But one question: if you are considering a SE/30, are you sure that you are
wanting to consider A/UX?  You do know that A/UX on an SE/30 is not supported
by Apple (maybe in the future, but not now), don't you?

> Chin-Ming Kuo
> cmkuo@cs.utexas.edu
> cs.utexas.edu!cmkuo

Maybe I made a mistake in my previous postings in referring to memory
using the '256Kx4' notation.  Maybe it should have been '4x256K'?  I don't
know...

I know the Macintosh creates some confusing situations, especially with
memory upgrades, so I completely understand why some people are concerned
with what they can and can't do.  I hope with the help of the people on the
net, we can prevent problems like those that occurred here at Ball State
last year:  2MB memory expansion kits were purchased, one for every Mac II
that was purchased earlier.  No one knew that you need 4M to upgrade, so
it ended up that every other person received their upgrade...  Case learned:
if you are not absolutely sure, make sure you ask someone.

Michael Niehaus
Apple Student Rep
Ball State University
UUCP: ..!{pur-ee,iuvax}!bsu-cs!mithomas
AppleLink: ST0374

sklein@cdp.UUCP (02/22/89)

256Kx1 chips are used in Macs.  256Kx4 chips are four bit wide, so you
only need two of these chips to get 256K bytes.

Chips are rated in K (thousand) BITS, while actual RAM is measured in KBYTES,
where 8 bits = 1 byte.

So, for a SIMM to have 256K bytes, it must have eight 256Kbit chips
on it.  Make sense?