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From: john@CAM.UNISYS.COM (John Dempsey)
Newsgroups: comp.sys.att
Subject: Missing: 5 MB on a Unix PC.
Message-ID: <381@sdcjove.CAM.UNISYS.COM>
Date: Mon, 26-Oct-87 20:06:32 EST
Article-I.D.: sdcjove.381
Posted: Mon Oct 26 20:06:32 1987
Date-Received: Thu, 29-Oct-87 20:03:10 EST
Reply-To: john@CAM.UNISYS.COM (John Dempsey)
Distribution: comp.sys.att
Organization: Unisys, R+D Distributed Architectures, Camarillo, CA
Lines: 45

I think I'm missing 5 MB of memory and would like to know why.  When I type:

	"du /"

I get "17366 /" as the result.  According to the du manual page, the value 
returned by the du command is the number of blocks used.  Then when I type:

	"df"

I get "4046 blocks" (i.e., the number of blocks free).  Well,

	17366 + 4046 = 21,412 blocks

	21,412 blocks * 512 bytes/block = 10,962,944 bytes

If I have a 20 MB drive (which I do), what happened to my other 9 MB?  I'm
assuming that 1 block == 512 bytes (see below).  Does anyone know how much 
memory a 20 MB drive holds formatted and unformatted?

In AT&T's UA/Administration under System Information menu, it says:

	"Disk Space Free:  13% or 2023k"

2023k is half of 4046 blocks (cf., "df" output and the calculation that 
(4046 blocks * 512 bytes/block = 2072k)).  But is the 13%, as is being reported,
really correct?  13% would be correct if there was 15 MB of memory available.  
That is:

	13% = 2023k/x
	x = 2023k/.13 = 15,561,530 bytes  (AT&T's number)

My question is: "Why is there roughly a 5 MB difference between AT&T's 
number (15.5 MB) and my number (10.9 MB)?"
--------
In checking to see if 1 block == 512 bytes, one of my directories contains:
	94,247 bytes 	(from running the "ls -la" command)
	191 .		(from the "du" command)
thus,
	191 blocks * 512 bytes/block = 97,792 bytes
which confirms that 1 block does equal 512 bytes.
------
Can someone with a 20 MB drive on a Unix PC type "du /" and "df", and tell me 
what the reported numbers are on their system.  Thanks a 5 MB!

							John Dempsey