Ooops.  I can't add.  I realized after I sent my reply that 
the .156 subnet should be .160.  Sorry about the confusion.  ;-)

John


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---- On Thu, 13 Dec 2001, John Neiberger ([EMAIL PROTECTED]) 
wrote:

> Hi Sarah,
> 
> Your calculations were correct at the beginning but somewhere 
> toward the end the train went of the track.  :-)
> 
> If you need at least five equal-sized subnets, you're going 
to 
> end up with three unused subnets.
> 
> You already figured out that by using two extra bits of 
> subnetting you'd get four subnets.  Add one more bit and 
you'll 
> have eight usable subnets.
> 
> In your last example, you have overlapping subnets:  
> 
> 1=65.85.105.0 255.255.255.128 
> 2=65.85.105.0 255.255.255.192
> 
> You must not overlap subnets when doing VLSM.  In the end you 
> should have the following subnets available:
> 
> 65.85.105.0
> 65.85.105.32
> 65.85.105.64
> 65.85.105.96
> 65.85.105.128
> 65.85.105.156
> 65.85.105.192
> 85.85.102.224
> 
> These all have the same mask, a /27, or 255.255.255.224.
> 
> I can't draw it out in text, but one way to think of 
subnetting 
> is to start out with a big circle.  That's your /24.  Draw a 
> line down the center and you have two /25s.  Draw another 
line 
> across the middle and you now have four /26s.  Take one of 
> those /26s and divide it in half and you now have three /26s 
> and two /27s.  You could then further divide the /27 into 
> two /28s.  
> 
> This makes it very simple and helps you not to overlap 
prefixes 
> on accident.
> 
> HTH,
> John
> 
> ________________________________________________
> Get your own "800" number
> Voicemail, fax, email, and a lot more
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> 
> 
> ---- On Thu, 13 Dec 2001, Sarah Parker 
([EMAIL PROTECTED]) 
> wrote:
> 
> > Hello Everyone,
> > 
> > I am working on a small IP address project and trying
> > to figure out VLSM.
> > 
> > Since I am not very good and do not have much
> > experience with IP addressing, I wanted to send this
> > to make sure what I have is correct or if I am really
> > wrong on this one. 
> > Thanks in advance for any feedback or corrections!!
> > 
> > This is a new network-
> > Current IP Address=65.85.105.0
> > Mask=255.255.255.0
> > 
> > I need a total of  5 subnets.
> > 
> > What I did
> > Took 65.85.105.0, 255.255.255.128 to subnet into  2
> > networks,
> > This gave me
> > Subnet 1= 65.85.105.0, hosts 1-126, broadcast  127
> > Subnet 2=65.85.105.128, hosts 129-254, broadcast 255
> > 
> > Took 65.85,105.128 255.255.255.192 to subnet into 4
> > subnets
> > This gave me
> > Subnet 1=65.85.105.0. hosts 1-62, broadcast 63
> > Subnet 2=65.85.105.64, hosts 54-126, broadcast 127
> > Subnet 3=65.85.105.128, hosts 129-190, broadcast 190
> > Subnet 4=65.85.105.192, hosts 193.254, broadcast 255
> > 
> > So this would give me to use on the network
> > 1=65.85.105.0 255.255.255.128 (17 mask?)
> > 2=65.85.105.0 255.255.255.192 (18 mask?)
> > 3=65.85.105.64 255.255.255.192
> > 4=65.85.105.128 255.255.255.192
> > 5=65.85.105.192 255.255.255.192
> > 
> > 
> > Did I do this correctly? This is based on using subnet
> > zero.
> > 
> > I am using a public class A but for security reasons I
> > did change the actual real address.
> > 
> > Thanks again for everyones feedback.
> > 
> > 
> > __________________________________________________
> > Do You Yahoo!?
> > Check out Yahoo! Shopping and Yahoo! Auctions for all of
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> > or bid at http://auctions.yahoo.com
> [EMAIL PROTECTED]
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