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 http://www.ureach.com/reg/tag ---- 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 > your unique holiday gifts! Buy at http://shopping.yahoo.com > or bid at http://auctions.yahoo.com [EMAIL PROTECTED] Message Posted at: http://www.groupstudy.com/form/read.php?f=7&i=29164&t=29160 -------------------------------------------------- FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
