Ok, looks like there are some issues. Listed are your results, and comments for each:
1=65.85.105.0 255.255.255.128 (17 mask?) This is a /25, it will allow you to use from 1-126, 127 is broadcast. 2=65.85.105.0 255.255.255.192 (18 mask?) This was already used in the original /25, you cannot use it twice, you now must start at 128, because that is where the first /25 ends. 3=65.85.105.64 255.255.255.192 This was also used in the original /25. 4=65.85.105.128 255.255.255.192 This one is ok, /26, gives you from 129-190 to use, 191 is broadcast. 5=65.85.105.192 255.255.255.192 This one is ok, /26, gives you from 193-254 to use, 255 is broadcast. So, in the end you would have the following: 65.85.105.0/25 65.85.105.128/26 65.85.105.192/26 How many subnets do you want? Do they each have to have a certain number of hosts? -----Original Message----- From: Sarah Parker To: [EMAIL PROTECTED] Sent: 12/13/2001 11:15 PM Subject: Help with IP Addressing/VLSM- work project [7:29160] 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 Message Posted at: http://www.groupstudy.com/form/read.php?f=7&i=29166&t=29160 -------------------------------------------------- FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
