You almost got it, but it does not seem correct. When you break the 65.85.105.128 255.255.255.192 into 4 subnets, you cut from the 65.85.105.128, not the 65.85.105.0. Also, you cannot cut 4 subnets out of a 65.85.105.128 with a 255.255.255.192 mask. You would only get 2 more. Since you have a 65.85.105.128, you only have 128 hosts left. A 255.255.25.192 mask will give you 64 host subnets. 128/64 is 2. A VLSM lets you subnet a subnet while routing is still sane. Anyway, you have to further slice it using a 255.255.255.224.
65.85.105.128 with 32 host subnets will give you the final 4 since 128 hosts divided by 32 is 4 subnets. I put the masks in both bitcount and dotted decimal notation. Subnet 1 = 65.85.105.0/25 (255.255.255.128) Subnet 2 = 65.85.105.128/27 (255.255.255.224) Subnet 3 = 65.85.105.160/27 (255.255.255.224) Subnet 4 = 65.85.105.192/27 (255.255.255.224) Subnet 5 = 65.85.105.224/27 (255.255.255.224) At 11:15 PM 12/13/01 -0500, Sarah Parker 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 -Carroll Kong Message Posted at: http://www.groupstudy.com/form/read.php?f=7&i=29162&t=29160 -------------------------------------------------- FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
