12-09-2005 05:04 AM - edited 03-03-2019 11:11 AM
I thought I understood VLSM until I got to the situation below. What is the procedure/proccess in assigning IP address 192.168.24.0/22 to following networks?
RouterA is connected to 3 other wan routers, B,C,D via serial.
routerA has 50 hosts on its lan.
routerB has 400 hosts on its lan.
routerC has 50 hosts on its lan.
routerD has 200 hosts connected to its lan.
Thanks.
Said
Solved! Go to Solution.
12-09-2005 07:26 AM
Hi,
you could use:
router A: 192.168.24.0/25 (126 hosts)
router B: 192.168.26.0/23 (508 hosts)
router C: 192.168.24.128/25 (126 hosts)
router D: 192.168.25.0/24 (254 hosts)
Make sure your routing protocoll can handle this.
Regards
Martin
12-09-2005 07:21 AM
You need to use a subnet length that will satisfy the minimum requirements.
Here's one example you could follow:
RouterA 192.168.24.0/26 (provides 64 addresses)
RouterB 192.168.26.0/23 (provides 512 addresses)
RouterC 192.168.24.64/26 (provides 64 addresses)
RouterD 192.168.28.0/24 (provides 256 addresses)
Hope this helps,
12-09-2005 07:28 AM
Hi Harold,
so I was half as fast as you were ... are you a double CCIE? ;-)
Regards
Martin
12-09-2005 07:45 AM
LOL!! I am indeed but I don't know if it comes and reinforces your theory.
Cheers,
12-10-2005 10:18 AM
BTW: I just noticed I made a mistake :o( I went beyond the /22 assignment in my previous posting.
Here's a correction:
RouterA 192.168.27.0/26 (64 addresses).
RouterB 192.168.24.0/23 (512 addresses).
RouterC 192.168.27.64/26 (64 addresses)
RouterD 192.168.26.0/24 (256 adresses).
Hope this helps,
12-09-2005 07:26 AM
Hi,
you could use:
router A: 192.168.24.0/25 (126 hosts)
router B: 192.168.26.0/23 (508 hosts)
router C: 192.168.24.128/25 (126 hosts)
router D: 192.168.25.0/24 (254 hosts)
Make sure your routing protocoll can handle this.
Regards
Martin
01-10-2007 03:18 PM
um, how do you get 508 hosts. (2^9)-2=510
have i missed something here?
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