IP address design question

Unanswered Question
Oct 21st, 2009

I am tasked with revamping an IP address scheme for a company that has a hodge podge of arbitrarily picked network addresses in place.

The network itself consists of 4 WAN locations. The largest of the WAN sides needs to have 15 networks with 170 hosts.

I have proposed the following solution to re-address the network:

Notes about IP address restructure

Glen Allen will use a /20 network boundary. The available networks in this boundary will be networks 192.168.0.0 to networks 192.168.15.0.

Louisa will have a /20 network boundary. The available networks in this boundary will be networks 192.168.16.0 to 192.168.31.0.

Marsh Run will have a /21 network boundary. The available networks in this boundary will be networks 192.168.32.0 to 192.168.39.0.

Rock Springs will have a /21 network boundary. The available networks in this boundary will be networks 192.168.40.0 to 192.168.47.0.

What I want to know is:

Will I still be able to use Class C masking within for each network number (ex. 192.168.1.0/24, 192.168.2.0/24, etc.) and then still be able to summarize at the WAN routers using the CIDR as in my proposal above?

I have this problem too.
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Jon Marshall Wed, 10/21/2009 - 10:32

Kevin

"Will I still be able to use Class C masking within for each network number (ex. 192.168.1.0/24, 192.168.2.0/24, etc.) and then still be able to summarize at the WAN routers using the CIDR as in my proposal above?"

Yes, no problem with that. Summarise out to the WAN and then once traffic reaches the correct location the more specific routes will route to the right place.

Jon

Giuseppe Larosa Wed, 10/21/2009 - 10:33

Hello Kevin,

>> Will I still be able to use Class C masking within for each network number (ex. 192.168.1.0/24, 192.168.2.0/24, etc.) and then still be able to summarize at the WAN routers using the CIDR as in my proposal above?

Yes, it is enough to use a modern classless routing protocol so if you don't use RIPv1 or IGRP you are fine.

Hope to help

Giuseppe

Kevin Melton Wed, 10/21/2009 - 17:28

I was reading some documentaton today that said

"Each /8 network supports a maximum of 224 -2 (16,777,214) hosts per network".

Where does the 224 -2 come from?

viyuan700 Wed, 10/21/2009 - 18:29

"Where does the 224 -2 come from?"

Every IP address is 32 bit and have 2 parts network and host. /8 is a A class network and there are 8 network bits and 24 host bits. So A class network have 2 ^24 -2 host (We are subtracting 2 because first address is network & last is broadcast address)

8.0.0.0/8 here

8.0.0.0 is network address

Host are from 8.0.0.1-8.255.255.254

(2nd octed can have 256 options, similarly 3rd and 4th octet hence 256*256*256=16,777,216 -2 host)

8.255.255.255 is broadcast address.

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