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/30 subnet question

  • 04-01-2011 5:41pm
    #1
    Registered Users, Registered Users 2 Posts: 1,214 ✭✭✭


    Hi all.

    I have a repeater router set up , but not sure if I can figure out the routing of ip's correctly. It's running RouterOS from Mikrotik.

    This is what I hope to setup:

    Router 1 (AP ip 192.168.2.5/30)
    > Router 2 (client side 192.168.2.6/30 AP side 192.168.3.1/24)

    Can the AP side of router 2 be allocated a /24 subnet without any problems,or will I have to stay in the /30 range?


Comments

  • Closed Accounts Posts: 4,564 ✭✭✭Naikon


    Hi all.

    I have a repeater router set up , but not sure if I can figure out the routing of ip's correctly. It's running RouterOS from Mikrotik.

    This is what I hope to setup:

    Router 1 (AP ip 192.168.2.5/30)
    > Router 2 (client side 192.168.2.6/30 AP side 192.168.3.1/24)

    Can the AP side of router 2 be allocated a /24 subnet without any problems,or will I have to stay in the /30 range?

    Router 2's mask can be anything you want, considering the 3 adrresses you have assigned all have different network prefix values. the value after the slash is totally up to you. So for the first subnet, we know we have 30 bits in the subnet mask field to indicate the first 30 bits represent the network portion, and the remaining bits are for the hosts. To make it a bit easier to visualise: 30 bits for our network prefix translate into binary as the following, where each group of 8 bits represents 1 out of 4 octets:
    Decimal  scale - left to right
              128 64 32 16 8  4 2 1
    
    11111111.11111111.11111111.11111100
    
    first section/octet - 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
    second section/octet - 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
    third section/octet - 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255
    fourth section/octet - 128 + 64 + 32 + 16 + 8 + 4 + 0 + 0 = 252
    

    So from the above, basically, all the bits bar 2 are turned on for
    the network prefix, which means that we only have the 2 + 1 positions
    in the last octet for computers/hosts. This means you can only have two
    hosts if you use /30. The rest makes up the network address prefix.

    So our netmask is 255.255.255.252 or /30 for 192.168.3.*

    192.168.1.3. 192.168.3.1 and 192.168.3.2 are the only addresses
    you can use if you assign /30 for this subnet. Not much. Honestly,
    just use /24 as it's used on most routers by default for most subnets.
    Again, because each address network prefix is unique, you can use
    /24 or /30 for each and every one of them.

    Hope this does not come across as overly complex, I am not the best
    at explanation, but the minimum you should take away here is that
    you can use any subnet value provided the first 3 octets/numbers don't
    appear in another network address for the other NIC's.


  • Registered Users, Registered Users 2 Posts: 6,007 ✭✭✭Moriarty


    Router 1 (AP ip 192.168.2.5/30)
    > Router 2 (client side 192.168.2.6/30 AP side 192.168.3.1/24)

    Can the AP side of router 2 be allocated a /24 subnet without any problems,or will I have to stay in the /30 range?

    That'll work fine, just make sure that anything that needs it is aware that any traffic for 192.168.3.0/24 needs to be routed through 192.168.2.6.

    AP = ? :)


  • Registered Users, Registered Users 2 Posts: 1,214 ✭✭✭JabbaTheHut


    Thanks for the info guys. I wasn't too sure if it was possible or not.

    Moriarty. AP is Access Point also running RouterOS


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