/23 subnetting question

Discussion in 'Cisco' started by guest, Apr 20, 2007.

  1. guest

    guest Guest

    hi,

    can someone please help me out? i am having difficulty with this
    mask.can i subnet a class A address with 23 bits of subnetting? e.g.
    10.20.0.0/23 is given to me and i need to subnet it to accomodate one
    subnet with 200 hosts, one subnet with 150 hosts, one subnet with 70
    hosts and two subnets with 50 hosts each... i know i can use
    10.20.0.0/24 to accomodate the first subnet with 200 hosts, what about
    the rest?

    thanks
     
    guest, Apr 20, 2007
    #1
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  2. guest

    Trendkill Guest

    A /23 has only 508 hosts total (255 * 2 = 510, minus network/broadcast
    addresses). With only a single /23, you can't service 520 hosts as
    you are asking. Second, and because networks can only be split into
    networks of specific sizes, you can't create networks for exactly that
    many hosts. You could do a /24 for the 200 requirement, or perhaps a /
    25, a /26, and a /30, which would give you 128 + 64 + 16. The first
    option would 'waste' 54 addresses, and the latter aren't going to be
    in the same broadcast domain. In short, get more than a /23, and try
    to consolidate your requirements or figure out more appropriate IP
    architecture. There is nothing wrong with a /23 or /22 network with
    all hosts in the same broadcast domain, especially since most modern
    networks are gig speed or faster......

    A close suggestion:
    /24 = 255 hosts (10.20.0.0 255.255.255.0)
    /25 = 128 hosts (10.21.0.0 255.255.255.128
    /26 = 64 hosts (10.21.128.0 255.255.255.192)
    /27 = 32 hosts (10.21.192.0 255.255.255.224)
    /27 = 32 hosts (10.21.224.0 255.255.255.224)

    This would be as close as you can get without requiring more networks
    and/or more network space.......
     
    Trendkill, Apr 20, 2007
    #2
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  3. guest

    Trendkill Guest

    And technically these are 2 addresses off each....for network/
    broadcast addresses. So 254, 126, 62, etc.
     
    Trendkill, Apr 20, 2007
    #3
  4. guest

    guest Guest

    i am only restricted to 10.20.x.x/23, only the third octet can be
    played around with can i use

    10.20.0.0/24- 200 hosts
    10.20.2.0/24 150 hosts
    10.20.4.0/26- 62 hosts
    .... and so on
     
    guest, Apr 20, 2007
    #4
  5. guest

    Trendkill Guest

    I'm sorry, I was typing as a /15 above....not /23......same thing
    applies but I was 1 octet off............

    So you can use any 10.20.x.x/23? I guess I didn't pick up on that
    before. Why can't you segment beyond /23 as proposed? You would
    easily meet your requirements by splitting up the /23 as I had
    suggested (slight modifications to meet your exact requirements, and
    using a small portion of the second /23 (10.20.2.0) for the last 120.

    24 = 255 hosts (10.20.0.0 255.255.255.0)
    /24 = 255 hosts (10.20.1.0 255.255.255.0)
    /25 = 128 hosts (10.20.2.0 255.255.255.128)
    /26 = 64 hosts (10.20.2.128 255.255.255.192)
    /26 = 64 hosts (10.20.2.192 255.255.255.192)

    That would do you fine............
     
    Trendkill, Apr 20, 2007
    #5
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