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  1. Junior Member
    Join Date
    Oct 2006
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    India
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    MCSA 2003,Wireless#,CWNA,CWSP
    #1

    Default Class B address with/25 subnet mask.

    Hello friends,

    can you have a look into this and explain me how it works.
    I thought the valid subnets had to be 256-128. how come its 256-255??? is it correct,if yes why?
    172.16.0.0 = Network address
    255.255.255.128 = Subnet mask
    _ Subnets? 2e9 = 512.
    _ Hosts? 2e7 – 2 = 126.
    _ Valid subnets? Okay, now for the tricky part. 256 – 255 = 1. 0, 1, 2, 3, etc., for the third
    octet. But you can’t forget the one subnet bit used in the fourth octet. Remember when I
    showed you how to figure one subnet bit with a Class C mask? You figure this the same
    way. (Now you know why I showed you the 1-bit subnet mask in the Class C section—to
    make this part easier.) You actually get two subnets for each third octet value, hence the
    512 subnets. For example, if the third octet is showing subnet 3, the two subnets would
    actually be 3.0 and 3.128.
    _ Broadcast address for each subnet?
    _ Valid hosts?
    The following table shows how you can create subnets, valid hosts, and broadcast addresses
    using the Class B 255.255.255.128 subnet mask (the first eight subnets are shown, and then the
    last two subnets):
    Subnet 0.0 1.0 2.0 3.0 ... 254.0 255.0
    First host 0.1 1.1 2.1 3.1 ... 254.1 255.1
    Last host 0.254 1.254 2.254 3.254 ... 254.254 255.254
    Broadcast 0.255 1.255 2.255 3.255 ... 254.255 255.255
    Subnet 0.0 0.128 1.0 1.128 2.0 2.128 3.0 3.128 ... 255.0 255.128
    First
    host
    0.1 0.129 1.1 1.129 2.1 2.129 3.1 3.129 ... 255.1 255.129
    Last
    host
    0.126 0.254 1.126 1.254 2.126 2.254 3.126 3.254 ... 255.126 255.254
    Broadcast
    0.127 0.255 1.127 1.255 2.127 2.255 3.127 3.255 ... 255.127 255.255



    Thanks in advance
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  3. Junior Member
    Join Date
    Dec 2007
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    Hayden Lake, ID
    Posts
    26

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    CCNA, A+, MCP: 70-270, 70-290, 70-291
    #2
    Well first of as a general rule when determining Subnets and Hosts numbers.

    Hosts=2^n-2

    However, with Subnets it depends. If its a classfull routing protocol that your subneting for, one that does not support VLSM or CIDR such as RIP or IGRP, for determining the # of Subnets you use 2^n-2. However if its a Classless routing protocol or one that Does support CLSM and CIDR such as EIGRP or OSPF the equation you use is 2^n subnets.

    I'm a little confused as to what you mean by 256-128 or 256-255. The number of valid subnets is dependent on What Class address your using and How many bits your Hosts are borrowing.

    With a 255.255.255.128 subnet with a Class B address the First two octets or 16 bits are reserved for the Network. This means that there are 9 bits for the Subnet (the full 3rd octet + 1 bit), and 7 bits for the host. This means that every subnet Increments every 2^7 hosts or 128 Hosts. 128-2 = 126 assignable Hosts.

    so you have 2^9 or 512 Subnets and 2^7 or 128 Hosts before removing wire and broadcast subnets/addresses.

    If you had a 255.255.255.224 224 is 128+64+32 which is 3 bits borrowed into the 4th octet. This would leave 5 bits for hosts and 8+3 or 11 for subnets. So 2^11 or 2048 subnets, and 2^5 or 32 Hosts per subnet.

    In general, I always just determine the Number of Hosts required to determine the number of Host bits required. This number is then your Incremental Number for your wire address of your subnets. And the number subnets is just whats left over minus the part allocated to the Network.
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  4. Junior Member
    Join Date
    Oct 2006
    Location
    India
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    MCSA 2003,Wireless#,CWNA,CWSP
    #3
    Thank you...that makes it easy. I had missed my subnetting class and was trying to catch up with somethings which i had missed.
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