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  1. Junior Member
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    #1

    Default help with subnetting

    Hello, i am a sophomore at a junior college for Computer Networking and was wondering if anyone could lend me a hand on how to subnet, iv actually been learning how to for sometime but nothing seems to stick and i am completely lost. The main part i get lost at is bit borrowing, i can see an Ip and break it down to binary, but after i find the prefix line for whatever it is i can remember how the 01
    11
    part goes into play,
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  4. Senior Member kMastaFlash's Avatar
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    #3
    Great ebook for $1. Pretty sure $1 won't break the bank

    https://www.amazon.com/CCNA-Success-...rds=subnetting
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  5. Junior Member
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    #4

    Default confused still

    im having difficulty with the subnet landing part, example, when you say this address must sit on this subnet, not really sure how you got that, i understand the rest though as far as getting the ip and a /19 next bound is 24 and the subtraction from 24-19=5 and you add by 5, but im just lost after that.
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    #5
    Quote Originally Posted by Ungadunga911 View Post
    im having difficulty with the subnet landing part, example, when you say this address must sit on this subnet, not really sure how you got that, i understand the rest though as far as getting the ip and a /19 next bound is 24 and the subtraction from 24-19=5 and you add by 5, but im just lost after that.

    I hope this is more of a help than more of a confusion, but here goes...

    Using your /19 example, let's put in a Host IP: 172.16.12.23/19

    To abstract the Network that this Host is using...

    Note *Boundaries = 8.16.24.32*

    24-19 = 5

    2**5 = 32. So, 32 is the 'steps' that Network IPs will take, starting at zero. As we only have 8, it's easy to list them.

    ID Network IP
    1 172.16.0.0 << Remember, we start at zero
    2 172.16.32.0 << An increment of 32.
    3 172.16.64.0 << ditto...
    4 172.16.96.0
    5 172.16.128.0
    6 172.16.160.0
    7 172.16.192.0
    8 172.16.224.0

    We can't add another 32 as that would be 256, too high.

    As the Host IP example is 172.16.12.23, this has to be apart of Net ID 1, the host range of that network being...

    172.16.0.1 to 172.16.31.254, with a Broadcast of 172.16.31.255.

    Like I say, I hope this is of help.

    Regards,

    Rob.
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  7. Senior Member 2URGSE's Avatar
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    #6
    Here is how I do it:

    172.16.12.23/19

    Convert /19 to decimal: 255.255.224.0, The special octet is the 3rd one. (255 is 8 bits, so the first 2 put you at 16 and then you need 3 more)

    So count 128, 192, 224.

    Magic number trick: 256-224 = 32

    Also, the address class is B so 16 bits are reserved for the network. Take 19-16 = 3, 2^3 = 8, so a total of 8 subnets.

    You are working with increments of 32, starting at zero

    172.16.0.0
    172.16.32.0
    172.16.64.0
    .
    .
    .

    If they ask you how many hosts per subnet, subtract 19 from 32, which is 13. (2^13) - 2 = 8190 hosts/subnet

    Don't get discouraged if you don't get these correctly in the beginning, it takes some time to master subnetting to the point you can do it quickly in your head.
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