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Subnetting and IP addressing

Discussion in 'Network Infrastructure' started by Xenophon, Apr 24, 2008.

  1. Xenophon

    Xenophon Bit Poster

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    As I did so well in that area today would it be of any use to anyone if I put some brief notes in a post of how I work these out? Is this the right place for such a post. Has this been done elsewhere? I dont want to repeat if done by someone else already?
    Thanks
    :D
     
    Certifications: A+, N+, 70-210, 70-270, MCSA 2003
    WIP: CCNA or 70-293 (not sure yet)
  2. Leehaa

    Leehaa Gigabyte Poster

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    Ooooh - please do - anything you know would be greeeat! We have a rather different subnet scheme where I have gone to...ANYTHING YOU CAN SUGGEST THAT WILL HELP ONE GET TO GRIPS A BIT MORE WITH IT ALL WOULD BE FANTASTIC!!

    aPOLOGIES FOR THE SHOUTING - I GOT OVER ENTHUSIASTIC ABOUT IT ALL AND ACCIDENTLY KNOCKED THE CAPS, BUT CAN'T BE BOTHERED TO TAKE IT OFF AS KNACKERED AND OFF TO BED!
     
    Certifications: MCP, MCDST, ITIL v3, MBCS, others...
    WIP: BSc IT & Computing, RHCE
  3. Xenophon

    Xenophon Bit Poster

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    Sub-netting Checklist Notes

    1. Add 1 to the Minimum Hosts per subnet required number. Convert this value to Binary. Count Bits to give you the number of required bits for the Subnet Mask value
    Final Max host figure is 2 (to the power of the number of Host bits) -2
    -2 due to not being able to use all 1’s or 0’s for host addresses
    2. Subtract 1 to the Minimum subnets required. Convert to Binary. Count bits for required subnet bits of Subnet mask.
    Max Subnets = 2 (to the power of number of subnet bits)
    On a modern router/network since RFC 1812 (CISCO) Network ID’s can be all 0’s or all 1’s
    3. Memorise – Write out all 8 Binary to decimal SM positions ie:
    128 = 10000000 /17 /25
    192 = 11000000 /18 /26
    224 = 11100000 /19 /27
    240 = 11110000 /20 /28
    248 = 11111000 /21 /29
    252 = 11111100 /22 /30
    254 = 11111110 /23 /31
    255 = 11111111 /24 /32

    Bit Positions 128 64 32 16 8 4 2 1

    This can be done just before you start the exam on the paper or wipe board you are given. You can of course use the calculator to keep working these out but I find it easier to have them there in front of me.
    If you do not understand what these numbers mean you need to do some more basic work on understanding IP Addresses and subnetting and then come back to this again.

    4. First available Subnet network ID is the Network ID given in the scenario. All subsequent subnet ID’s increment by last bit binary to decimal value of SM Subnet position. Eg
    Network ID = 192.168.56.0 SM 255.255.255.240
    240 = 1 1 1 1 0 0 0 0
    1’s are the subnet bits (4) the 0’s are the host bits (4) Bit position 16 (in decimal) is the bit where subnet bits start. So in this scenario Network ID’s increment by 16.
    5. Host Ranges – 1st available address (all 0’s and one 1) to last available address on subnet (all 1’s and one 0) Eg
    Network ID (SM 255.255.224.0) Host Ranges
    172.16.0.0 172.16.0.1 – 172.16.31.254
    172.16.32.0 172.16.0.1 – 172.16.63.254
    Just remember that the end host range is always one below the start of the next Network ID 31 and 32 in the above example.

    Well thats about it, if you have access to the 70-291 CBT Nuggets (Videos) then the expalination there is pretty much what the above is based on.
    Hope this helps some of you.
    Any questions and I will try and come back to this thread and answer. If I cant I am sure someone in the forum will.
     
    Certifications: A+, N+, 70-210, 70-270, MCSA 2003
    WIP: CCNA or 70-293 (not sure yet)

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