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Thread: ipv6 subnetting

  1. #1
    Join Date
    Nov 2009
    Posts
    1

    ipv6 subnetting

    Need some help understanding why A is the correct answer...

    Question states you need to configure IPV6 addresses on all computers in network. A global address prefix is assigned as 3FFA:FF2B:4D:B000::/41. You have to assign a subnet to four departments. Which subnetted address prefix will you assign to the fourth department?

    A. 3FFA:FF2B:4D:C800::/43
    B. 3FFA:FF2B:4D:B400::/43
    C. 3FFA:FF2B:4D:C000::/43
    D. 3FFA:FF2B:4D:F000::/45
    E. None of the above.

    The correct answer they give is A ,but having difficulty understanding. Looking at http://technet.microsoft.com/en-us/l...6997.aspx#ECAA
    I determined F = 0xB000, n = 4, s = 2^s = 4 s=2, f = ?, i = ? (looks like they increment by 1536(decimal) 0x600(hex) but not understanding how they get that increment?

  2. #2
    Join Date
    Mar 2010
    Posts
    1

    Re: ipv6 subnetting

    Not sure if you ever received a reply to this question but for what its worth here it goes:
    The question actually is wrong. I believe this may be a typing error or something simillar.
    The question really shoud state the following:

    Question states you need to configure IPV6 addresses on all computers in network. A global address prefix is assigned as 3FFA:FF2B:4D:B000::/51. You have to assign a subnet to four departments. Which subnetted address prefix will you assign to the fourth department?

    A. 3FFA:FF2B:4D:C800::/53
    B. 3FFA:FF2B:4D:B400::/53
    C. 3FFA:FF2B:4D:C000::/53
    D. 3FFA:FF2B:4D:F000::/55
    E. None of the above.

    Review the question using these figures and with reference to the Microsoft posting you mentioned earlier you would find that:

    F = B000 (the hexadecimal value of the subnet ID being subnetted otherwise known as the starting subnet)
    n = 4 (number of subnets required)
    s = 2 (the number of bits you are using for subnetting)
    m = 51 (the prefix length of the address prefix being subnetted)
    f = m - 48 OR 51 - 48 = 3

    i = 2^16-(f+s)

    i is the incremental value between each successive subnet ID.


    Hence i = 2^16 - (3 + 2) or 2^11 = 2048 (800 Hex)

    Hence first subnet is F = B000
    Second subnet = B000 + 800 = B800
    Third Subnet = B800 + 800 = C000
    Fourth Subnet = C000 + 800 = C800

    The net subnet prefix of course if the existing prefix (51) plus the number of bits required to add 4 new subnets which was 2 OR
    51 + 2 = 53

    This is how they arrived at the answer A. For those who are need more information on the formular used, pls go to
    http://technet.microsoft.com/en-us/l...6997.aspx#ECAA

    There really is a lot of misinformation on IPV6 subnetting out there and its already a challenging topic on its own without this added complication of typo's and a whole load of guys who are copying the wrong answers without a clue of how this works and passing on the answers as their own!!!

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