ipv6 advertise route

Posted on 2012-09-19
Last Modified: 2012-10-04
I have customer A IP address as 2001: 0410:0001::/48 and customer B as 2001:0410:0002::/48. Both customers are going to an ISP and the ISP will advertise the route as 2001:0410::/35. I understand the CIDR business. But I don't understand 35 bits of /35 in relation to the 2001:0410::
I understand the 48 bits from /48. But how do you get /35 out of 2001:0410::

Question by:biggynet
    LVL 19

    Accepted Solution

    Each block of 4 hex digits is 16 bits.
    2001 is 16 bits
    0410 is 16 bits
    2001:0410::/32 is 32 bits
    2001:0410:0 is 36 bits
    2001:0410:00 is 40 bits
    2001:0410:000 is 44 bits
    2001:0410:0000 is 48 bits.

    2001:0410::/35 is

    IPv6 address      2001:0410:0000:0000:0000:0000:0000:0000       (2001:410::)
    Prefix length      35
    Network start      2001:0410:0000:0000:0000:0000:0000:0000       (2001:410::)
    Network end      2001:0410:1fff:ffff:ffff:ffff:ffff:ffff       (2001:410:1fff:ffff:ffff:ffff:ffff:ffff)
    Netmask      ffff:ffff:e000:0000:0000:0000:0000:0000       (ffff:ffff:e000::)
    LVL 17

    Assisted Solution

    Doing the last two rows in hex is really confusing.
    Doing them in binary makes it clearer, but 128 bits is really long...............................

    That why subnetting on "nimble" (48/44/40/36...) makes things much easier to read - and much easier to do reverse DNS for.

    Note that the next network, after 2001:0410::/35 is 2001:0410:2::/35.

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