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ipv6 advertise route

Posted on 2012-09-19
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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::

Thanks
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Question by:biggynet
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bevhost earned 1000 total points
ID: 38416164
Each block of 4 hex digits is 16 bits.
so
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::)
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by:pergr
pergr earned 1000 total points
ID: 38416974
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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