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Author Topic: Routed /48 subnetting  (Read 3383 times)

fonestar

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Routed /48 subnetting
« on: April 17, 2010, 12:27:29 AM »

Just want to make sure I understand /48 subnetting.  If I have been assigned a routed /48 by HE that means I can do whatever I want with the fourth octet?

My assigned ip is 2001:470:e853::/48 so for example I could use one subnet as an inside address of:

2001:470:e853:a::/64

and another...

2001:470:e853:b::/64

and another

2001:470:e853:c::/64

Do I have this right?  I have sixteen bits to do whatever I want with my /48?
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jimb

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Re: Routed /48 subnetting
« Reply #1 on: April 17, 2010, 02:22:35 AM »

Exactly. 

You "own" those 16 bits.  Well, actually, the 80 bits, but because IPv6 pretty much mandates 64 bit LANs you can't really do much w/ the lower 64.

65,536 subnets with ~18.4 billion billion hosts each.  And it will be one of the more common, mundane allocations for IPv6.  :)
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fonestar

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Re: Routed /48 subnetting
« Reply #2 on: April 17, 2010, 01:05:51 PM »

So my above addressing was correct and I could assign one of those addresses to my inside interface?
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jimb

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Re: Routed /48 subnetting
« Reply #3 on: April 17, 2010, 07:37:08 PM »

So my above addressing was correct and I could assign one of those addresses to my inside interface?
Yes.

It works exactly like IPv4 CIDR addressing/subnetting except:

  • Way more bits.
  • Uses hexadecimal instead of decimal (easier than decimal IMHO).

I typically start addressing from 0 though, not a, b, c, so my subnets would be like 2001:db8:1234::/64, 2001:db8:1234:1::/64, 2001:db8:1234:2::/64 ...  2001:db8:1234:ffff::/64.
« Last Edit: April 17, 2010, 07:39:25 PM by jimb »
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