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I am new to IP v6 and I am looking to translate some existing private IPv4 addresses into v6 address assignment. Can someone please help me to answer/explain the questions below?

If I have an v4 address of:

  1. 10.10.0.0/22
  2. 10.10.1.0/22
  3. 10.10.2.0/22
  4. 10.10.3.0/22
  5. 10.10.8.0/20
  6. 10.20.1.0/24

What will the new v6 address to be?

I have been looking online @ http://www.subnetonline.com/pages/subnet-calculators/ipv4-to-ipv6-converter.php

or other sites,

Seems like they translated it directly to be:

  1. fe80::a0a:0 /118
  2. fe80::a0a:100 /118
  3. fe80::a0a:200 /118
  4. fe80::a0a:300 /118
  5. fe80::a0a:800 /118
  6. fe80::a14:100 /120

Can someone please explain to me how we get to /118 from either "/22 or /24" (1. and 5)

In addition,

I would like to create the new private address based on the Unique local address "fc00::/64"

How do I expand from there?

Does the following example look correct?

10.10.1.0/24 => fc00::100

10.10.2.0/24 => fc00::200

10.10.3.0/24 => fc00::300

10.10.4.0/24 => fc00::400

10.10.5.0/24 => fc00::500

10.10.6.0/24 => fc00::600

10.10.7.0/24 => fc00::700

10.10.8.0/24 => fc00::800

10.10.9.0/24 => fc00::900

10.10.10.0/24 => fc00::a00

10.10.11.0/24 => fc00::b00

10.10.12.0/24 => fc00::c00

10.10.13.0/24 => fc00::d00

10.10.14.0/24 => fc00::e00

10.10.15.0/24 => fc00::f00

10.10.16.0/24 => fc00::1000

10.10.65.0/24 => fc00::4100

Any help is greatly appreciated it!!

Thanks,

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1 Answer

Unless you're doing something highly unusual and bizarre, the smallest subnet you should use is /64. Anything smaller risks breaking various IPv6 necessities like neighbor discovery and the nice-to-have stateless autoconfiguration. Might even break DHCPv6, though I haven't tried that myself.

To be crystal clear, the tool to which you linked appears to be generating bad data, and I wouldn't use it. (Some of their other tools look fine, though.)

As for "translating" IPv4 to IPv6, you generally don't. Instead, you run dual-stack, where each device talks both IPv4 and IPv6, with independent addresses for each protocol. The addresses themselves are provided by stateless autoconfiguration, DHCPv6, and/or privacy extensions, or (typically for servers) assigned manually.

So let's say you get a block of addresses from your ISP, 2001:db8:1234:5600::/56. This gives you 256 subnets to work with, which is more than sufficient for a small office environment. If you need more than 256 subnets, you can get a /48 without too much difficulty, giving you 65536 subnets. That's enough for all but the largest deployments.

If you can tell us more about what you're actually trying to do, we may be able to provide better answers.

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Thank you for answering Michael, I will start with /64 for the subnet. Can you please explain me a bit on why should 2001:db8:1234:5600::/56 give us 256 subnets? :) –  Cute Puppy Sep 4 '12 at 21:40
    
2^(64-56) == 256. –  Michael Hampton Sep 4 '12 at 21:42
    
I see, thank you Michael –  Cute Puppy Sep 4 '12 at 22:00
    
Hi Michael, Can you please help me to look into this quickly and see if it makes sense at all?? Made some edit on the topic –  Cute Puppy Sep 5 '12 at 13:58
    
I already answered that question. –  Michael Hampton Sep 5 '12 at 19:12
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