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:
- 10.10.0.0/22
- 10.10.1.0/22
- 10.10.2.0/22
- 10.10.3.0/22
- 10.10.8.0/20
- 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:
- fe80::a0a:0 /118
- fe80::a0a:100 /118
- fe80::a0a:200 /118
- fe80::a0a:300 /118
- fe80::a0a:800 /118
- 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,