IPv6 Address Planning

Kateel Vijayanandakvijayan@cisco.comWim Verrydtwverrydt@cisco.com

IPv6 Workshop ManchesterSeptember 2013

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The Powerpoint version of this material may be reused and modified only with written authorization

Using any part of this material is allowed if credit is given to 6DEPLOY

The PDF files are available from www.6deploy.eu

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• Mail to: martin.potts@martel-consulting.ch

• Or: bernard.tuy@renater.fr

Contents

1. IPv6 Subnetting: Step by Step

2. Recommendations

3. Address Planning Example

Preparing an IPv6 addressing plan is not a trivial task

Needs timely planningAll remote network points and existing

topologies need to be considered Look at your IPv4 Addressing Plan

If you don’t have one, build one! But, keep in mind:

Aggregation = YESConservation = NO

Introduction

Why do we perform subnetting? IPv4: conserve address space IPv6: planning and optimization for

routing or security VLSM vs SLSM – there’s no point to do

VLSM in IPv6 Subnets vs hosts – number of hosts is

irrelevant in v6 There’ll rarely be a need to expand a /64

subnet!

IPv4 subnetting concepts to FORGET!

For a given IPv6 prefix ‘P’ and prefix length ‘L’ List all the sub-prefixes of length ‘L’ therein Break ‘P’ into N subnets

Repeat for each sub-prefix as required

The generic IPv6 subnetting problem

Derived from total number of desired subnets

Range of hexits that define each individual subnet

The difference between each subnetID The individual

subnets

Generic IPv6 subnetting procedure

An ISP with operations in 10 cities just got a 2001:db8:: /32 allocation from its RIR.

Subnet this prefix equally between the 10 cities.

Subnetting Example

Number of subnets: N = 10 Subnet bits required (s): 2s ≥ 10 , sb = 4 (to the nearest integer)

Thus, to subnet 2001:db8::/32 to cover 10 subnets:

We’ll need to use 4 bitsThose 4 bits give us 24 = 16 subnets (we’ve 6 spare subnets)Prefix length of each subnet is /36 (i.e 32 + 4 = 36)

We calculateNumber of interesting hexits = sb/4 = 1Block:

Subnetting example: analysis

First subnetID[Decimal]: a1= 4096(1-1) = 0 (0x0) | from

an=(n-1)dFirst subnet: 2001:db8:0000::/36

Last subnetID [Decimal]: a16 = 4096(16-1) = 61440 (0xf000)[Hex]: a10= 1000(10-1) = 1000(f) = 0xf000Last subnet: 2001:db8:f000::/36

Verify your answer using subnet toolse.g. sipcalc 2001:db8::/32 –v6split=36

Subnetting example: analysis

sipcalc 2001:db8::/32 –v6split=36 | grep NetworkNetwork - 2001:0db8:0000:0000:0000:0000:0000:0000 -Network - 2001:0db8:1000:0000:0000:0000:0000:0000 -Network - 2001:0db8:2000:0000:0000:0000:0000:0000 -Network - 2001:0db8:3000:0000:0000:0000:0000:0000 -Network - 2001:0db8:4000:0000:0000:0000:0000:0000 -Network - 2001:0db8:5000:0000:0000:0000:0000:0000 -Network - 2001:0db8:6000:0000:0000:0000:0000:0000 -Network - 2001:0db8:7000:0000:0000:0000:0000:0000 -Network - 2001:0db8:8000:0000:0000:0000:0000:0000 -Network - 2001:0db8:9000:0000:0000:0000:0000:0000 -Network - 2001:0db8:a000:0000:0000:0000:0000:0000 -Network - 2001:0db8:b000:0000:0000:0000:0000:0000 -Network - 2001:0db8:c000:0000:0000:0000:0000:0000 -Network - 2001:0db8:d000:0000:0000:0000:0000:0000 -Network - 2001:0db8:e000:0000:0000:0000:0000:0000 -Network - 2001:0db8:f000:0000:0000:0000:0000:0000 -

Subnetting – subnets using sipcalc

Ensure that all prefixes fall on nibble boundaries

Plan a hierarchical scheme to allow for aggregation Site: any logical L3 aggregation point

(POP, building, floor) Region: a collection of sites Autonomous System

Use same prefix lengths for all prefixes of the same level (SLSM)

Recommendations for planning

Global IPv6 address hierarchy

Conceptual view of an ISP network

Select your largest SITE

Proceed as follows Estimate the number of end-networks in it

now Adjust for growth in 5 years Round to nearest nibble boundary

(maxSITEsize) 2^(4n) = 16, 256, 4096, 65535, …

Estimating the needs of SITEs

Try to align allocation units to nibble boundaries

Round up your estimates to 2n where n is a multiple of 4

[16, 256, 4096, 65536 etc]Ensure your prefixes fall on the following

nibbles: /12, /16, /20, /24, /28, /32, /36, /40, /44,

/48, /52, /56, /60, /64 Working with nibble boundaries

Greatly simplifies address planningProvides room for expansion at each level of the

network hierarchy

About nibble boundaries

Consider the range of addresses for 2001:db8:3c00::/40 [first] 2001:db8:3c00:0000:0000:0000:0000:0000

[last] 2001:db8:3cff:ffff:ffff:ffff:ffff:ffffEasy to see that differentiating hexits range from 0-f

Consider the range of addresses for 2001:df8:3c00::/42

[first] 2001:db8:3c00:0000:0000:0000:0000:0000 [last] 2001:db8:3c3f:ffff:ffff:ffff:ffff:ffff

You’ll have to calculate the differentiating hexits

Nibble boundary alignment example

“End-prefix” is the prefix given to a network that connects to each site e.g customer network Estimate the number of #SITEs in your largest region (round to nibble boundary) Calculate the number of end-site prefixes: N = #regions x #SITEs x maxSITEsize

Finding the total number of end prefixes required

Calculate number of subnet bits required to give us N prefixes:

Allocation size (what you request from the RIR) is 48 – s [if assigning /48s per end-site] 52 – s [if assigning /52s per end-site]

Calculating your allocation size

An ISP has operations in 10 provinces.

The largest province has 50 POPs, the largest of which has about 2700 customers.

Estimate the IPv6 addressing needs of this ISP.

IPv6 address planning – Example

We know Number of regions: #regions = 10 [round to 16] Number of sites: #SITEs = 50 [round up to 256] maxSITEsize = 2700 [round up to 4096]

We calculate Total number of end-network prefixes required is

N N=16 x 256 x 4096 = 16,777,216 Number of subnet bits required:

s=log16,777,216/log2 = 24.

AP example – analysis and solution

Allocation size: 48 – 24 = 24 [Assuming /48s to end-sites] 52 – 24 = 28 [Assuming /52s to end-sites]

Thus the ISP needs to request a /24 or /28 from its service region RIR

AP example – analysis and solution

/32 for LIRs is just the minimum size according to most RIR policies

If you can show that you need more, you usually can get more!

Do NOT start with /32 [or /48] and try to fit in.

INSTEAD analyse your needs and apply based on them.

IPv6 address planning – a few clarifications

RFCs recommend /64 for all subnets Even p2p and loopbacks DO allocate a /64 for all links …but, DO configure what makes operational

sense (e.g /127 for p2p and /128 for loopbacks)

Understand what breaks if you use longer prefix lengths

IPv6 address planning – a few clarifications

While performing IPv6 address planning, forget conservation

Paradigm change: moving to SLSM

Tools like sipcalc are useful

It’s fairly quick to reach some numbers if you have all the details available

Conclusion

Questions

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