ipv6 address planning kateel vijayananda [email protected] wim verrydt [email protected] ipv6...
TRANSCRIPT
![Page 1: IPv6 Address Planning Kateel Vijayananda kvijayan@cisco.com Wim Verrydt wverrydt@cisco.com IPv6 Workshop Manchester September 2013](https://reader030.vdocument.in/reader030/viewer/2022032707/56649e215503460f94b0df9d/html5/thumbnails/1.jpg)
IPv6 Address Planning
Kateel [email protected] [email protected]
IPv6 Workshop ManchesterSeptember 2013
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Copyrights
This slideset is the ownership of the 6DEPLOY project via its partners
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
Looking for a contact ?
• Mail to: [email protected]
• Or: [email protected]
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Contents
1. IPv6 Subnetting: Step by Step
2. Recommendations
3. Address Planning Example
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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
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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!
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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
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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
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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
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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
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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
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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
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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
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Global IPv6 address hierarchy
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Conceptual view of an ISP network
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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
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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
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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
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“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
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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
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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
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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
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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
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/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
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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
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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
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Questions
32