1. IPv6 Address Planning Kateel Vijayananda kvijayan@cisco.com Wim Verrydt wverrydt@cisco.com IPv6 Workshop Manchester September 2013

2. 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: martin.potts@martel-consulting.ch Or: bernard.tuy@renater.fr

3. Contribs & updates Mukom Akong Tamon, AfriNIC Carlos Friaças, FCCN 11/2012 03/2013 This slideset includes mostly slides from AfriNIC. Please also visit http://learn.afrinic.net

4. Contents 1. IPv6 Subnetting: Step by Step 2. Recommendations 3. Address Planning Example

5.  Preparing an IPv6 addressing plan is not a trivial task  Needs timely planning  All 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 = YES  Conservation = NO Introduction

6.  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!

7.  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

8. Find subnet bits (s) Find Subnet hexits Find Subnet ID increm ent (B) Enume rate subnet IDs 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

9.  The prefix lengths of the mother and sub- prefixes - L and L’ are known. s = L – L’ Ex: breaking a /32 into /56s requires 56 – 32 = 24 bits  Only the number of desired subnets is known  Ex: breaking a /36 into 700 networks needs Step 1: Finding the subnet bits (s)

10.  These are to distinguishing hexits of each subnet  Knowing number of subnet bits ‘s’ (from step 1)  Knowing that 1 hexit (char) = 4 bits (0 … F), then  Number of subnet hexits = s/4 (round up)  Ex: Breaking 2001:db8:c000::/36 to 700 subnets  s = log 700 ÷ log 2 = 9.81 ≈ 10  # subnet hexits = 10/4 = 2.5 ≈ 3  Each of the subnets will be like: 2001:db8:cHHH::/46 Step 2: Finding the # of subnet hexits 10 = L – 36  L = 36 + 10

11.  This is the difference between consecutive subnetIDs  % = MOD  Breaking 2001:db8:c000::/36 in to 700 subnets  sh = 3 (calculated in step 2)  L’ = 46 (/36 original length + 10 bits of subnetting)  Format 2001:db8:cHHH::/46 (calculated previously) Step 3: Finding the Increment/Block (B) Increment (B) = 4

12.  At this point you know the general subnet format  Taking the subnetIDs only, these form an arithmetic progression with following characteristics  Common difference d = block (B)  Initial term = 000  Any term of the progression is  Substituting for d = B and initial term = 000  The nth term is: Step 4: Enumerating the subnetIDs

13.  Recap data  sh = 3 (calculated in step 2)  L’ = L + s = 36 + 10 = 46 (step 1 found s=10)  Format 2001:db8:cHHH::/46 (calculated after step 2)  B = 4 (0x4) (calculated in step 3)  First subnetID  [Decimal]: a 1 = 4(1-1) = 0 (0x0)  First subnet: 2001:db8:c000::/46  Last subnetID  [Decimal]: a 1024 = 4(1024-1) = 4(1023) = 4092 (0xFFC)  [Hex]: a 400 = 4(400-1) = 4(3ff) = FFC  Last subnet: 2001:db8:cffc::/46 Breaking 2001:db8:c000::/36 in to 700 subnets

14.  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

15.  Number of subnets: N = 10  Subnet bits required (s): 2 s ≥ 10, sb = 4 (to the nearest integer)  Thus, to subnet 2001:db8::/32 to cover 10 subnets:  We’ll need to use 4 bits  Those 4 bits give us 2 4 = 16 subnets (we’ve 6 spare subnets)  Prefix length of each subnet is /36 (i.e 32 + 4 = 36)  We calculate  Number of interesting hexits = sb/4 = 1  Block: Subnetting example: analysis

16.  First subnetID  [Decimal]: a 1 = 4096(1-1) = 0 (0x0) | from a n =(n-1)d  First subnet: 2001:db8:0000::/36  Last subnetID  [Decimal]: a 16 = 4096(16-1) = 61440 (0xf000)  [Hex]: a 10 = 1000(10-1) = 1000(f) = 0xf000  Last subnet: 2001:db8:f000::/36  Verify your answer using subnet tools  e.g. sipcalc 2001:db8::/32 –v6split=36 Subnetting example: analysis

17. sipcalc 2001:db8::/32 –v6split=36 | grep Network Network- 2001:0db8:0000:0000:0000:0000:0000:0000 - Network- 2001:0db8: 1 000:0000:0000:0000:0000:0000 - Network- 2001:0db8: 2 000:0000:0000:0000:0000:0000 - Network- 2001:0db8: 3 000:0000:0000:0000:0000:0000 - Network- 2001:0db8: 4 000:0000:0000:0000:0000:0000 - Network- 2001:0db8: 5 000:0000:0000:0000:0000:0000 - Network- 2001:0db8: 6 000:0000:0000:0000:0000:0000 - Network- 2001:0db8: 7 000:0000:0000:0000:0000:0000 - Network- 2001:0db8: 8 000:0000:0000:0000:0000:0000 - Network- 2001:0db8: 9 000:0000:0000:0000:0000:0000 - Network- 2001:0db8: a 000:0000:0000:0000:0000:0000 - Network- 2001:0db8: b 000:0000:0000:0000:0000:0000 - Network- 2001:0db8: c 000:0000:0000:0000:0000:0000 - Network- 2001:0db8: d 000:0000:0000:0000:0000:0000 - Network- 2001:0db8: e 000:0000:0000:0000:0000:0000 - Network- 2001:0db8: f 000:0000:0000:0000:0000:0000 - Subnetting – subnets using sipcalc

18.  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

19. Global IPv6 address hierarchy

20. ASN Region #1 Site #1 Site #2 Site #n Region #2 Site #1 Site #2 Site #n Region #n Site #1 Site #2 Site #n Conceptual view of an ISP network

21.  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

22.  Try to align allocation units to nibble boundaries  Round up your estimates to 2 n 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 planning  Provides room for expansion at each level of the network hierarchy About nibble boundaries

23.  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:ffff  Easy 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

24. “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

25.  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

26.  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

27.  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

28.  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

29.  /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

30.  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

31.  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

32. Questions 32