Thoughts on IPv6 /64 scanning and NDP cache size

December 24, 2013

Today I was talking with some friends about the possibility to make a DOS attack against an IPv6 router/switch if I was in the same /64 subnet by simply sending IPv6 NDP Packets to fill the neighbour cache on the router. But the question I was thinking than about was how many packets can I send e.g. over an 1Gbit link per second? How many entries will the neighbour cache need to hold if the timeout is e.g. set to 120 sec? How long would it take to scan the whole /64? So I sat down and looked at the questions.

How man packets can I send in one direction send over an 1Gbit Ethernet link?

The amount of packets which can be sent over a link depends on the size of the packets. The smallest ones used for calculation are 64byte in the IP world.  We need to put that into a Ethernet frame which adds up to 84 octets Details can be found here. Which leads to following formula:

1000MbitPerSec / 8 Bits  / 84 OctetsPerFrame= 1.488.095 FramesPerSec

As only one packet can be in a frame we can send 1.488.095 packets per second (often called: pps), which is also often called line speed or wire speed. The calculation is true for pure Ethernet, but I changes if you use VLAN Tags, QinQ or MPLS … in these cases take a look at this article.

How many entries will the neighbour cache need to hold if the timeout is e.g. 120 sec?

So now we know how many packets a can send at most and forget that we need some additional bytes for the NDP, which makes it easy to set the limit for the neigbour cache of our router.

1.488.095 PacketsPerSecond * 120 SecondsTimeout = 178571400 entries = 178  Million Entries

Lets say that this is only a RAM problem and everything else would work. Each entry contains a least the IP address and the MAC address. (There would be an optimization possible in only to store the host part of the IP address). An IPv6 address has 128bit = 16byte and the MAC address has 48bit = 6byte which leads to a total of 22byte per entry. A router needs 3,6Gbyte of RAM to store this table … not impossible but not common also. 😉

How long would it take to scan the whole /64?

And as bonus question we talk on how long it would take to scan that many IP addresses. First we need to get the amount of IP addresses a /64 can hold.

2^64 = 18.446.744.073.709.551.616 = 1,844674407×10¹⁹ IP Addresses

We know that we can scan 1.488.095 IP addresses per second which leads to

1,844674407×10¹⁹IPaddresses / 1488095 packetsPerSec / 60/ 60 / 24 / 365 = 393081 years

Ok not practical. But wait … we need only to scan for /48 IP addresses as the host part is derived from the MAC … this makes only 2,814749767×10¹⁴ IP addresses

2,814749767×10¹⁴IPaddreses / 1488095 packetsPerSec / 60/ 60 / 24 / 365 = 6 years

Much smaller but still too long for my spare time. 😉


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  1. Nice article! And merry xmas 🙂

    Comment by Bernd — December 24, 2013 #

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