Assuming that he do not know port number and must try 20 - 40 ports, it takes 200 * 10 = 2000 seconds to resert a single session... Useless except a very special cases 9such as a big community decided to knock down SCO, for example).
At 05:09 PM 20/04/2004, Richard A Steenbergen wrote:
party to know which side won the collision handling. Therefore you need 262144 packets * 3976 ephemeral ports (assuming both sides are jnpr,
again
worst case) * 2 (to figure out who was the connecter and who was the accepter) = 2084569088 packets to exhaustively search all space on this one single Juniper to Juniper session. Now, lets just for the sake of argument say that the router is capable of actively processing 10,000 packets/sec of rst (a fairly exagerated number) and still have this be considered a tcp attack instead of a straight DoS against the routing engine. This will still take 208456 seconds, or 57.9 hours. <snip> I dont understand why the large differences in claims
http://www.ietf.org/internet-drafts/draft-ietf-tcpm-tcpsecure-00.txt
says Modern operating systems normally default the RCV.WND to about 32,768 bytes. This means that a blind attacker need only guess 65,535 RST segments (2^^32/(RCV.WND*2)) in order to reset a connection. At DSL speeds this means that most connections (assuming the attacker can accurately guess both ports) can be reset in under 200 seconds (usually far less). With the rise of broadband availability and increasing available bandwidth, many Operating Systems have raised their default RCV.WND to as much as 64k, thus making these attacks even easier.
Also, with the various 'bots' at peoples disposal, why the assumption the attack would not be distributed.
---Mike