Date: Wed, 13 Dec 2006 23:30:43 -0800 From: Michael Smith <mksmith@adhost.com> To: Dan Nelson <dnelson@allantgroup.com> Cc: "N. Harrington" <drumslayer2@yahoo.com>, freebsd-questions@freebsd.org Subject: Re: How does one bond two interfaces together to share bandwidth? Message-ID: <F0CEF52E-0ADD-4518-88D3-78404A40985A@adhost.com> In-Reply-To: <20061214011324.GF79418@dan.emsphone.com> References: <20061214010124.29818.qmail@web34502.mail.mud.yahoo.com> <20061214011324.GF79418@dan.emsphone.com>
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Hello: On Dec 13, 2006, at 5:13 PM, Dan Nelson wrote: > In the last episode (Dec 13), N. Harrington said: >> I am trying to figure out how to bond or combine 2 interfaces >> together. Such that they each share traffic. >> >> I have tried one way, however when I use it I seem to have an odd >> broadcast occuring on my switch. Such that I am seeing incoming >> traffic hit some other ports on the switch. Can someone confirm if I >> am doing it correctly? Perhaps I have a switch issue? Do I also need >> to bond the ports together on the switch? Sadly the switch they are >> connected to does not support port bonding. Does that matter? I have >> not seen any mention of that being required. > > If the remote switch doesn't support it, only outgoing traffic will be > split across both ports. Incoming traffic will probably come in on > the > first port that came up, or the switch may decide that there's a > routing loop (or other misconfiguration) because the same MAC address > is seen on both ports, and disable one of the ports (or even both). > Most managed switches should support it; they may call it trunking. Both sides need to support EtherChannel which is 802.3ad (although Cisco does have a proprietary variant (go figure)). If only one side is set to channel and the other side is not, the non-channeled side will detect a loop and set one of the ports into blocking state; that is, if it's Spanning Tree aware. If it's a consumer-grade switch or hub, the network will do the functional equivalent of a Bill the Cat face and fall over most dramatically. Regards, Mike
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