>   The Ethernet Bundling Controller (EBC) performs an X-OR operation >on 
>the last two bits of the source MAC address and the destination >MAC 
>address. This operation yields one of four results: (0 0), >(0,1), (1 0), 
>or (1 1). Each of these values points to a link in the >Fast EtherChannel 
>bundle.

That really makes sense.

I always pictured this process as the first frame in a stream being sent 
down the first wire, and the second frame in the stream being down the 
second wire, third frame being sent down the first wire.. ect..

The way you explain it makes more sense because this way the switch wont 
have to think about which wire it last used to send a frame, but instead it 
does a simple calculation. So in essence, if you had 8 wires in an 
EtherChannel, but most of your traffic was going only between 2 servers  ( 
Server1  <----- Switch-----> Server2), then only 1 or 2 of the links would 
be used based on the X-OR calculation? So 6 of the wires in the bundle would 
be wasted? Is this accurate?

Also, I've never really thought about it, but is it possible to have a 
number of wires in an EtherChannel that are not a power of 2? I.E. 3,5,6,7? 
It wouldn't seem possible if each wire was assigned a value ((0,0), (0,1), 
(1,0), (1,1)). If you had 3 wires in an EtherChannel, then the 1,1 value 
would not be used, and to me, I would think that the Switch would get 
confused if all the values didn't have an assigned value.


One more thing guys! Don't let the hype about fool-duplex full you! If you 
have a full-duplex 100Mbps connection, you dont REALLY get 200Mbps 
bandwidth, you just get 100Mbps in both directions simultaneously.


Fred Danson





>From: Priscilla Oppenheimer <[EMAIL PROTECTED]>
>Reply-To: Priscilla Oppenheimer <[EMAIL PROTECTED]>
>To: "AndyD" <[EMAIL PROTECTED]>, [EMAIL PROTECTED]
>Subject: Re: Equal cost switching
>Date: Mon, 19 Feb 2001 11:21:24 -0800
>
>At 08:14 AM 2/19/01, AndyD wrote:
> >Thanks for all your help.  The way I understand it now is that with 
>multiple
> >vlans using different root bridges, you can have different vlans 
>splitting
> >the bandwidth - some going in one direction, some in the other.  But if 
>one
> >link goes down, STP will then shift all to the good link.  This gives you
> >some load balancing and also redundancy.
>
>Yes, and thanks to the people who provided the details. This is often a
>good solution. It has been shown to scale to even very large switched 
>networks.
>
> >It looks like you need to go to
> >layer 3 switching to do any load balancing other than this.  And
> >etherchannel is another option for aggregating bandwidth.  But someone 
>said
> >with etherchannel using 4 full duplex 100 mbp ports will not give 800 
>mbps
> >of throughput?  I always thought that in theory that was the case??
>
>It's "statistical" load balancing, according to Cisco. The operation that
>determines which link in a Fast EtherChannel to use is quite bizarre, and
>does not provide precise load balancing. It provides load sharing. Think of
>it like a complex highway system. Adding new highways distributes the load,
>but it doesn't usually balance the load very precisely.
>
>The division of traffic across a Fast EtherChannel is based on
>source/destination pairs, which is usually not very balanced. There are
>usually some big talkers and receivers. The Ethernet Bundling Controller
>(EBC) performs an X-OR operation on the last two bits of the source MAC
>address and the destination MAC address. This operation yields one of four
>results: (0 0), (0 1), (1 0), or (1 1). Each of these values points to a
>link in the Fast EtherChannel bundle.
>
>Priscilla
>
> >   Since
> >the data is transmitted on different wire pairs, if the sender and 
>receiver
> >transmit at the same time, why isn't 800 mbps possible????
> >
> >Thanks again !!
> >
> >
> >
> >""Peter Van Oene"" <[EMAIL PROTECTED]> wrote in message
> >[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> > > Per my other post, STP prevents looping traffic in general, not simply
> >broadcasts.
> > >
> > > Pete
> > >
> > >
> > > *********** REPLY SEPARATOR  ***********
> > >
> > > On 2/19/2001 at 6:50 AM Kenneth wrote:
> > >
> > > >Jason is right. This will defeat the purpose of Spanning Tree of 
>creating
> >a
> > > >single path to a destination. The primary reason this was designed 
>was to
> > > >prevent broadcast loops.
> > > >
> > > >If you want to force it to use 2 paths to one destination, use
> > > >port-channelling which statically load-balances traffic going out of 
>two
> > > >ports. Statically meaning it creates a list of source-destination MAC
> > > >address pairs and these pair will communicate from a specific port
> > > >configured to be part of the port-channel. This is in contrary to 
>Dynamic
> > > >load-balancing where each packet will go out of each port of the
> > > >port-channel.
> > > >
> > > >With this in mind, if 4 ports are configured for 100 Mbps full-duplex
> > > >port-channels, this doesn't mean it provides an 800Mbps link.
> > > >
> > > >
> > > >
> > > >
> > > >"AndyD" <[EMAIL PROTECTED]> wrote in message
> > > >96p2uk$rt5$[EMAIL PROTECTED]">news:96p2uk$rt5$[EMAIL PROTECTED]...
> > > >> Spanning tree is supposed to choose the one best switched path.  
>But if
> > > >you
> > > >> set up two equal cost paths, will it use both?  Is there a way to 
>force
> >it
> > > >> to use the bandwidth from both paths?
> > > >>
> > > >> Thanks!
> > > >>
> > > >>
> > > >> _________________________________
> > > >> FAQ, list archives, and subscription info:
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>[EMAIL PROTECTED]
> > > >>
> > > >
> > > >
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> >
> >
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>
>________________________
>
>Priscilla Oppenheimer
>http://www.priscilla.com
>
>_________________________________
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