[c-nsp] Load Balancing of Unequal Ethernet Bandwidth

Ben Steele illcritikz at gmail.com
Mon Feb 16 01:32:29 EST 2009


Woops meant to reply all in case someone else wants to chime in.

On Mon, Feb 16, 2009 at 4:59 PM, Ben Steele <illcritikz at gmail.com> wrote:

> You could do this with variance in eigrp, just add variance 2 into the
> eigrp config and it will load balance on a 2:1 ratio, if your links are
> equally matched in terms of latency you can look at enabling per-packet load
> sharing on the 2 egress interfaces to get an even more granular
> distribution, this can wreck some havoc with unequal paths and out of
> sequence packets though, however if equally similar in characteristics then
> performance is usually very good.
> Ben
>
>
> On Mon, Feb 16, 2009 at 4:01 PM, Andy Saykao <
> andy.saykao at staff.netspace.net.au> wrote:
>
>>  Is it possible to aggregate and then load balance unequal ethernet
>> circuits like so:
>>
>> I have two ethenet circuits on my Cisco router. Both have equal costs to
>> the next hop.
>>
>> Ethernet Circuit #1- 200M
>> Ethernet Circuit #2 - 100M
>>
>> Can I aggregate both ethernet circuits so that the total amount of
>> bandwidth available to the next hop is is 300M?
>> Can I then load balance it so both circuits are equally utilized?
>>
>> For example...
>>
>> * If I have 150M of traffic flowing to the next hop then the router
>> would spread the load across both links like so:
>>
>> 100M through Ethernet Circuit #1.
>> 50M through Ethernet Circuit #2.
>>
>> * The formula to use for this would be something like:
>>
>> Utilization / Total Bandwidth = percentage of utilization required per
>> link
>> 150/300 = 0.5
>>
>> 0.5 x bandwidth of Ethernet #1 = 0.5 x 200 = 100M
>> 0.5 x bandwidth of Ethernet #1 = 0.5 x 100 = 50M
>>
>> * If there was a total of 250M of traffic flowing to the next hop, and
>> applying the formula above, the router would work out that the load
>> distributed across both ethernet links would be:
>>
>> 166M through Ethernet Circuit #1.
>> 84M through Ethernet Circuit #2.
>>
>> Any ideas???
>>
>> Thanks.
>>
>> Andy
>>
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>
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