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why lag require same port speed ?

Deepak Sukumar
Level 1
Level 1

I am always confused why all members in the LAG must be of same bandwidth . What prevents us from aggregating an 1G and 10 G .What is reason for avoiding

5 Replies 5

Mark Malone
VIP Alumni
VIP Alumni

Hi they need to have same setup so they can be bundled into 1 logical working link , there are minimum requirements for it to work and speed and duplex are part of that , as well it would make load balancing in the hash algorithm a lot more difficult if there is multiple speed types bundled together acting as one link

Hi Mark,

Thanks for the reply. 

What does multiple speed types have to do with hashing algorithm . I am of the opinion hashing depends on the traffic ( IP, MAC ) sent and not on the speeds .

Hi

yes exactly it doesn't currently but it would have to if there are different speed links im assuming as my opinion on it or it would be a waste of 10gb to a 1gb link if they were both working at same rate  , if unequal cost is in place you would want it calculating on speed which would make it a more complicated algorithim like eigrps version

I think Mark may have touched on the most likely reason, consider the impact to the hashing algorithm.

Consider a dual Etherchannel of a FE and a 10g link.  You would want to load balance 1:100.

Thomas brings up a good point too, as a flow directed to the the FE link would be limited to 100 Mbps, while if it was directed to the 10g, it could push its traffic 100x faster.  Such variable performance, would likely not be desired.

Consider what would happen if a single TCP session was initiated to perform a large data backup out of a data center. If the LAG consisted of a single 10G interface and a second 1G interface, the TCP session could potentially be limited by the 1G link depending on the load balancing algorithm.

I doubt the feature will ever be considered.

If you want to use unequal cost links, I would look into MPLS-TE (MPLS Traffic Engineering). It may have a practical use in your situation.

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