How L2/L3 Etherchannel does Load-Balancing

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How L2/L3 Etherchannel does Load-Balancing

Considering the diagram above ,  assuming that there are many PCs in 192.168.1.0/24 network and many PCs in 192.168.2.0/24 network
What is the best method to choose for load-balancing in this case (Source-Destination IP, Source Destination Mac,etc....) ?

another case is , Let's say in network 192.168.2.0/24 there are separate servers  that procide services to  PCs in the network 192.168.1.0/24
in this case what would be the best Load-Balancing Method ?

Thank you
jskfanAsked:
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atlas_shudderedSr. Network EngineerCommented:
In both cases IP to IP would be most appropriate as your MAC addresses are always going to be the MACs of the routers jsk.  If the EC connections where switch to switch to host then it would be best to use MAC to MAC.
mikecrIT Architect/Technology Delivery ManagerCommented:
I agree with atlas_shuddered, normally you'll always want to do ip to ip unless everything is on a local layer 2/3 network and won't talk over a router.
jskfanAuthor Commented:
OK ...if we have eliminated the Routers , then Load -Balancing on Mac address would be the best method ?
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atlas_shudderedSr. Network EngineerCommented:
Yes, if the routers were not part of the diagram, by MAC would be your best method.

Like noted above, you want to consider what the broadest scope of aggregation is going to be.  If the path is all layer 2 with no intermediary layer 3 transit points, you want to use MAC as the MAC addresses will reflect the end host MACs through the entire path.

However, if the path has a layer 3 device that must be transited, you have to account for the fact that the MAC address seen at the far end of the path is going to be that of the layer 3 transit - a single MAC.  In this case, the IP addresses will remain unique end to end for the end hosts, so use IP method.

The reason why these points are significant is due to the way the traffic will be allocated onto the port-channels.  Each unique MAC to MAC or IP to IP pair will be allocated separately onto the channel links.  If you have only one address pair, then all that traffic will be allocated onto a single physical link in the port-channel leading to a saturation event of that one physical link and underperformance of the port channel overall.

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jskfanAuthor Commented:
so, for instance, if client1 sends traffic it will go over  link1
if client2 sends traffic, it will go over different link
and so on
atlas_shudderedSr. Network EngineerCommented:
Theoretically anyway.  The algorithm will make the determination of what traffic goes across which link.  Base is round robin but it is offset by load so this can diff the traffic as individual links are loaded.
jskfanAuthor Commented:
Thank you
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