etherchannel

[Alim670]

If in ether channel 1 bit use for 2 link and 2 bits is use for 4 links and 3 bits use for 8 links so tell me how many bits will be use for 3 ,5,6 and 7 links. i do not know how to calculate value for these links or which link will be use for forwarding frame.

please let me know if anyone knows.

[Seb Rupik]

Hi there,

Great questions! 

I had always assumed the bit bucket distribution would follow this pattern:

num of links		bit-buckets		
2			1
3			2
4			2
5			3
6			3
7			3
8			3

If you had less bit buckets than you had links then the load-balancing algorithm would be prevented from using member links. In the case where you have more buckets than links, the 'extra' bucket contents are distributed at one end of the bundle.

Number of Ports in the EtherChannel	Load Balancing
8					1:1:1:1:1:1:1:1
7					2:1:1:1:1:1:1
6					2:2:1:1:1:1
5					2:2:2:1:1
4					2:2:2:2
3					3:3:2
2					4:4

https://www.cisco.com/c/en/us/support/docs/lan-switching/etherchannel/12023-4.html

Cheers,

Seb.

[Andrew Khalil]

Hello @Alim670,

This is a pretty good question, as a lot of people get confused from the algorithm being calculated by the means of etherchannel, 

Let me first review 1 concept upon it, we can discuss your question, the concept is the load balance method you would like to choose for your traffic through the port channel,

The methods can be one of the following:

src-ip

dst-ip

src-mac 

dst-mac

and for these methods, the algorithm is going to check the most right bits (according to the number of links) to choose what link is going to be used, for example if the method used is src-ip,

and the ip address is 10.1.10.1 so the last bit is 1, and your channel consists of 2 physical links, so the 1st link is going to be used, and so on! 

but also you can use these methods:

src-dst-ip

src-dst-mac

and for these methods, the algorithm is going to compare the most right bits by the XOR operation,

for example if the method used is src-dst-ip, and the src ip is 10.1.10.1 while the dst ip is 10.1.10.2,

and you use 3 links, so the algorithim will compare the 2 most right bits,

for the src 01 while for the dst 10, 01 XOR 10 = 11, which is the most right 3rd bit, so the third link will be the one carrying the data.

Now, after understanding that concept, and according to the 1st table @Seb Rupik mentioned (which is answering your question) let's take an example and test it,

Let's say the we have 3 links, and the used method is src-ip, while the src-ip is 10.1.10.38

The bits of the most right octet will be  0 0 1 0 0 1 1 0

And considering that maximum links can be used for ethernet-channel are 8 so we are interested only in the most right 3 bits: 1 1 0

Now, let's assume that you weren't aware of table and you want manually to test what link will be used, you can consider to use all the most right 3 bits, so the answer would be = 6 (for sure it's wrong answer because you know that physically we have only 3 cables) so now, you will need to use the last 2 bits only, so the answer would be = 2, which means the 2nd link will be used! 

That's it, the algorithm behavior is logic enough to be tested manually, 

Finally, I would like to advice to use the following command to test what link would be used directly from the command line:

#test etherhannel load-balance interface port-channel 1 ip 10.1.10.38

and you have to be carefully to choose the load balancing method according to your topology as well as to be careful when you plan your ip addressing, because all are affecting what link would be chosen to carry the data!

I apologize for the prolongation, just wanted to make it clear and simple, i hope it helps,

Please don't forget to rate all helpful responses and mark solutions! 

Bst Rgds,

Andrew Khalil