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johnp123
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Suggestions on how to balance (very) unequal WANS

Fri Sep 21, 2018 11:15 am

Have 4 WANs balanced using PCC on a RB951Ui-2HnD routerboard. Three WANs are 10/1 (down/up) Mbps and the forth is 50/50 Mbps (i.e. 10/1, 10/1, 10/1, 50/50). The system is a small WISP with roughly 75 internet users (connections mainly between 2.5 and 5Mbps)

The balancer is currently using PCC, with 5x more PCC rules on the 50/50 WAN. This setup works but obviously is not taking full advantage of upload speed on the 50/50. I am looking for guidance on a getting a better setup and a few (greatly appreciated) links that would help to learn.

Thank you -
John
 
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Re: Suggestions on how to balance (very) unequal WANS

Tue Oct 09, 2018 7:48 pm

Excellent question and one I looked at awhile ago and gave up but am willing to revisit and perhaps attempt on mine (which is basically failover based only at the moment).

Perhaps this presentation may give you some ideas.
https://mum.mikrotik.com/presentations/US12/tomas.pdf

Also the presentation here although the slides are in English they are a tad sparse at the end and of course one has to be able to speak Thai......
https://www.youtube.com/watch?v=8UKQzZzLczE

What I got out of it was, exclude https from calculations 443, exclude servers from pcc, fix mikrotik dynamic DNS whatever that meant, use public DNS servers,
Move rules over PCC, I wasn't sure if that meant, have pCC rules first? or last?

Finally, the way to load balance was to create the number of PCC style connections appropriately.
(lets say you have 3 Wans, 10MBs, 20MBs, and 30MBs (down to keep it simple) 1/6, 2/6 and 3/6 fractionwise.
then one would have 0/5 for wan1, 1/5 and 2/5 for wan2 and 3/5 4/5 and 5/5 for wan 3..........
Basically associating/tuning the number of pcc classifiers based on the number of Wans and to differentiate or account for the differing bandwidths.

What I am not sure of in my example is if order is important and if it would be better then to state
3/5 , 1/5, 4/5 , 0/5, 2/5, 5/5 (to not only divide the load proportionally in total but to spread out the classifiers to have an even distribution based on capacity.

(anyone that is willing to explain the lecture in real terms that would be grand)

Perhaps in your case 10-10-10-50=80 0/5 1/5 2/5 {3/5 4/5 5/5}
If order is important 3/5 0/5 4/5 1/5 5/5 2/5 or perhaps 3/5 4/5 5/5 0/5 1/5 2/5

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