2 Replies Latest reply on Oct 8, 2014 11:44 AM by levi

    3448 Asymmetric WAN load-balance failover

    abcsupport New Member

      I've read through the "Configuring IP Load Sharing" and "Configuring WAN Failover with Network Monitor" documents as well as various posts in this forum, but I've not found a good solution for my needs.  I have a single 3448 with dual ISPs.  One ISP has 50/5 bandwidth (cable modem) and the other has 12/1 (DSL).  I would like to know the best approach to load-balance across both ISPs if possible.  I believe the standard per-destination round-robin load-balance will not work as it will fill up the 12/1 pipe and the 50/5 pipe will sit underutilized.  This is a one-way load-balance between the inside NAT'd clients and the Internet.  There are no other destinations or AOS devices in the mix.  There is no need to load-balance incoming WAN-based traffic at this time.

       

      Is there a way to alter the "bandwidth" or "speed" parameters of the two 100BaseT interfaces that would properly allow load-balance to function, WAN failover with Network Monitor to function and both WAN links to maximize their bandwidth?

       

      Thank you.

        • Re: 3448 Asymmetric WAN load-balance failover
          Employee

          abcsupport - Thanks for posting your question on the forum!

           

          Unfortunately, at the time of this post, there is no way for AOS devices to load balance based on bandwidth. As you probably read in the AOS load sharing guide, Configuring IP Load Sharing in AOS - Quick Configuration Guide, load-sharing is done based on connections as opposed to bandwidth. Specifically, it can load share based on a per-packet basis or per-destination/per-session basis.

           

          Please do not hesitate to let us know if you have any further questions.

           

          Thanks,

          Noor

          • Re: 3448 Asymmetric WAN load-balance failover
            levi Employee

            abcsupport:

             

            I went ahead and flagged the "Correct Answer" on this post to make it more visible and help other members of the community find solutions more easily. If you don't feel like the answer I marked was correct, feel free to come back to this post to unmark it and select another in its place with the applicable buttons.  If you have any additional information on this that others may benefit from, please come back to this post to provide an update.  If you still need assistance, we would be more than happy to continue working with you on this - just let us know in a reply.

             

            Thanks,

             

            Levi