1. Yes the topology is forced using "iw mpath set"

2. The 49% packet loss was when the nodes were far appart, on the setup of the 
second email they are right next to each other and I have about no packet loss.

3. Since the route are forced I don't think the path selection is changing.


I looked in the 802.11-2012 spec at section 9.32.4.2 and it seems that as long 
as a node address is in address 3 field it will accept the frame so if there is 
5 hops to reach the destination, the node could receive up to 5 time the same 
frame and will accept all of them. However section 9.32.7 says it should drop 
the duplicates but right now it doesn't.

Could someone point me to the source code where section 9.32.7 (Detection of 
duplicate MSDUs/MMPDUs) is implemented?

François Gervais
Firmware Designer

________________________________________
From: Yeoh Chun-Yeow <[email protected]>
Sent: April 20, 2015 9:45 PM
To: Francois Gervais; [email protected]
Subject: Re: Mesh duplicate (DUP!) echo reply

On Tue, Apr 21, 2015 at 5:37 AM, Francois Gervais via Devel
<[email protected]> wrote:
> I have more information on this.
>
> My mesh test setup is composed of 3 nodes like so:
>
> NodeA --> NodeB --> NodeC
>

How you force the above topology? Using "iw"?

> - All node are close enough so they see each other's packets.
> - NodeA has a forced route to reach NodeC through NodeA
> - NodeC has a forced route to reach NodeA through NodeB
>
> When NodeA sends a packet to NodeC, NodeC receives both the packet 
> NodeA-->NodeB and NodeB-->NodeC. If this is a ICMP request it will think it 
> received 2 requests and it will answer to both.

How frequency the path selection is changing? How strong the signal
strength between each nodes?

49% packet loss is too much for only 3 nodes. What 802.11 devices are
you using? Secured mesh or non secured mesh?

---
Chun-Yeow
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