On Mon, Aug 01, 2016 at 10:02:04AM -0400, Vanderpool, Clyde wrote:
> The left side of the diagram is synchronized first, with the Ground Gateway
> assuming the Grand Master role. Then the right side of the diagram is
> brought up. On the right side link 1-A (which has no direct connection to
> the left side) and the Air Gateway will always pick link 3-A as it's parent
> boundary clock (despite the order in which they are synced to the LAN). I
> find this strange due to the fact that the path to the Grand Master needs
> to go over a radio (3-A-->3-G-->Ground Gateway) instead of going through a
> direct Ethernet connection (2-A-->2-G-->Ground Gateway).
Don't run PTP over wireless. The performance will be awful. You are
better off using NTP over wireless.
> I then tried to take link 3-G offline. This caused links 1-A, 3-A, and the
> Air Gateway to pick 2-A as their parent boundary clock. This makes sense,
> as 2-A offers the only connection to the Grand Master. When I bring back
> link 3-G, link 3-A picks it as it's parent boundary clock. This makes
> sense to me as it requires less steps to get to the Grand Master. What
> does not make sense to me is that both link 1-A and the Air Gateway drop
> link 2-A as their parent and once again sync to link 3-A despite the fact
> that the steps removed from the Grand Master remains the same.
On the right hand side of your network, the PTP spanning tree depends
on the roles decided by the BMC for the two eth0 ports on nodes 2-A
and 3-A. You can follow the BMC algorithm in the file bmc.c.
Since both nodes have the same GM, one port will take the MASTER role
and the other will take the PASSIVE role (line 152). If the paths are
the same length, then you fall through to dscmp2(). Here the port
identities (MAC address) act as a tie breaker. That means the role
assignment should always turn out the same way.
When reading the dataset comparison algorithm, it is important to
remember where the fields of the dataset come from. For the case of
clock_best and port_best, the origin of the fields is as follows.
struct dataset {
UInteger8 priority1; // GM, from the announce message
struct ClockIdentity identity; // GM, from the announce message
struct ClockQuality quality; // GM, from the announce message
UInteger8 priority2; // GM, from the announce message
UInteger16 stepsRemoved; // announce message
struct PortIdentity sender; // sourcePortIdentity (of the
announce)
struct PortIdentity receiver; // parentDS.portIdentity (upstream
port ID)
};
BTW, using vde_switch I simulated a redundant network similar to your
setup. I discover a bug. Depending on the timing of the announce
messages, the port that should become PASSIVE may instead stay in the
MASTER state. I will post a fix as soon as I have it.
(Your issue is not related to the bug, but it is rather the expect behavior.)
> I was just wondering why the two machines with no direct connection to the
> Grand Master would consistently choose the longer pathway (time wise) to
> the Grand Master. Is there something about the Best Master Clock Algorithm
> that makes this so?
The time plays no role in the BMC algorithm.
HTH,
Richard
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