Re: [Tipc-discussion] Re: tipc multicast patch

[Jon M. <jon...@er...>]

See below.
Regards /jon

Ling, Xiaofeng wrote:


-----Original Message-----

From: Jon Maloy [ mailto:jon...@er...
<mailto:jon...@er...> ]=20

Sent: 2004=C4=EA4=D4=C220=C8=D5 1:09

To: Ling, Xiaofeng

Cc: Daniel McNeil; Guo, Min; Mark Haverkamp; tipc

Subject: Re: [Tipc-discussion] Re: tipc multicast patch





My comments below.

/jon



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In TIPCv1, what I understand is 2 processes on one node can not open the
same port name sequence,

This has changed in TIPCv2. Now you can bind more than one socket to the
same=20

sequence even on the same node. This may be useful if we want "weighted"
load sharing,=20

e.g. we may have 1 binding on one node, and 2 bindings on a second,
leading the=20

second node to take twice the load of the first one for this particular
function.

on two or more node, only one node will get the a message sent to this
port name, that can be treated as a load

balance. As for multicast, maybe this rule can also be applying. Of
cause,this also depends on application mode.

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No. The semantics of multicast behaviour must be absolute and
predictable. Statictical=20

load sharing applies  to unicast only.

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So, in TIPCv2, any process binding to the same instance can take the
multicast receiver role?

For example,

on node 1

A, bind(17777, 20)

B, bind(17777, 20)

no node 2

C, bind(17777, 20)

D, bind(17777, 20)



if sendto name (17777, 20, 0),=20

 only one process will receive the message.

Exactly so. And if you with the third argument mean 'domain =3D 0' you
will know that each of the four sockets will receive the same number of
calls,=20
following a round-robin schedule.




if sendto nameseq (17777, 20, 20),=20

 all the processes will received the message.

Is this description correct?

Yes.














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