Hi Joe,
          Thank you very much fro your guidance and help.
I would want to ask some more things.
I am completely aggree with the information you have given.
As per my current requirement the need is as follows:

    B        C
     \        /
       \    /
         A
        /   \
      /       \
    D         E
 

Node A will be Master node and all other nodes are remote nodes. The requirement is that from all remote nodes (B,C,D,E)  data should be replicated to the master node (A). In this case master node A is not going to forward the data to other remote nodes.
      So in above case the data propagation paths will be from remote nodes to master node.
        With the numbers to nte nodes A to E as 10,20,30,40,50
        i.e.
       Original     Local     Target
        20             20           10
        30             30           10
        40             40           10
        50             50           10

If I defined the paths as above what is happening is that at each remote node end,records are getting transfered in to the input queue but they are not getting processed further as if all the nodes are getting deadlock.
        But if I add atleast one path in the  reverse way. i.e. from master node to remote node
     i.e.
        Original     Local     Target
        20             20           10
        10             10           20         //extra added path from master to remote
        30             30           10
        40             40           10
        50             50           10
    with above paths replication setup works fine and data from remote nodes gets transfered to master nodes.
   
    So is there any way that will allow me to avoid  this extra path from master to remote node.

Thanks

-Ajay







On Mon, 2005-08-29 at 18:49, Abbate, Joseph M wrote:
Hi Ajay,

You wrote ...

> For both the scenarios all other remote nodes the records are
transfered to
> input queue but remains there only and not getting processed further. 
> 
> So if any one can give me solution for this problem..... 
 
Even though the master will only be the target of replication, it may
the conduit through which other nodes are maintained in sync.  There are
two basic scenarios which I'll try to depict graphically first:
 

     B   C      B---C
      \ /       |\ /|
       A        | A |
      / \       |/ \|
     D   E      D---E
 
The first approach may be called a "star" configuration, while the
second may be called a "mesh".  In the star, an update made in B is
propagated to the other four nodes via the master, so B's update travels
from B to A and then from A to C, D and E.  In the mesh, B's update is
copied to A but it does nothing with it.  B's update is also sent
directly to C and D and, depending on actual network topology, desired
redundancy, etc., can be sent to E directly or via one of the other
three nodes (A, C or D).  In the star diagram, you can visualize that
the lines connecting A to the other nodes have arrows at both ends.  In
the mesh, the lines around the perimeter have arrows at both ends, but
the lines going to A only have arrows in that direction.
 
To replicate in either scenario you have to ensure that your paths are
fully specified.  To achieve this, imagine yourself in each node and
ensure that an update reaches all of the desired targets.  For example,
let's number the nodes A through E as 10, 20, 30, 40 and 50,
respectively.  If an update occurs at B, it has to reach the other four
nodes, so the paths for a star configuration are
 
  Orig   Local   Target
   20      20      10       Copy to the master
   20      10      30       Master sends on to other three
   20      10      40
   20      10      50
 
For the mesh, the paths could be (assuming propagation to E occurs via
C)
 
  Orig   Local   Target
   20      20      10       Copy to the master
   20      20      30       To C
   20      20      40       To D
   20      30      50       To E via C
 
For completeness, it is recommended that you add paths from A to the
outer nodes even if they will be seldom or never used.  So you should
end up with 20 paths (or n * (n - 1) in the general case without
redundancy).
 

Joe Abbate 
Senior Software Engineer 
Computer Associates 
[EMAIL PROTECTED]
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