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Adding/Removing Node

Here is my scenario:

We have an application deployed on 2 nodes (1 server in each node)
We need to add another node to distribute loading, but without disrupting the existing servers (the service cannot be disrupted).

To achieve this, we follow,
  • Update the ice-deploy.xml with the new node
  • application update ice-deploy.xml (this creates the server in the new node - but it is still disabled)
  • server patch <servername>
  • server enable <servername>

Do we have other/better approaches ?

Thanks,
Jeeva

Comments

  • Removing Node

    Here is my scenario:

    We have an application deployed on 3 nodes (1 server in each node)
    We need to remove a node (to shift to a new machine), but without disrupting the existing other 2 servers (the service cannot be disrupted).

    To achieve this, we follow,
    • server disable <server-to-be-removed>
    • server stop <server-to-be-removed>
    • Update the ice-deploy.xml by removing the node, which has the <server-to-be-removed>
    • application update ice-deploy.xml (this removes the server)
    Do we have other/better approaches ?

    Thanks,
    Jeeva
  • benoit
    benoit Rennes, France
    Hi,

    When you update the application to add a server, the added server is enabled by default so if the server activation mode is on demand it might get started right away before the patch. You should instead change the server activation mode to manual in order to prevent the server from being activated. You can eventually update the application again after the patch to change the server activation mode back to on-demand.

    The steps to remove a server are fine.

    Other options to add/remove servers are to use the IceGrid administrative GUI or write a small IcePy script that directly talks to the IceGrid administrative interface programatically.

    Cheers,
    Benoit.
  • Adding Node without disrupting existing server

    Changing the server activation mode to “manual” brings down the existing nodes. So, if disruption is not desirable, leave it as “always”