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Cisco Unity Connection 8.5 Cluster License

fuhrersk8
Level 3
Level 3

                   Hello guys,

     I'm in the process of ordering an additional server for clustering a Unity Connection 8.5 implementation. The server part number is the same as the one that the client already has  MCS7825I5-K9-UCC2. My questions are about licensing:

1. Does it is necessary to buy an additional Unity Connection software (UNITYCN8-K9) with user licenses for this new server? The client does not need additional users.

2. Does it is necessary an additional license for this server in order to be configured as a cluster?

Thanks in advanced guys!

3 Replies 3

Chris Deren
Hall of Fame
Hall of Fame

Are you looking to add HA server to the cluster?

If so the license is free now and the part number is for example:

UNITYCN8-HA-VMWARE

This is for VMWare, you would need one for your hardaware.

Chris

Rob Huffman
Hall of Fame
Hall of Fame

Hi Nephtali,

The users licenses are tied to the Publisher license only so you are

good to go there (no additional user licenses )

As Chris nicely noted (+5 bud!) These are zero dollar $$ parts now I believe;

L-UNCN8-ADD-HA-PT     Unity Connection 8.x Add HA Ports xx (where xx= the required number of ports)

L-UNITYCN8-HA-PAK     Unity Connection 8.x HA PAK     1

Cheers!

Rob

"Show a little faith, there's magic in the night" - Springsteen

Thanks to both! As noted in the ordering guide, the part numbers are as follows to add HA:

L-UNITYCN8-LIC-UPG
Unity Connection 8.x, SW Feature License Add-ons
1USD 0.0021 Days
L-UNITYCN8-HA-7825
Unity Connection 8.x HA for 7825 - eDelivery
0.001USD 0.0021 Days
Included: L-UNITYCN8-HA-PAK
Unity Connection 8.x HA PAK
0.001USD 0.0021 Days
CON-ESW-LUNITYCN
ESSENTIAL SW Unity Connection 8.x, SW Feature License
1USD 0.00


My questions now are the following:

1. Does it is still necessary to order the server separately (MCS7825I5-K9-UCC2)?

2. Does it is still necessary to add the same amount of ports as Publisher or no ports are needed to add HA?

Thanks again to both?

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