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Make priority for Skill group

Nguyen KeThuc
Level 1
Level 1

Hi all,

I am using Cisco ICM 7.2. In my script, I have 2 skillgroup ( Sk1 and Sk2). Now, I am using Queue To Skill Group node and Skill Group node for call routing to agent. I want All incoming calls are served by Sk1 before Sk2. If all agents are busy in Sk1, the calls will overflow to Sk2. Could you show me how to do this?

Thuc

Thanks,

1 Accepted Solution

Accepted Solutions

Gergely Szabo
VIP Alumni
VIP Alumni

Hi,

you can do this:

Select -> Skill Sk1

|

|

V

Select -> Skill Sk2

|

|

V

Queue to Skill Sk1

|

V

Queue to Skill Sk2

This way it's always the Sk1 skill group tried before Sk2.

And when there are no agents available in either skill group, the call will be queued to Sk1 and Sk2 and the first available agent (from any skill group) will pick them up.

G.

View solution in original post

10 Replies 10

Gergely Szabo
VIP Alumni
VIP Alumni

Hi,

you can do this:

Select -> Skill Sk1

|

|

V

Select -> Skill Sk2

|

|

V

Queue to Skill Sk1

|

V

Queue to Skill Sk2

This way it's always the Sk1 skill group tried before Sk2.

And when there are no agents available in either skill group, the call will be queued to Sk1 and Sk2 and the first available agent (from any skill group) will pick them up.

G.

Hi Gergely,

Thank you for reply. In Select node, which   agent algorism should we set?

Thuc


Hello Thuc,

We use the solution Gergely proposed. We use LAA (Longest Available Agent) to select the "right" agent.

HTH

Hi, exactly, LAA is the best choice. By the way, I think the other algorithms don't work in an UCCE environment, only ICM.

G.

Thanks everybody.

V

Queue to Skill Sk1

|

V

Queue to Skill Sk2

In this step, agents in Sk1 always  pick incoming calls before Sk2, right? Ex: I assume that  Sk1 has 2 available agents, Sk2 also has 2 available agents. If call comes in, the  first agent in Sk1 will pick it. then If second call comes in, the second agent in Sk1 will pick it. After that the third call comes in,the  first agent in Sk2 will pick it and so on. Am I correct?

Thuc

Hi,

well, yes and no. It's a bit different. You need to see that as a whole, including those select+skill group nodes, because those are actually sending the calls to agents if they're available.

So again: if an agent in Sk1 is available: Select->Skill Sk1 will send the call to him. We don't care about Sk2, we don't even get to the point to see whether there's someone available in it or not.

If there's no available agent in Sk1 but there are agents available in Sk2, then the Select->Skill Sk2 will route the calls to the agents.

If there's no agent available in Sk1 or Sk2, the Queue nodes will take care of keeping an eye of the skill groups, but of course, they'll route the call to an agent regardless of the skill group since the call is being queued to both skill groups.

G.

Hi Gergely Szabo,

So, Why do we need to separate into two queue nodes? One queue node with two rows, right?

You said that Queue node will take care Sk1 and Sk2 and route call to agent regardless of any skill groups? If I want to always select the available agents in Sk1 before Sk2 ( in Queue node step), Is it possible?

Thanks

Thuc

Hi,

again, if you are talking about available agents, they'll be taken care of by the Select -> Skill Group combination.

Queue nodes kick in only when the Select -> Skill Group combinations fail (meaning there's no available agent).

Cisco tells you to write your calls this way.

Of course, if for some reason you want to use only queue nodes, that's fine, but to be honest with you, I would not do that.

In my sketch I used two queue nodes only to make it more clear.

G.

Greetings,

Most important thing to remember is that when you route all calls through queues directly, your reporting is affected. It means that you will always have queued calls as opposed to calls which are routed to a free agent directly.

Your internal business or your customer might get upset to see they always have queued calls.

Regards,

Hi everybody,

Thank you so much for your help.

Thuc

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