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use 2104 WLC as a Guest Anchor with 5500 WLC

John Doe
Level 1
Level 1

Hello,

 

I want to know, even if the 2104 is EOS, if I can use it as a Guest Anchor with 5500 WLC ?

I guess it can works but not supported. Is it right ?

Do I need to run the same release for both 2104 and 5500 WLC to make it work ?

 

Thank you.

Regards,

2 Accepted Solutions

Accepted Solutions

No sir, the 2106 never supported being the anchor for a Guest WLAN. It was actually a feature added to the 2504 in 7.4 code.

 

http://www.cisco.com/c/en/us/support/docs/wireless/4400-series-wireless-lan-controllers/107458-wga-faq.html#qa4

 

HTH,

Steve

HTH,
Steve

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View solution in original post

Yes it can initiate just not be the termination

 

HTH,

Steve

HTH,
Steve

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View solution in original post

7 Replies 7

Stephen Rodriguez
Cisco Employee
Cisco Employee

So do you mean the 2106 or the 2504?

 

the 2106 is EOL/EOS

http://www.cisco.com/c/en/us/products/collateral/wireless/2100-series-wireless-lan-controllers/end_of_life_notice_c51-691053.html

 

the 2504 is still valid, and can be used as a Guest Anchor. Just be aware of the 500 user limitation, and the throughput limitation of 500Mbps per port, so only a 2G if you LAG all ports.

 

HTH,
Steve

HTH,
Steve

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Yes I was talking about the 2106, sorry.

No worries. See my previous post for the answer.

 

Thanks,

Steve

HTH,
Steve

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Ok so : 

- 2504 it works for up to 500 users.

- 2106 it's not supported because of the EOL/EOS.

 

So I have too more questions :

- Does it work with 2106 even if it is not supported for Cisco ?

- Does it need any additionnal license ?

 

Thanks you.

No sir, the 2106 never supported being the anchor for a Guest WLAN. It was actually a feature added to the 2504 in 7.4 code.

 

http://www.cisco.com/c/en/us/support/docs/wireless/4400-series-wireless-lan-controllers/107458-wga-faq.html#qa4

 

HTH,

Steve

HTH,
Steve

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Can you confirm that the 2106 can initiate tunnel to a Guest Anchor WLC ?

Thanks.

Yes it can initiate just not be the termination

 

HTH,

Steve

HTH,
Steve

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