Server Fault is a question and answer site for system and network administrators. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm having troubles implementing access-lists on my asa 5510 (8.2) in a way that makes sense for me.

I have one access-list for every interface i have on the device. The access-lists are added to the interface via the access-group command.

let's say I have these access-lists

access-group WAN_access_in in interface WAN
access-group INTERNAL_access_in in interface INTERNAL
access-group Production_access_in in interface PRODUCTION 

WAN has security level 0, Internal Security level 100, Production has security level 50.

What i want to do is have an easy way to poke holes from Production to Internal. This seams to be pretty easy, but then the whole notion of security levels doesn't seam to matter any more. I then can't exit out the WAN interface. I would need to add an ANY ANY access-list, which in turn opens access completely for the INTERNAL net. I could solve this by issuing explicit DENY ACEs for my internal net, but that sounds like quite the hassle.

How is this done in practice? In iptables i would use a logic of something like this. If source equals production-subnet and outgoing interface equals WAN. ACCEPT.

share|improve this question

Right, the security levels no longer matter for evaluation of ACLs - the implicit "accept traffic bound to a lower security net" is overridden when there's an access-group assigned to the interface.

You'll want something along these lines:

access-list WAN_access_in extended permit tcp host wan_allowed_host any host inside_dest_host http
access-list WAN_access_in extended deny ip wan_net inside_net
access-list WAN_access_in extended permit ip any any
share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.