BGP Conditional Advertisement – it let the impression of a very complex task. Actually you will find it very easy to implement once you understand it.

In simple words I would say  that BGP conditional advertisement it’s a feature that let you advertise a prefix to one of your neighbors if mandatory condition is satisfied. To implement this feature you must configure at least two of the following three maps:

– Advertise-map -> this contains the selected prefixes which will be advertised if the mandatory condition is satisfied
– Exist-map -> condition to be satisfied (e.g. advertise the prefixes in the “advertise-map” only if the prefixes in this condition exist)
– Non-exist-map -> condition to be satisfied (e.g. advertise the prefixes in the “advertise-map” only if the prefixes in this condition doesn’t exist)

Let’s take an example. We have the following topology:

We have here a simple topology, with eBGP peering between R1 – R2 and R2 – R3. Each router has a Loopback interface with the following IP addresses:

R1 – L0 – 1.1.1.1 /24
R2 -L0 – 2.2.2.2 /24
R3 – L0 – 3.3.3.3 /24

These interfaces are advertised into BGP and they have full reachability:

Task

An easy task would ask you to solve something like if 1.1.1.0 /24 is NOT in the BGP table of R2, then R2 should advertise its 2.2.2.0 /24 to R3.

Solution

Let’s analyze this request a little bit. The prefix to be advertised or not, depending of the satisfaction of the condition, is 2.2.2.0 /24. According to what I said before about the necessary “maps” for BGP conditional advertisement, this prefix will go into the “advertise-map”. R2 is the only router you need to configure to accomplish this task.

OK, we have the advertisement map. What about the condition? The task request that 2.2.2.0 /24 should be advertised if 1.1.1.0 / 24 does NOT exist in the BGP table of R2. Using logical deduction we can say that “non-exist-map” is what we need to configure.

Let’s add everything together in BGP:

Verification

On R2 check the BGP table:

We have the 1.1.1.0 /24 prefix in the BGP table. According to our task, 2.2.2.0 /24 should be advertised if 1.1.1.0 /24 does NOT EXIST. By analogy, if 1.1.1.0 /24 EXIST, the 2.2.2.0 /24 should NOT be advertised. You see? It’s just a tricky words game.

We advertise only one network. The 2.2.2.0 /24 is not advertised to R3:

To check that the conditional advertisement really works, stop R1 from announcing 1.1.1.0 /24 in BGP. You can just shutdown the interface.
Check if the L0 of R1 is in the routing table / BGP table of R2:

1.1.1.0 /24 is not in the BGP table of R2, then 2.2.2.0 /24 should be advertised to R3:

On R3:

Remember that usually in the exams, the tasks related to BGP Conditional Advertisement are more word tricks than complex. As an example, base on the above topology, resolve the following:
– If 1.1.1.0 /24 and 2.2.2.0 /24 are both in the BGP table of R2, both prefixes should be advertised to R3
– If 1.1.1.0 /24 is not in the BGP table of R2, then 2.2.2.0 /24 should not be advertised to R3
– If 1.1.1.0 /24 and 2.2.2.0 /24 are not in the BGP table of R2, then none of them should be advertised to R3

If you don’t understand the trick, let me know and I will explain.


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