Split Horizon Behaviour with EIGRP

[Wajih Rehman]

Hello,

I see a strange behavior with split horizon with EIGRP and wants some clarification.

Please see the topology ( attach picture). I am advertising 111.111.111.X to EIGRP. Router-2 Learns two route ; one from R3 and the other from R1 where it is directly connected.

As the Feasibility condition is not satisfied , one of the route through R3 is NOT a FS but it appears in show ip eigrp topology all-link command.

However, the same subnet (111.111.111.X) is being advertise to R4 as well. But it does not send this route to R2. When I disable split horizon , R4 sends this update to R2. Which makes three routes in R2 topology table. 

My question is why split horizon allows R3 to send the route to R2 but restricts R4 not to send it?

Some display from R4 and R2 are attach below:

From R2:
R2#show ip eigrp topology all-links
IP-EIGRP Topology Table for AS(100)/ID(0.0.0.2)

Codes: P - Passive, A - Active, U - Update, Q - Query, R - Reply,
r - reply Status, s - sia Status

P 1.2.1.0/24, 1 successors, FD is 28160, serno 1
via Connected, FastEthernet0/0
P 1.3.1.0/24, 2 successors, FD is 30720, serno 70
via 1.2.1.1 (30720/28160), FastEthernet0/0
via 2.3.2.3 (30720/28160), FastEthernet2/0
P 2.3.2.0/24, 1 successors, FD is 28160, serno 2
via Connected, FastEthernet2/0
P 1.4.1.0/24, 1 successors, FD is 2172416, serno 68
via 1.2.1.1 (2172416/2169856), FastEthernet0/0
via 2.3.2.3 (2174976/2172416), FastEthernet2/0
via 2.4.2.4 (20514560/20512000), FastEthernet1/0
P 2.4.2.0/24, 1 successors, FD is 28160, serno 3
via Connected, FastEthernet1/0
P 111.111.111.8/29, 1 successors, FD is 156160, serno 72
via 1.2.1.1 (156160/128256), FastEthernet0/0 --> No routes from R4
via 2.3.2.3 (158720/156160), FastEthernet2/0
P 111.111.111.0/29, 1 successors, FD is 156160, serno 69
via 1.2.1.1 (156160/128256), FastEthernet0/0 --> No routes from R4
via 2.3.2.3 (158720/156160), FastEthernet2/0
P 111.111.111.16/29, 1 successors, FD is 156160, serno 71
via 1.2.1.1 (156160/128256), FastEthernet0/0 --> No routes from R4
via 2.3.2.3 (158720/156160), FastEthernet2/0


From R4:
R4#show ip eigrp topology all-links
IP-EIGRP Topology Table for AS(100)/ID(0.0.0.4)

Codes: P - Passive, A - Active, U - Update, Q - Query, R - Reply,
r - reply Status, s - sia Status

P 1.2.1.0/24, 1 successors, FD is 30720, serno 67
via 2.4.2.2 (30720/28160), FastEthernet0/0
via 1.4.1.1 (20514560/28160), Serial3/0
P 2.3.2.0/24, 1 successors, FD is 30720, serno 68
via 2.4.2.2 (30720/28160), FastEthernet0/0
via 1.4.1.1 (20517120/30720), Serial3/0
P 1.3.1.0/24, 1 successors, FD is 33280, serno 83
via 2.4.2.2 (33280/30720), FastEthernet0/0
via 1.4.1.1 (20514560/28160), Serial3/0
P 2.4.2.0/24, 1 successors, FD is 28160, serno 65
via Connected, FastEthernet0/0
via 1.4.1.1 (20517120/30720), Serial3/0
P 1.4.1.0/24, 1 successors, FD is 20512000, serno 1
via Connected, Serial3/0
via 2.4.2.2 (2174976/2172416), FastEthernet0/0
P 111.111.111.8/29, 1 successors, FD is 158720, serno 92
via 2.4.2.2 (158720/156160), FastEthernet0/0
via 1.4.1.1 (20640000/128256), Serial3/0
P 111.111.111.0/29, 1 successors, FD is 158720, serno 90
via 2.4.2.2 (158720/156160), FastEthernet0/0
via 1.4.1.1 (20640000/128256), Serial3/0
P 111.111.111.16/29, 1 successors, FD is 158720, serno 91
via 2.4.2.2 (158720/156160), FastEthernet0/0
via 1.4.1.1 (20640000/128256), Serial3/0

[Vidyadhar Evani · Accepted solution]

Hi,

"The split horizon rule states:

• Never advertise a route out of the interface through which you learned it." 

In your case, R4 router has two possible ways to learn 111.111.111.x network.

1. Using serial link to R1

2. Ethernet link to R2.

EIGRP topology table of R4 shows both of them. Looks like R4 chose, the route learnt from Ethernet link to R2 as best route (successor) based on composite metric and installed in routing table. Since it selected best route from R2, It will not advertise the same route back to R2 (as per split horizon rule ). you can issue " show ip route 111.111.111.x" to check with route is installed in routing table at R4.

Note: Please rate the post if this is helpful !!

HTH

Cheers,

Vidy

[Peter Paluch]

Hi Vidy,

Please allow me to join.

First, I agree with your assessment of the situation. Because of the slow serial link from R4 to R1, R4 chose R2 as its next hop, so it does not advertise the route back to R2. R3 decided to use the direct link to R1 as opposed to a detour through R2, so R2 is not a next hop, and R3 happily advertises the network to R2.

I would like to point out that the definition of the Split Horizon rule as you have stated it is often quoted in many sources but it is incorrect. Split Horizon does not affect all interfaces (which is what your formulation suggests), rather, it applies only to next-hop interfaces. Think of this: A router can learn about a network possibly through all of its interfaces. If Split Horizon truly prevented a router from advertising a route through an interface through which the route was learned, the router would be unable to advertise that network to any other router in its neighborhood, and as a result, no router would be able to choose this router as the next hop. However, usually, only one interface - or a subset of interfaces - will be chosen as the next hop interface(s). Split Horizon only tries to prevent advertising a route back to its own next hop, so applying to other interfaces does not make sense - in fact, it is harmful. So in reality, the Split Horizon rule should read:

Never advertise a network out the interface used to reach that network, that is, the next-hop interface.

Would that make sense?

Best regards,
Peter

[Wajih Rehman]

Thanks Peter and Vidy for your help and wonderful explanation.