Router# show ip eigrp topology
EIGRP-IPv4 Topology Table for AS(1)/ID(192.168.80.28)
via 192.168.80.28 (46251776/46226176), Ethernet0/0
via 192.168.81.28 (46251776/46226176), Ethernet0/1
via 192.168.80.28 (307200/281600), Ethernet0/1
via 192.168.81.28 (307200/281600), Ethernet0/0
via 192.168.80.31 (332800/307200), Serial 0
Please advise on the answer and how do you get the answer.
Thank you in advanced.
the possible answer depends on other details that you haven't provided.
However, a possible answer is 2 or more , that is the number of LAN interfaces that appears as outgoing interfaces/next-hops.
The serial interface is a separate broadcast domain / IP subnet but it is point to point link.
Nothing can be said about the remote IP subnets we can guess that IP subnet 192.168.90.0/24 belongs to a LAN interface of another router.
This can be counted as the third broadcast domain.
To be noted 192.168.81.0/24 is the IP subnet of connected interface 0/1
So I would choice 3 as the possible answer the two local connected LAN interfaces + remote IP subnet 192.168.90.0/24 (but it could be associated to a loopback or to a serial interface so there is no 100% security it is a broadcast domain).
If the question wants to enforce the concept of routers as broadcast firewall, each interface is a separate broadcast domain so the answer could be 4 including the serial interface even if sending broadcast packets over a point to point link is not meaningful.
Hope to help
Thank you for your reply.
In fact the correct answer provided to me is 3 broadcast domains.
As what you mentioned, I suppose 3 broadcast domains will have 3 seperate subnets.
In this case the 3 subnets are 192.168.80.0; 192.168.81.0; 192.168.90.0.
Will you agree with me?
I agree that the practical answer is 3 and that is equivalent to the number of LAN segments = broadcast domains
As I noted in a test, the answer could have been 4 if we equate a broadcast domain = IP subnet (including also the serial interface in the count)
Hope to help
I'm using the LAN concept to define the 3 broadcast domains for this EIGRP example. I'm unsure why do you include the serial interface (192.168.80.31) as one of the broadcast domain, as I see this interface in the same subnet as 192.168.80.28. Kindly seek your explaination.
I didn't notice that.
the example is wrong, if you try to use an IP address overlapping with another interface the router should refuse it unless both interfaces are serial interfaces (known fact that allows to configure a second serial interface on branch router to be used in case the first has problems just having someone moving the serial cable to the second interface, this is handy).
if you attempt to configure two LAN interfaces and one serial interfaces like in the example the router should complain if on eth0 we use a 192.168.80.0/24 subnet.
Hope to help
Thank you for your clarification.
You could be right as the Serial interface and ethernet interface should not belong to the same subnet.
I guess it may be a typo to make the Serial interface as its own subnet. If this is the case, I suppose then the correct answer will be 3 broadcast domains.