Can you tell me how many broadcast domains exist for this EIGRP Topology Table?

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May 30th, 2010
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Router# show ip eigrp topology

EIGRP-IPv4 Topology Table for AS(1)/ID(192.168.80.28)
 
Codes: P - Passive, A - Active, U - Update, Q - Query, R - Reply,
       r - Reply status, s - sia status
P 192.168.90.0 255.255.255.0, 2 successors, FD is 0
        via 192.168.80.28 (46251776/46226176), Ethernet0/0
        via 192.168.81.28 (46251776/46226176), Ethernet0/1
        via 192.168.80.31 (46277376/46226176), Serial 0
P 192.168.81.0/24 255.255.255.0, 1 successors, FD is 0is 307200
        via Connected ethernet1
        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.

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Giuseppe Larosa Sun, 05/30/2010 - 05:03
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Hello Andy,

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

Giuseppe

macky1313 Mon, 05/31/2010 - 02:00
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Hi Giuseppe,


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?


Andy

Giuseppe Larosa Mon, 05/31/2010 - 02:16
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Hello Andy,

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

Giuseppe

macky1313 Mon, 05/31/2010 - 07:48
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Hi Giuseppe,


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.

Thanks!

Giuseppe Larosa Mon, 05/31/2010 - 13:08
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Hello Andy,

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

Giuseppe

macky1313 Mon, 05/31/2010 - 21:56
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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.

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