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Community Member

EIGRP feasible successors

Hello, i've got three questions concering the

"show ip eigrp topology" command.

Sample Output:

--------------

P 172.16.90.0 255.255.255.0, 2 successors, FD is 0

via 172.16.80.28 (46251776/46226176), Ethernet0

via 172.16.81.28 (46251776/46226176), Ethernet1

via 172.16.80.31 (46277376/46251776), Serial0

Question 1:

-----------

Does this mean, that the first routes (.80.28 and .81.28)

are the successors, and the last (.80.31) is

the feasible successor??

Question 2:

-----------

Why is (in the first line) the FD = 0.

I think it should be 46251776??

Shouldn't it?

Question 3:

-----------

The feasible condition states, that the AD must be lower than the FD.

Is this true, or can it be equal to the FD of the successor?

Because Route No. 3, the AD=FD (of the successor)

Thanks for your efforts!

5 REPLIES
Community Member

Re: EIGRP feasible successors

An FD of 0 is not possible. 46251776 is the FD

The AD must, must be less than the FD. Equal doesn't work.

NS

Community Member

Re: EIGRP feasible successors

Thank you.

But what does the first line

"2 successors, FD is 0" mean?

Why "FD is 0" ??

When you state "The AD must, must be less than the FD. Equal doesn't work.",

what type is the third route??

Isn't it a feasible successor?

This output is directly from

http://www.cisco.com/en/US/docs/ios/12_3/iproute/command/reference/ip2_s2g.html#wp1042641

This says, that the "show ip eigrp topology"

shows only routes that are feasible successors .

Community Member

Re: EIGRP feasible successors

FD is 0 is wrong.

And no, the third route is not a feasible successor, because its advertised distance is not LESS THAN the current best metric to the network (the FD, which isn't 0).

NS

Community Member

Re: EIGRP feasible successors

Thank for your efforts.

It is very important for be, because i am going to take the BSCI exam for CCNP.

OK. So am I finally right in thinking, that the "Fd is 0" is an IOS Bug? (I have seen it more often than just in this cisco document)

So i conclude, that this route does only have 2 successors.

But why is the third line displayed, although it is not a feasible successor?

quotation from cisco.com

"The show ip eigrp topology command can be used without any keywords or arguments. If this command is used without any keywords or arguments, then only routes that are feasible successors are displayed."

OK. Another Output from cisco

-----------------------------

P 172.16.81.0 255.255.255.0, 1 successors, FD is 307200

via Connected, Ethernet1

via 172.16.81.28 (307200/281600), Ethernet1

via 172.16.80.28 (307200/281600), Ethernet0

via 172.16.80.31 (332800/307200), Serial0

So in this example, which is my successor (first line indicates that there is only 1) ?

via 172.16.81.28 or via via 172.16.80.28

or is the router load-balancing?

I asumme the .80.31 route is not a successor, but why displayed?????

Community Member

Re: EIGRP feasible successors

I sincerely doubt its an IOS bug, but likely a myth that has been propagated around as a result of this exam. I guess it could be an IOS bug from back in the day.

That second example isn't much better.

You are the successor for that route. You are connected to it via Eth1. You will never use the path via 81.28, cause it is out Eth1. Eth0 would be your feasible successor. Serial0 gets left out.

These aren't great examples.

NS

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