Pls help - Wrong Feasible Succesors in EIGRP Topology Table

[kfarrington]

Please can someone tell me if I am going crazy.

The way I understand EIGRP to work out feasable sucessors is the following :-

Feasible distance. This value is used in the feasibility condition check. If the neighbor's reported distance (the metric after the slash) is less than the feasible distance, the feasibility condition is met and that path is a feasible successor. Once the software determines it has a feasible successor, it does not have to send a query for that destination.

Please look at the last entry in the topology table below (pls note the FD of 572672) :-

140.202.153.81 (Serial1/0), from 140.202.153.81, Send flag is 0x0

Composite metric is (3549696/3037696), Route is External

The only thing that is the same with the 3 sucessors are the last entry (I suppose it's an FS) is the external metric of 2816.

router>sh ip eigrp top 154.202.0.0 255.255.192.0

IP-EIGRP topology entry for 154.202.0.0/18

State is Passive, Query origin flag is 1, 3 Successor(s), FD is 572672

Routing Descriptor Blocks:

140.202.125.10 (Hssi4/0/0), from 140.202.125.10, Send flag is 0x0

Composite metric is (572672/19200), Route is External

Vector metric:

Minimum bandwidth is 44247 Kbit

Total delay is 20110 microseconds

Reliability is 255/255

Load is 1/255

Minimum MTU is 1500

Hop count is 2

External data:

Originating router is 154.202.200.165

AS number of route is 202

External protocol is EIGRP, external metric is 19200

Administrator tag is 202 (0x000000D4)

140.202.47.109 (Fddi5/0), from 140.202.47.109, Send flag is 0x0

Composite metric is (572672/570112), Route is External

Vector metric:

Minimum bandwidth is 44247 Kbit

Total delay is 20110 microseconds

Reliability is 255/255

Load is 12/255

Minimum MTU is 1500

Hop count is 2

External data:

Originating router is 154.202.1.165

AS number of route is 202

External protocol is EIGRP, external metric is 2816

Administrator tag is 202 (0x000000D4)

140.202.143.223 (Fddi6/0), from 140.202.143.223, Send flag is 0x0

Composite metric is (572672/570112), Route is External

Vector metric:

Minimum bandwidth is 44247 Kbit

Total delay is 20110 microseconds

Reliability is 255/255

Load is 12/255

Minimum MTU is 1500

Hop count is 2

External data:

Originating router is 154.202.1.165

AS number of route is 202

External protocol is EIGRP, external metric is 2816

Administrator tag is 202 (0x000000D4)

140.202.153.81 (Serial1/0), from 140.202.153.81, Send flag is 0x0

Composite metric is (3549696/3037696), Route is External

Vector metric:

Minimum bandwidth is 1024 Kbit

Total delay is 41010 microseconds

Reliability is 255/255

Load is 161/255

Minimum MTU is 1500

Hop count is 3

External data:

Originating router is 154.202.1.166

AS number of route is 202

External protocol is EIGRP, external metric is 2816

Administrator tag is 202 (0x000000D4)

router>

[donewald]

All three of your FS metric requrements are met. The only entry not meeting this metric requirement is 140.202.153.81. Your output from "show" command shows 3 Feasible Sucessors, which you have.

Note the following (below) topology (3)entries all report less than FD of 572672

State is Passive, Query origin flag is 1, 3 Successor(s), FD is 572672

140.202.125.10 (Hssi4/0/0), from 140.202.125.10, Send flag is 0x0

Composite metric is (572672/19200), Route is External

140.202.47.109 (Fddi5/0), from 140.202.47.109, Send flag is 0x0

Composite metric is (572672/570112), Route is External

140.202.143.223 (Fddi6/0), from 140.202.143.223, Send flag is 0x0

Composite metric is (572672/570112), Route is External

Not an FS for this route:

140.202.153.81 (Serial1/0), from 140.202.153.81, Send flag is 0x0

Composite metric is (3549696/3037696), Route is External

Hope this helps you,

Don

[kfarrington]

Hi Don, Thx for the very quick reply.

What I am not sure about is why this route is actually in the topology table as it does not pass the feasibilty condition check.

.

I would expect the router to query for this route on loss of the 3 sucessors, and the router on s1/0 to reply to the query?

Is this assumption correct?

Many thx indeed.

[donewald]

If the three feasible successors were to go away you would send queries out all interfaces that there were EIGRP neighbors. Not just s1/0.

Hope this helps you,

Don

[kfarrington]

Hi Don,

I was going from the text on the cisco web site that says

The reported distance from Router Two is 46277376, which is higher than the feasible distance - so this path is not a feasible successor. If you were to look in the topology table of Router One at this point (using show ip eigrp topology), you would only see one entry for Network A - through Router Four. (In reality there are two entries in the topology table at Router One, but only one will be a feasible successor, so the other will not be displayed in show ip eigrp topology; you can see the routes that are not feasible successors using show ip eigrp topology all-links).

http://www.cisco.com/warp/public/103/eigrp-toc.html#feasibleandreported

Just a tad confused. I am assuming that all potential routes are shown in the topology and the statement above is incorrect?

Regards,

Ken

[pbarman]

Ok, what I feel is, when you issue the command "sh ip eigrp top 154.202.0.0 255.255.192.0" the detailed information of this route is shown here including the successor, feasible successor and the other entries that are not feasible successors.

If you issue only "sh ip eigrp topology" you'll see only the successors and feasible successors, in your case the 3 successors only (the last one will be omitted). And this is what the text in CCO says.

[kfarrington]

Hi,

This is absolutley correct.

My misunderstanding of the CCO text.

Many thx for the reply :)

Ken