EIGRP Topology Table Question

If anyone has got 5 minutes could you answer this question and explain how you reached your answer. This question is really puzzling me and I can't remember learning anything in the Academy which would help me achieve this answer... I probably have forgotten it!

I'm really struggling with understanding this big time. Cheers guys.

Based on the topology table that is shown in the exhibit and assuming that variance is not configured for EIGRP, which route or routes should appear in the routing table?

Router#show ip eigrp topology
IP-EIGRP Topology Table for AS (100)/ID(192.168.8.21)

Codes: P- Passive, A – Active, U – update, Q – Query, R- Reply,
R – reply status, s – sia status

P 192.168.2.0/24, 1 successors, FD is 2707456
Via 192.168.8.22 (2707456/2195456), Serial0/0
Via 192.168.8.18 (3815424/281600), Serial0/2
P 192.168.8.20/30, 1 successors, FD is 2169856
Via Connected, Serial0/0
P 192.168.8.16 /30, 1 successors, FD is 3789824
Via Connected, Serial0/2
P 192.168.8.24/30, 1 successors, FD is 2681856
Via 192.168.8.22 (2681856/2169856), Serial0/0
Via 192.168.8.18 (4301824/2169856), Serial0/2

A. D 192.168.8.20 (2707456/2195456), Serial0/1
B. D 192.168.2.0/24 [90/2707456] via 192.168.8.22, 00:27:50, Serial0/0
[90/3815424] via 192.168.8.18, 00:27:50, Serial0/2
C. D 192.168.2.0/24 [90/3815424] via 192.168.8.18, 00:27:50, Serial0/2
D. D 192.168.8.24/30 [90/2681856] via 192.168.8.22, 00:27:50, Serial0/0
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Comments

mattipler

No takers!?!?!?! It is a tricky one... got me completely baffled.

dtlokee

mattipler wrote:

If anyone has got 5 minutes could you answer this question and explain how you reached your answer. This question is really puzzling me and I can't remember learning anything in the Academy which would help me achieve this answer... I probably have forgotten it! I'm really struggling with understanding this big time. Cheers guys.

Based on the topology table that is shown in the exhibit and assuming that variance is not configured for EIGRP, which route or routes should appear in the routing table?

Router#show ip eigrp topology
IP-EIGRP Topology Table for AS (100)/ID(192.168.8.21)

Codes: P- Passive, A – Active, U – update, Q – Query, R- Reply,
R – reply status, s – sia status

P 192.168.2.0/24, 1 successors, FD is 2707456
Via 192.168.8.22 (2707456/2195456), Serial0/0
Via 192.168.8.18 (3815424/281600), Serial0/2
P 192.168.8.20/30, 1 successors, FD is 2169856
Via Connected, Serial0/0
P 192.168.8.16 /30, 1 successors, FD is 3789824
Via Connected, Serial0/2
P 192.168.8.24/30, 1 successors, FD is 2681856
Via 192.168.8.22 (2681856/2169856), Serial0/0
Via 192.168.8.18 (4301824/2169856), Serial0/2

A. D 192.168.8.20 (2707456/2195456), Serial0/1
B. D 192.168.2.0/24 [90/2707456] via 192.168.8.22, 00:27:50, Serial0/0
[90/3815424] via 192.168.8.18, 00:27:50, Serial0/2
C. D 192.168.2.0/24 [90/3815424] via 192.168.8.18, 00:27:50, Serial0/2
D. D 192.168.8.24/30 [90/2681856] via 192.168.8.22, 00:27:50, Serial0/0
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Looks like "D" to me.

The topology table for EIGRP lists all the routes and the total number of sucessors and feasiable successors the router knows about. What you're looking for is the:

"P 192.168.8.24/30, 1 successor, FD is 2681856"

This line is a route entry, it tells us:
P = Passive, meaning the router is not looking for a new successor (this is what it should be)
192.168.100.0/24 is the prefix
1 successor = the number of best routes (if there are 2 it would say "2 successors")
FD = the feasiable distance which is used to see if a newly learned route meets the feasiablity condition

Via 192.168.8.22 (2681856/2169856), Serial0/0
Via 192.168.8.18 (4301824/2169856), Serial0/2

what we see here is the 2 possible routes. The first one has the lowest FD. Now the 2 numbers after the route are the (Fesiable Distance/Advertised distance). In this case the 2 routes have the same advertised distance, but the FD is higher on the second route so it is not the best route. Serial0/0 indicates the exit interface.

mattipler

Cheers for replying DTLOKEE. Really appreciate it mate as I've been banging my head against walls trying to get my head around this one.

So you saying in this type of question look for the lowest feasible distance...

A. D 192.168.8.20 (2707456/2195456), Serial0/1
B. D 192.168.2.0/24 [90/2707456] via 192.168.8.22, 00:27:50, Serial0/0
[90/3815424] via 192.168.8.18, 00:27:50, Serial0/2
C. D 192.168.2.0/24 [90/3815424] via 192.168.8.18, 00:27:50, Serial0/2
D. D 192.168.8.24/30 [90/2681856] via 192.168.8.22, 00:27:50, Serial0/0

Oh yeah... "D" was the correct answer.

I "kinda" understand what your saying but not 100%.

gabrielbtoledo

I think B - C - D why?
A is wrong because it learned through serial 0/0, not serial 0/1
B is ok because that route was learned through serial 0/0 in that IP
C is ok because that route was learned through serial 0/2 in that IP
D is ok because that route was learned through serial 0/0 in that IP

I hope I got ir right, otherwise I will have to revise my studies.

mattipler

There's only one answer and it's D

I sort of understand it.

Don't think it's got anything to do with the interface it was learned through but rather the feasible distance of the successor route... the lower the better. Might be wrong though.

NeonNoodle

Let me take a stab at this:

mattipler wrote:

If anyone has got 5 minutes could you answer this question and explain how you reached your answer. This question is really puzzling me and I can't remember learning anything in the Academy which would help me achieve this answer... I probably have forgotten it! I'm really struggling with understanding this big time. Cheers guys.

Based on the topology table that is shown in the exhibit and assuming that variance is not configured for EIGRP, which route or routes should appear in the routing table?

Router#show ip eigrp topology
IP-EIGRP Topology Table for AS (100)/ID(192.168.8.21)

Codes: P- Passive, A – Active, U – update, Q – Query, R- Reply,
R – reply status, s – sia status

P 192.168.2.0/24, 1 successors, FD is 2707456
Via 192.168.8.22 (2707456/2195456), Serial0/0
Via 192.168.8.18 (3815424/281600), Serial0/2
P 192.168.8.20/30, 1 successors, FD is 2169856
Via Connected, Serial0/0
P 192.168.8.16 /30, 1 successors, FD is 3789824
Via Connected, Serial0/2
P 192.168.8.24/30, 1 successors, FD is 2681856
Via 192.168.8.22 (2681856/2169856), Serial0/0
Via 192.168.8.18 (4301824/2169856), Serial0/2

A. D 192.168.8.20 (2707456/2195456), Serial0/1

wrong interface

B. D 192.168.2.0/24 [90/2707456] via 192.168.8.22, 00:27:50, Serial0/0
[90/3815424] via 192.168.8.18, 00:27:50, Serial0/2

topology table says 1 successor. also the metrics are different and variance is not being used

C. D 192.168.2.0/24 [90/3815424] via 192.168.8.18, 00:27:50, Serial0/2

metric is greater than the other possible route

D. D 192.168.8.24/30 [90/2681856] via 192.168.8.22, 00:27:50, Serial0/0

the metric is lower than the other metric and it is the only route which is indicated by the topology table and it is the correct interface so D it is

NeonNoodle

double post

dtlokee

mattipler wrote:

There's only one answer and it's D

I sort of understand it. Don't think it's got anything to do with the interface it was learned through but rather the feasible distance of the successor route... the lower the better. Might be wrong though.

yeah the FD determines what route will be put into the routing table(this becomes the metric for the route, this also assumes that the EIGRP route has the lowest administrative distance, that is there's no static routes to the same destination). If there is more than 1 route with the same AD (or the variance value is set to somthing other than 1) then multiple routes may be entered in the routing table.

for example if you had variance = 2 and:

P 192.168.8.24/30, 2 successors, FD is 2681856
Via 192.168.8.22 (2681856/2169856), Serial0/0
Via 192.168.8.18 (4301824/2169856), Serial0/2

Both routes would be in the ip routnig table because you take the lowest FD (2681856) and multiply by the variance (2 * 2681856 = 5363712). Now the 5363712 is greater than the FD of the other route (4301824) so the router would enter both in the routing table and would load share across the links in a ratio of 2681856:4301824.

rjbarlow

mattipler wrote:

P 192.168.8.24/30, 1 successors, FD is 2681856
Via 192.168.8.22 (2681856/2169856), Serial0/0
Via 192.168.8.18 (4301824/2169856), Serial0/2

D. D 192.168.8.24/30 [90/2681856] via 192.168.8.22, 00:27:50, Serial0/0

I would want to add that the FD is the distance inclusive of the local cost of the link, while the Advertised Distance (or Reported Distance), is the cost as neighbor routers advertised the route. Then, this router has a s0/2 link slower then s0/0.