EIGRP Questions

Would someone mind explaining question 3. Answer option D pls? There doesnt appear to be a route to 172.20.2.0/24 as far as I can see. It isnt mentioned anywhere in the topology.

TIA.

I think 172.20.20.0 /24 in the output is a typo and it should be 172.20.2.0 /24 however I doubt myself as Im still learning it all.

There is another typo in Q4. 72.20.2.0 (answer C) I think should be 172.2.2.0

Question 10 was on the ICND 2 exam yesterday.

SPORTY did you took 640-816 OR the CCNA exam?

Walter, I took the 640-816 exam.

Question 4, C option should be 172.20.20.0/24

Question 4:
I believe the Advertised Distance (AD) 205122000 should be 20512200. Otherwise it is greater/larger than the Feasible Distance (FD) 20514560, thus excluding it as a feasible successor.

I need deeper explanations in Q3 and Q4.anyone.huhu

@stephen

your totally right man its a typo… what is even more confusing to me is that when i linked to the example on the other site… the first route in that one has 2 successors yet neither of them is one of the answers… :-/

@stephen, @phalex: Yes, it is a typo. Thanks for your detection, I updated it.

Hi Guys, great site, quick question regarding Q9, if the successor route is in active status, would the router not follow answer b and forward on the packet? and then if the successor route fails, then follow answer c?

On Q 3 + 4 the output mentions 172.20.20.0 and the answers refer to 172.20.2.0 I think either the output or the answers are still incorrect or am I missing something?

Thanks

Mr. Guy , Ur right .. its either one of the two !

@ICDN Newbie In this particular scenario, there is no backup path (feasible successor) to forward data on and the successor is in the active state–thus there are no routes available. I think what’s throwing you off is that “active” here means the path is being calculated and unusable.

Thanks whizkid, it was the active status that was throwing me off.

Q4. how can C. 172.20.2.0 /24 be a feasible successors when the IP does not even exist? the question doesnt make sense..

yep, typo. should’ve been 172.20.20.0

i know what is eigrp and how it works.But i dont understand on what basis the answers are chosen.Can anyone hlep me to understand which part of the output is considered to add the routes in the RT(routing table).

I am mentioning Q3 and Q4

check out the EIGRP examples in this module…it explains it perfectly…http://www.digitaltut.com/route-eigrp-simlet

Thanks guys! I was trying to figure how they were getting the 172.20.2.0

re:q4, i thoght that the conditions for a FS is that the RD or AD has to be less than the FD, yet all the answers A,C,F have the FD as equal to the RD. Any ideas?

To meet the Feasibility condition, an alternative route’s RD has to be less than OR equal to the FD…

The feasibility condition is a sufficient condition for loop freedom in EIGRP-routed network. It is used to select the successors and feasible successors that are guaranteed to be on a loop-free route to a destination. Its simplified formulation is strikingly simple:

If, for a destination, a neighbor router advertises a distance that is STRICTLY lower than our feasible distance, then this neighbor lies on a loop-free route to this destination.

From Wikipedia, the free encyclopedia

can somebody plz explain question 8? thanks much!!

@ron
A. network 172.26.168.128 area 478
B. network 172.26.0.0
C. network 172.26.168.0 area 478
D. network 172.26.168.128 0.0.0.127

A. ospf uses “area”. wrong answer
B. eigrp is a mixture of rip and ospf: very easy to configure and very good at routing. this command enables routing for the entire class B network of 172.26.0.0/16 which includes the 2 networks that have to be advertised. correct.
C. yet again… ospf uses “area”. wrong
D. yes, eigrp can use wildcards (ospf feature), but this wildcard would advertise only subnetwork 172.26.168.128/26 which is part of 172.26.168.128/25. this would not advertise the 172.26.169.0/26 network

answer: C

@9tut – Q4 has the typo in the topology. You could fix that or change the answer in C back to 172.20.20.0/24

can anyone expain question 4 and question 9 to me. i dont understand them at all.

Don’t let Question 3 and 4 confuse you, the feasibility rule is that the AD of the feasible successor must be LESS THAN the FD of the successor.
Look at:
http://www.digitaltut.com/route-eigrp-simlet

Regarding question 4, page 394 of CCNA ICND2 Official Exam Certification Guide 2nd edition by Wendell Odom states “[…] feasibility condition, in which an alternative route’s RD is less than or equal to the FD […]”. However, on page “Introduction to EIGRP” http://www.cisco.com/en/US/tech/tk365/technologies_tech_note09186a0080093f07.shtml#feasible “the neighbors that have an advertised metric less than the current routing table metric are considered feasible successors”. Does anyone know the correct answer?

Page 394 – paragraph before example 10-4 – end of 2nd line – text “…alternative route’s RD is less than or equal to…” – delete “or equal to”. So the definitive answer is RD has to be LESS THAN FD, which means Question 4 is wrong having no possible answers.

The Cisco documentation on EIGRP states that the reported distance (RD) of the neighbor has to be strictly LESS than the current feasible distance, in order for it to qualify for the role of feasible successor.
(Source: http://www.cisco.com/en/US/tech/tk365/technologies_white_paper09186a0080094cb7.shtml#loopfree)

Furthermore, if you check their example here, you can clearly see that an RD which is EQUAL to the feasible distance is disregarded and the neighbor is not installed into the topology database as a feasible successor.

Hence, at Q4 none of the answers qualify as suitable for the feasible successor role because each backup route from the topology database has an RD EQUAL to the feasible distance of the successor for that route.

For rr64519 – route status passive=”stable route” active=”recalculating route”

re: Q4, the neighbours RD to the same subnet must LESS THAN the best calculated metric FD for that same subnet, hence i don’t think any of the answers qualify as being the correct ones.

Q4 on the exam.

@Sarpich, you can’t have ‘ show ip eigrp topology all-links’ command questions on the ICND2 exam. EIGRP on switches is the part of CCNP track — unless the question was just one of those trial questions

Question 10.
Why does a certain well known IT certification material provider picks option C as the correct answer?
One would think fewer hops would be the best route.

Passed ICND2 today with 944!! I had question #6 on my test

Hi WKC i left you a comment on OSPF1-congratulations btw

@Bernie U are right i saw the same question 10 they choose answer C xallas could pls u help us out? but wht am thinking again is that the question is saying the company is using EIGRP routing protocols and the other question does not specified any routing protocols maybe that was the reason I guess so? Xallas pls help us out

Hi all. Do you have questions on the exam to do the math to work out metrics? I’m struggling with the maths tbh :/

Sorry to be a pain but i’m also confused by this which is in Odoms book on pg 385: http://i.imgur.com/Oogaw.png

Why is the route through C chosen once the bw is set to 64. The Route straight to A from B still has the lowest metric (without going into the deep maths) 64 compared to 1544+1544 = 3088 ?

Thanks.

OK guys this is very confusing. Does the feasible successor have to have a FD less than, equal to or both to the Feasible Distance?

@kevin

condition for feasible successor route is : RD(reported distance) to a particular route must be LESS THAN( NOT EQUAL TO or GREATER THAN) the FD(feasible distance) to that particular route.

@RICK

formula for eigrp metric calculation is = ((10^7/least BW) + (cumulative delay)) * 256

that page says that, router os defaults BW in serial connection = 1544Mbps even if you attach 64Kbps cable or 2Mbps cable. so, for correct metric calculation to be done by eigrp, always configure BW of interface matching to the actual speed of link.

now, in the above figure, BW is left to default even if the A—B link is using 64Kbps of link. so the metric eigrp calculated by the above formula is

Least Bw = 1544Kbps, Delay for FE on A = 100 microsec. DElay for Serial link = 20000microsec.

NOTE: formula uses delay in Tens of microsec. i.e value of delay in formula will be 10 and 2000 for FE and serial respectively

B—A > ((10^7/1544)+(2000+10))*256 = 2172416
B–C—A > ((10^7/1544)+(2000+2000+10))*256 = 2684416

so it will choose B—A path, due to leaving default BW on interface and using less BW in actual.

in second figure, when BW is corrected on interface to 64Kbps, eigrp calculates correct metrics as :

B—A > ((10^7/64)+(2000+10))*256 = 40514560
and B—C—A > 2684416

so now it will choose high BW route B—C—A i.e correct route to pass packets through.

HOPE THAT HELPS…

Question 4 how can F. 10.69.0.0/24 be considered a feasible successor? The AD is equal to the FD of the successor 156160 – I thought it had to be less?

Thanks Anon… i think 😮

In fact, all the answers to Question 4 are wrong if the AD should be less than the FD of the successor.

The example here is correct though: http://www.digitaltut.com/route-eigrp-simlet

#confused

Just to clarify one more time

Trust me – the feasibility condition states the RD must be LESS than the FD of the successor route. Set it up in packet tracer – it doesn’t get installed into the topology table if the RD = FD

Q4 should be changed, or removed.

questions 1 & 2 were on my exam..and i also passed. Many thanks guys/gals.

Q3 & 4 Since when does a switch have Routing Tables anyway:

“Switch1# show ip eigrp topology all-links”

I hope I don’t see that on the EXAM! And who’s got time to do all these calculations?

That being said, on #4 B,D,E are eliminated because they are directly connected.

Oops, D & E are eliminated, A should be correct… B is higher cost.

very frustrated by Q3 and Q4, but i think i finally understand it and wanted to help everyone else who is stuck like i was.

Q4: answer ACF is correct. The reason why option E is not correct is because if you look closely at the command output you will recognize that both routes to 172.20.3.128/128 have an FD that is equal to the current FD(30720), aka these routes are already installed as successors (as shown in Q3) and therefore are no longer eligible as feasible successor.

hope this helps.

Q3-4: I would still like someone to explain how this output came from a switch! What am I missing?

Passed with 944. none of these were on my exam.

@9tut
plz explain q 10
in CCNA eigrp 2 section u answered to q10 answer d ,here c ?? so which one is correct ?

@kratos
look at the output given

“D 192.168.10.64/26 [90/2195456] via 192.168.10.9, 00:03:31, Serial0/0”
it will reach 192.168.10.64/26 (the subnet were interested on) via 192.168.10.9, which is part of 192.168.10.8/30, which is the subnet between R3 and R1.
answer is from R3 to R1

can any one help me about q10 i think( c) correct

@jose
look at the routing table. there is only 1 entry for the destination subnet and it is via an interface that has an IP on the subnet that connects R3 to R1.
D is correct

Passed ICND2 today with 986/1000 🙂

none of these were there

Question 4 is completely screwed as none of the answers meet the feasibility condition.

Please sort it out so no body else gets confused, thanks

With regards to question 10, wouldn’t the decision be made to go from R3 – R1 based on the fact that 192.168.10.8/30 is directly connected via Serial 0/0 and directly connected routes have an administrative distance of 1 therefore making all other routes irrelevant?

Your logic obviously works by the way. I was just wondering if the problem could be looked at from that point of view also.

@Jason, yes 192.168.10.8/30 is directly connected. What does that have to do with the 192.168.10.192/26 network?

Passed today, did not have any of these questions though. I did however, have some from the CCNA EIGRP section, so study that as well.

thanks 9tut you are great pioneer. you comprehend well in your tutorial.
thanks lots,

Q4 – as it’s written, the only correct answer is E. It’s the only one who’s Advertised Distance is less than the Feasible Distance of the successor.

@Bob I agree.

Hi Bob! My understanding for Q4 is that it asks which way to the destination includes the feasible successor route. Answer E, 172.20.3.128/25 has two routes that happen to be both successors, as indicated in the topology table. (I take it’s after configuring unequal-cost load balancing.) Therefore no need for a feasible successor route (backup route) for this case.

But I agree with you that from the way Q4 is showing, NONE has conditions to become feasible successors, as one of each has AD=FD!!!!! I believe that’s the case. As I further look at the Cisco official material, as FD is successor metric, it’s equal to the left number in bracket. That is, that route is the successor route. The one below it should be feasible successor (AD, the metric on the right side of bracket, should be less than FD)

The above link provided by 9tut, http://www.digitaltut.com/route-eigrp-simlet makes more sense than this Q, as feasible successor AD is indeed lower than FD!!!!!!!!!!!

A similar question like Q10 today, but of course all IP addresses are different, and one even has two successor routes.

Q10 is C not D

@PittsMan, Q10 is D, because R1 LAN can be learnt only by EIGRP as it’s not directly connected to R3. And there is exact route for it on R3 routing table, via R3 S0/0. As R1 advertises R1 LAN via EIGRP, R1 will route the traffic to its local LAN.

C is right only when there are two successor routes for R1 LAN, in R3 routing table. It’ll show that R1 LAN can be reached via R3 S0/0 and R3 S0/1.

Q4 is correct. Only one path needs to be correct for the answer to be.

Q4 the answer is C,E,F allow me to explain.

You need to compare the second value not the Feasible Distance of the possible successor route for Feasibility Condition. (FD/Next-Hop-Router Distance) so we need to ensure the Next-hop router distance is less than our existing routes FD.

172.20.2.0 with FD 30720 has a successor route with (30720/28160) 28160 is less than FD 30720 hence valid successor.
172.20.3.128 with FD 30720 has a successor route with (30720/28160) 28160 is less than 30720 hence valid successor.
10.69.0.0 with FD 156150 has a successor route with (156160/128256) 128256 is less than 156160 hence valid successor.

You do not compare the first value which is the complete FD for the route, you compare the Distance to the next hop router second value against the active routes FD. Read the information around Feasibility Condition it is a very simple rule.

Wish I could remove a comment 🙂

The only correct answer is E. All the others I listed are the actual Successor link itself 🙂

2 hrs to go, I shouldn’t be posting comments, should be recapping!

Question 4 should be removed from the site. Its all mashed up.

Passed ICND2 Today

Q10 with different IP’s so the answer is different too.

i think Q10s answer in ecam collection is different

i think Q10s answer in exam* collection is different

What is the exact condition for Feasible Successor? some say that condition is FD>RD and some say that the condition is FD=>RD so what exactly is the condition? please explain!

Alo while working in GNS3 I notice that if FD=RD then the feasible successor don’t appear in the #show ip iegrp topology table why is that so??

any one who can help me out please 🙁

hello! any one can help i’m doing for my icdn2

nobody reply to my questions above:( no problem guys 🙂 today I cleared my ICND 2 with 1000/986. Finally done with my CCNA. Thanks 9TUT 🙂

@ born2win-congrats

“To qualify as a feasible successor, a router must have an AD less than the FD of the CURRENT successor route“

Please need help
I’ll take exam tomorrow icnd2 can any body send me lates exam
On sabih.noora@yahoo

Please answer me

Q1: Admin Distance correction:
+ Static route: 1
+ RIPv1 & RIPv2: 120
+ OSPF: 110
+ IGRP: 100 (not 90)
+ EIGRP: 90
Sources: Odom CCNA ICND2 Official Cert Guide, 3rd Ed, p353; Lammle CCNA Study Guide 7th Ed. p386

Question 4: Doesn’t need to be removed just fixed before it confuses people. My suggestion is reading up how a successor route and a feasible successor works. If you know how it works and how to apply it, then you will be just fine for the test.

Q.9 on the test

Q10 today!!

Hi! On question #7, what if the EIGRP maximum metric variance is valued at 2. Will the answer be the same? And what if it is 3?

*Hi! On question #2, what if the EIGRP maximum metric variance is valued at 2. Will the answer be the same? And what if it is 1?

1 and 6 were on my test today.

Taking the INCD2 on March 4th…
Thanks for all the great input and this informative site… this site is well worth my donation.
If anyone has the latest/newest dumps please email them to me @ ( unvmm@yahoo.com).
Thanks in advance!

Very Helpful

Has anybody seen this lab in a recent exam?

Today I am going to sit for my CCNA.
I am following this site for last one month.
Lets see how much it helps in real CCNA exam.
Once I come back home after my exam, I shall inform you all.