EIGRP Questions 4

Is answer A not also correct in question 10?

Yes. answer A is not correct because CISCO always tries to trick everyone; the keyword in here is “broadcast.” EIGRP doesn’t use a broadcast address as RIPv1 does. It uses the default multicast address of 224.0.0.10. I also was trying to figure out why A wasn’t the right answer for a while until I realized the “broadcast” word.

Q2: as far as Router C learns route to the flapping link through the summary (or being a stub router), answer E would be correct!??

Very great job Digitaltut !!!

@Da If you think Cisco is tricking you.. you should try a Microsoft test :D

Has anyone passed the exam recently?..

#3,4, 10 were on my test

Q10 C is also tricky. the key word is multicast. Replies are unicasts.

Hello Friends,

I just started to study the ROUTE test after taking almost two years off from passing the CCNA. I am planning to take the ROUTE test in two months. Are the Sims, Drag and Drop, Questions, and everything else on this site still valid for the ROUTE test?

Thank you all!

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in question 10,why not C?

@ furqan because replies are unicast and not multicast as in question it is stated replies are multicast .. so wrong statement .. thats why C is not the option

question 5, can someone explain?

never mind. I figured it out. I was confused with AD/FD.

Q3: How do we know router C knows an alternate path to the network? Isn’t the reply sent back in any case (whether reachable or unreachable)?

OK, no queries are sent out, but what if there’s EIGRP running only on one int?

Anyone can see distance 5 in the Q1 ?

Paulo,
sh ip route 1.1.1.0 on the router that does summary

is the answer for question # 4 is corect ? cheack it pls, cos it is summarised to the class full address how ?

pyar ki pungi :p

Can some one please explain question 5.
many thanks,

In Q.5- what 2 paths are not used by eigrp routing table—its not

So AD of 40 and 45 not being used
so C and D is ans

Q5

FD of R1-R3-R6 is15 and that multiplied by variance 2 is 30
the Metric of R1-R4-R6 is 40 and R1-R5-R6 is 45
there metric is higher then the metric of R1-R3-R6 which is 30 ( due to variance 2)

R1-R5-R6 does have a AD of 5 and could become a FS which is less the FD of R1-R3-R6 which is 15 but because of It’s higher metric of 45 it cant be use to for loadbalacing

R1-R4-R6 isn’t even a FS because its AD (20) is higher then the FD of R1-R3-R6 (15)

hope this clear it up…

cheers…

great work black-cisco

Q3:
In my opinion there is an error in explanation, but the answer is correct.

You say:
We notice that the “Queries received” number is increased so router C has been asked for a route. The “Replies sent” number is also increased -> router C knows an alternate path to the network.

I say:
We notice that the “Queries received” number is increased so router C has been asked for a route. The “Queries sent” numbers stays the same (5) in all three show commands, which tells us that this router knows an alternate path and therefore doesn’t need to send Queries further.

Hope I helped.

can some one help me with GNS3, Just passed my CCNA self study , preaparing for CCNP self study, I’m not from IT byground , your help appreciated Syedameenullah@yahoo.com

Q1 – Answer E is worng I think.
We will see a summary route with an administrative distance of 5 not a default route

Hi Guys!
Who can explain quest #7…….I think right answer will be A
AD for R2=10 FD=30 or 25 (I’m not sure)
FC=AD<FD so 10<30
FC is satisfied so answer I think A….

Hi,

no I think the given answer (B) is correct.

FD is 20 (R1 – R4 – R5) = 5 + 10 + 5
AD from R2 is 25 (R2 – R3 – R5) = 10 + 10 + 5

´cause AD>FD there´s no feasible successor

JJ,

AD of R2 is 25
FD of R1 is 20
25 >20 does not meet Feasibility Condition,
thus R2 cannot be Feasible successor.

Q1 – YEs Q1 is BCF!

Q1 – according to this, C is surely wrong: https://learningnetwork.cisco.com/message/237245#237245

q4 is wrong right ans are A & B

Which summary routes does R1 generate automatically (assuming auto-summarization is enable)?

R1 has four interfaces, with IP address 172.16.1.1/24, 172.16.2.3/24,172.16.5.1/24, and 10.1.1.1/24.
By summarizing 172.16.1.1/24, 172.16.2.3/24,172.16.5.1/24 you will get 172.16.1.0/22
while the summary for 10.1.1.1 is 10.0.0.0/8
i.e for 172.16……, the difference starts occurring in the third octet
00000001
00000010
00000101 the first different bit form the left marks the summary, therefore from the /16 we add 6 bits to the point where the difference is noted to give a /22

Don’t concentrate on R2

Q4
@deno when autosummarisation is enabled I think that EIGRP process create summary routers only to classfull prefix (eg. 10.0.0.0/8, 172.16.0.0/16, 192.168.0.0/24 etc.) so that summarisation in yours example must be done manually :)

EIGRP auto summarisation is done on major networks boundary so in question 4 on router R1 are networks 172.16.1.1/24, 172.16.2.3/24, 172.16.5.1/24, 10.1.1.1/24 and on router R2 are networks 172.16.1.2/24, 192.168.1.1/24. (and there is other networks)

In general when router R1 send updates to 172.16.X.0/24 neighbors only 10.0.0.0/8 will be summarised and when network updates are sent to neighbors 10.1.1.0/24, only 172.16.0.0/16 will be summarised. (similar is with router R2)
So router R1 generate summary routes
10.0.0.0/8
172.16.0.0/16

Q1 – E would be a correct answer, but it is incorrectly worded. It should state:
By default, the ip summary-address eigrp command generates a summary route with an administrative distance of 5.

agree with TW

Question 1

Should correct answer are B, C, F. ?

hi.

could you explain me the questio 6 and 7 please?

Thanks

hi please help me out …examcollection in not accessible any more if any one have that .VCE file from wheelman please send me …thanks a lot in advance.
my ID is “Prasiji10oct@gmail.com”

Hi All ,
I’m planning to take up and exams by november 1st week , but im not able to get the full version of VCE , can any one help !
Which dumps are the Best ??
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Q10 … when looking at Answer D, I’d think, by “only the affected routers”, they are referring to “these are unicasts” which is not true. On the other hand I’d think “Reply-Packets” do contain routing updates as they are replies to queries for missing routes. So, can anyone explain to me, why D is correct an E is wrong?

DaSteve, replay packets dont contain updates, they just reply to quries sent to them. Update packets contain updates, so change information, etc.

Thanks for all this. very usefull.

its good

Writin my exam next week

use the minimum number of network commands to advertise the network range 192.168.66.0/24-192.168.166.0/24 into EIGRP domain.

need help for this question . answer me on my email please. azizdia2003@yahoo.fr

thanks.

Since you need to learn the basic so First step, convert the network into binary

192. 168. 66. 0
11000000.101010000.01000010.00000000

192. 168. 166. 0
11000000.101010000.10100110.00000000

second step: we find what are the numbers of “COMMON BITs” between these 2 networks.
we line up the networks and then find out till where, the BITs are common.
We find that in this case the BITs are common till the end of second octet. See the line below,
telling that the bits are common (same) till the 16th bits. So in that case
192.168.0.0 will be our network and 16 will be our subnet mask.

11000000.101010000.|01000010.00000000
|
same Different
|
11000000.101010000.|10100110.00000000

so for your case the EIGRP summary address would be

ip summary-address eigrp ASnumber 192.168.0.0 255.255.0.0

I am curious how to see the AD=5 in EIGRP summary in question #1.

It says (in the explanation) that this value is only shown on the local router doing the summarization.

Q1.
Is this statement correct? Can any one help me?

“E. THE IP SUMMARY-ADDRESS EIGRP COMMAND GENERATES A DEFAULT
ROUTE WITH AN ADMINISTRATIVE DISTANCE OF 5″

Zeremahesh,

Yes, the summary-adress eigrp generates a AD of 5. You can verify this with GNS3 ;)

@new_student

You can check that through the command #show ip route [network] [network-mask]
where network network-mask -> summarized address

Hi all,

Which show command will display the two values used in the calculation of
the EIGRP metric?
A. show protocol
B. show ip eigrp interface
C. show interface
D. show ip eigrp neighbor

I think answer C is the correct but I found some tests that says answer B. Please help, I’ll take the exam next thursday.

Answer of Q-10 should includes c. Multicast on 224.0.0.10. Is it not correct?? Explain Please.

Digitaltut Sir or Ma’am. I just have one request.. Could you please add a little more separation between the question and answer? I like that the answers are right after the question but I like to be able to hide the answer until after I have considered the questions fully. As it is now, the answers are too close and I frequently scroll too much so I see the answer before I want to. No problem if it can’t be done but sure would help me a lot. Then again, you folks at the Tut have helped my education tremendously as is.

Thanks, muddy!

@Arpit

Hello packets- Unicats/multicats

Update:- unicast/Multicast

Query:-multicast

reply:- unicast

ack:- unicast

for summary metric….u can check it with “Show Run” also.

when you will look at the command in the end you will find 5. this is by default even u can give desirable metric int hat command.

else
show ip route
show ip eigrp topo

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Hi Lalo.

Q7. for selecting Feasible successor. The condition Must be AD(Advertised Distance) of Feasible Succesor must be less than the FD(Feasible distance) of the Current Successor.Here Successor FD is 15. None of the router will meets the desired criteria.

R1-R4-R5- Will be one route R4 will selected as succesor.

R1-R2-r3-R5 – will be the other route. which AD was 20(R2-R5).
The condition doesnot met – 20<15
So there wont be any feasible successor for this topology if succesor goes down that is R4 goes down. then route enter in toa active state. I Hope u r clear on Q7.

@ Lalo

Q8. first line Sh EIGRP toplogy indicates. its running EIGRP AS 200 So Answer A is correct.

for 192.168.1.48 there is only one successor so it load balance in three paths is incorrect so Answer B is incorrect.

R1 for 192.168.1.48 has only one successor which was 192.168.1.66 because its FD was 40537600 rest of two ips greater than this. so 192.168.1.66 was successor for 192.168.1.48. So answer C is correcct.
Answer D is correct 40512000 is the advertised distance (AD) via 192.168.1.66 to reach the 192.168.1.48/28 prefix.

Hi guys,
Is any book or materials/website that is good on objectives/theory? I greatly appreciate feedback. Thanks. firstroberts@gmail.com