EIGRP Questions

Question 4

EIGRP has been configured to operate over Frame Relay multipoint connections. What should the bandwidth command be set to?

A – the CIR rate of the lowest speed connection multiplied by the number of circuits
B – the CIR rate of the lowest speed connection
C – the CIR rate of the highest speed connection
D – the sum of all the CIRs divided by the number of connections

Answer: A

it seems to be logical, btw in Odom Route 642-902 book in Chapter 3 at paragraph EIGRP Bandwidth control noted:
>>General recomendation: Set the bandwidth of multipoint subinterfaces toaround the total CIR for all VCs assigned to the subinterface.

I a bit confused whom i will trust, can exam takers clear up this question?

from BSCI guide —> Configuring Bandwidth over a Hybrid Multipoint Network
If the multipoint network has differing speeds allocated to the VCs, a more complex solution is
needed. There are two main approaches:
■ Take the lowest CIR and multiply it by the number of circuits. Apply the product as the
bandwidth of the physical interface. The problem with this configuration is that EIGRP will
underutilize the higher-bandwidth links.
■ If possible, it is much easier to configure and manage an environment that has used
subinterfaces, where a VC is logically treated as if it were a separate interface. The
bandwidth command can be configured on each subinterface, which allows different speeds
on each VC. In this solution, subinterfaces are configured for each VC and the CIR is
configured as the bandwidth. Cisco recommends this as the preferred solution.

I always inspired by you, your opinion and way of thinking, again, thanks for this nice post.

- Norman

EIGRP Statc Route.

I want to put a static route to EIGRP ASN. I’m thinking of below sequence of commands. Is this correct?

router eigrp 100
network 192.168.1.0
exit

ip default-network 192.168.1.0
end

hi wats your myspace page

@question number 4:

its asking for the total bandwidth that should be assigned on the main interface itself. its not asking for the sub-interface’s bandwidth. But reading it for the first will definitely make a candidate somehow confused and might lead to choosing letter B. especially when rushing to beat the timer.

test developers who prepared this exams should consider that some of the candidate may not be good in analyzing deep english.

common man, taking such exam should test the technical knowledge not to test how strong your vocabulary is

real exam questions i suppose ?

@Confucius,

” taking such exam should test the technical knowledge not to test how strong your vocabulary is”

We should tell Cisco about that!

Thanks for good stuff

I am just enjoying looking at this site (as my final review) before taking route exam on Wed.

Superb blog post, I have book marked this internet site so ideally I’ll see much more on this subject in the foreseeable future!

Hi Guys,

I’ve just posted up my CCNP – Route nots for EIGRP, hope they are useful:

http://cisco-revision.com/?p=153
http://cisco-revision.com/?p=156

- Adam

I would like to give thanks for create this site to admin…..really it’s very helping side for who want to maintain carrier about network……….

check out good summary notes on packetlife.net

Scored 977/1000 today. All Labs same from this site except ip address. Question gogy 273 is valid. Going for switching now. Guys thanks for the co-operation extended and spl thanks to the admin of Digital Tut.

Hi Guys,
I just starting studying and had a doubt. The Official Certification Guide (by wendell Odom) says that for EIGRP Authentication the key number does not have to match for the two routers, but Jeremy Cioara (CBT Nuggets) says they do not have to match.
Could someone please explain who is correct.

Question 4

EIGRP has been configured to operate over Frame Relay multipoint connections. What should the bandwidth command be set to?

A – the CIR rate of the lowest speed connection multiplied by the number of circuits
B – the CIR rate of the lowest speed connection
C – the CIR rate of the highest speed connection
D – the sum of all the CIRs divided by the number of connections

Answer: A

This is right answer Because: In Frame-Relay all the neighbor share bandwidth equally, regardless actual CIR on each individual PVC.

Hi, need some help here

If variance is set to 2 and all other metric and K values are configured to their default routes, traffic from the internet is balanced across how many paths?

40 15
———B —— C —-
| 10 10 \ <-10 \ <- 15
Internet — A — D — E ——– H — Data Centre
| / <-10 / <- 20
———F —– B —-
20 20

How do you calc that? Sorry for the poor drawing

@ King
I can confirm this answer from ROUTE Training. In the Cisco manual for ROUTE, under BW Utilization Issues – Multipoint interfaces, it states:
“configure the BW to represent the minimum CIR multiplied by the number of circuits”
THis ensures that the circuits with the lowest CIR values are not overdriven.

Ken I was looking at that question too and It says that it should be 3 paths , but form my understanding it suppose to be 4 path , because variance 2 in this case is 60.. 2×30= 60
and we have

internet 10-10-10 data center =>A-D-E-H = 30
internet 20-20-20 data center=>A-F-G-H = 60
internet 20-20-10 data center=>A-F-E-H = 50
internet 40-10-10 data center=>A-B-E-H = 60

pls correct me if I am wrong

http://www.cisco.com/en/US/tech/tk365/technologies_tech_note09186a008009437d.shtml

you can see this question in http://www.7techs.com/bsci-exam?start=5 part 6,.

the summ of 10+10+10 is 30 x by 2 =60
ok

but there are from internet to data center

one path with 40 no valid > because the reported distance of neighbor B is 40, which is greater than the feasible distance (FD) so we have 3 path,. even when the sum give us 5 option, with 30, 40 and 60 respectively,. but the other 60 and 70 I suppose are not valid for that reason,. even when we have a tie,..
please guys comment about this question to arrive to the true,.
thanks

kingkong :-)

A->D->E->H = 30=FD of A (best route) =>valid (1)
A->B->E->H = 60, AD of B=20valid (2)
A->B->C->H = 70 > 60 => not valid
A->F->E->H = 40, AD of F=20valid (3)
A->F->G->H = 60, AD of F=40>FD of A=30 => not valid
=>3 valid paths (Answer C)

EH (10/0) -> best route from E to H
EBCH (40/30)-> reported 30 > 2×10 -> invalid to E
EFGH (50/40)-> reported 40 > 2×10 -> invalid to E
AD -> H (30/20) -> best route from A to H

BEH (20/10) -> E is best route from B to H
BCH (30/15) -> reported 15 valid to B
AB -> H (40/30) -> reported 30 valid to A
AB -> H (50/35) -> reported 35 valid to A

FEH (20/10) -> E is best route from F to H
FGH (40/20) -> reported 20 valid to F
AF -> H (40/30) -> reported 30 valid to A
AF -> H (60/40) -> reported 40 valid to A

So I think there are 5 (!) valid paths at all:
ADEH, ABEH, ABCH, AFEH and AFGH

Regarding to the answer possibilities there are 3 valid routes A can load ballance:
AD -> H, AB -> H and AF -> H

Question about GOGY. Following scenario:

How to avoid prefixes from RB or S0/0/0 entering the enterprise network?

RB ——– S0/0/0 RA ———- Enterprise network.

The answer (according to GOGY) is to use a mask exluding S0/0/0. I am pondering as to why using a distribute-list is not the correct answer. With a distribute list, the outcome would be more granular and there would be communication between RB and RA via EIGRP.

Anyone?

tsm

wher should u apply tthe distibute liust\\

think it u will get answer\

why we r using no autosummary in eigrp?
thank u

i hV ONE MORE DOUBT..PLZ HELP ME..
WHEN SUMMERIZATION IS configured why router creates a route pointed 2 nullo?
i know it 2 avoid loops..bt how?

EIGRP quest from gogy,

Which three statements are true about EIGRP route summarization? (choose three)

A. Manual route summarization is configured in router configuration mode when the router is configured for EIGRP routing.
B. Manual route summarization is configured on the interface.
C. When manual summarization is configured, the summary route will use the metric of the largest specific of the summary routes.
D. The ip summary-address eigrp command generates a default route with an adinistrative distance of 90
E The ip summary-address eigrp command generates a default route with an administrative distance of 5
F When manual summarization is configured, the router immediately creates a route that points to null0 interface.

gogy answ : BEF

is it true if ip summary-address eigrp command issued it would generates a default route with AD 5 ?

thx

@javas, yes, it`s true.
When you don’t set the metric in the ip summary-address command, they automatically put 5.

hai friends,

consider if router A advertise Eigrp summary address 10.1.0.0 255.255.0.0 ( for the routes 10.1.1.0/24 and 10.1.2.0/24 ) to router B.router B will install this route in its routing table.
consider if a packet the comes to router B with the destination address of 10.1.5.1 ,router B will forward the packet to router A (according to summary address ) but router A doesnot know about the network 10.1.5.0 so it will match the router A default route and routing loop or suboptimal forwarding may happen .To avoid such situvation router A will create a route pointing to NULL 0 at the time of eigrp route summarization .this is also known as discard route .so if a packet with address that not known to router A ( no spcific route to the deatination ) will match the null 0 and discard the packet ..

Eigrp summary address AD can be seen via sh ip route 10.1.0.0 255.255.0.0

output will show know via Eigrp distance 5

plz reply if anything wrong in my comments

cant understand why AD become 5 ?

In the Question Which condition must be satisfied before an EIGRP neighbor can be considered a feasible
successor?

Many Dumps have given the correct answer as The neighbor’s advertised distance must be less than or equal to the feasible distance of the current successor.

Few have given the Correct answer as The neighbor’s advertised distance must be less than the feasible distance of the current successor.

As per the link http://www.cisco.com/en/US/tech/tk365 /technologies_white_paper09186a0080094cb7.shtml#feasibleandreported

feasible successor is a path whose reported distance is less than the feasible distance (current best path)

What is correct.

i failed my TSHOOT exam. this OSPF questions all of them wrong and old. one question from this questions. pls change and update that questions!!!! 9tut!!!

dont rely on dumps only please…study seriously before taking the exam guy that important read the maximum book you can read and try to understand the concept tats better

Can someone help me with this question on feasible sucessors:

which two networks does core1 device have a feasible sucessors for:
a) 172.17.0.0/30
b) 172.17.1.0/24
c) 172.17.2.0/24
d)
e)
f)10.140.0.0/24

AF are the answer but I dont know why… the entire question is in the gogy. In fact the other 2 questions I just cant get either but I understand the rules about successors and feasible successors.

I just dont get the answers what are they looking for?

Please dis regard my previous question I find my answer in EIGRP SIMLET! thanks digtut! another donation coming sooon.

@Sergey for:
A->D->E->H = 30=FD of A (best route) =>valid (1)
A->B->E->H = 60, AD of B=20valid (2)
A->B->C->H = 70 > 60 => not valid
A->F->E->H = 40, AD of F=20valid (3)
A->F->G->H = 60, AD of F=40>FD of A=30 => not valid
=>3 valid paths (Answer C)

The current FD after applying variance is 60, so the only valid paths that have to be ADEH, AFEH.

Refer to what GOD says :) The metric of the entire path (the FD of the alternative route) must be lower than the variance multiplied by the local best metric (the current FD). In other words, the metric for the entire alternate path must be within the variance.

@fschris You need to check that question in this website,It’s in eigrp simlet, however, the AF is the correct answer cause they have just 1 FS.

thanks!

Admins, hello! here are having problems with your site. malware warning Write me. icq 67446567456

Pls can any one in d house give me a printable version of CCNP Route Official Certification Guide or any one. my contact is akin_joseph82@yahoo.com
Thanks

Can anyone give me the answer of P4S v4.11 question number 14 of EIGRP, from my understanding it could be 5 but answer shows 3,

confused

Qs no EIGRP17 of p4s 4.11, need discuss

@Azim
imo C should be correct. What u think?

@azim
last psot was to EIGRP 17

Q14: 3 is correct, because Router A only has 2 feasible successors in its eigrp topology
A B E H and A F E H which now go into routing table because of variance.
Keep in mind that AD has to be less than FD to be a feasible successor.
Variance do not change behaviour what what will be in eigrp topology table, it changes behaviour what will be in routing table.

I agree with Marco. There are just three paths possible considering the RD.
A-D-E-H : Best with a metric of 30 which will enter the routing table.
A-F-E-H : Feasible successor with a metric of 40 enters the topology table.
A-B-E-H : Feasible successor with a metric of 60 enters the topology table.

The rest of the paths won’t make it to the topology table as feasible successors because their reported distances (AD) is equal to the FD.

Hence a variance of 2 will allow the paths through A-F-E-H and A-B-E-H to be entered into the routing table in addition to the Successor A-D-E-H (total of 3 paths)
Best metric 30×2= 60 which will include paths A-F-E-h and A-B-E-H with metrics of 40 and 60 respectively.

hope that clears that question.
mail.sidney@gmail.com

Hi Sidney:

Thanks for clarification for Qs-17, can you give me a solution about QS-14, i suspect it might be ‘C’

P4s 4.11 279: Question-40

10.0.0.0/8 or 10.10.10.0/24 which one will be in Routing table of R3 where AUTO-SUM is OFF on R1 ?

P4s 4.11 279: Question-54

confusing,,,,
EIGRP cannot add any route into Routing Table from Redistribution until METRIC values are SET. How could ‘C’ became the correct answer for STATIC routes here ?

Hi Azim,

Please paste the entire question as i do not have the p4s dumps.

You are true I think Azim for Q14 P4s 4.11 : for me the answer is B

Which condition must be satisfied b4 an EIGRP neighbor can be considered a feasible successor?

A. The neighbor’s advertised distance must be less than or equal to the feasible distance of the current successor.
B. The neighbor’s advertised distance must be less than the feasible distance of the current successor.

Which one is correct? I think the right ans: is B. Am i right??

Yes it’s B

The question asks…. “traffic from the Internet to the data center will bo load balanced across how many paths?”

The question also states… “If the wariance value is configured as 2 on ALL routers…”. This means that the variance is set also on router B and that means if router B receives packects from the Internet which are destinated to the DC it will load balance across 2 paths:
B>C>H and B>E>H. Both of these paths will be put into router’s B routing table because both of them meet successor/feasible successor requirements.

So, the corect answer is 4:
-A-B and then:
-C-H
-E-H
-A-D-E-H
-A-F-E-H

regards

just wanna say a big thank u to this site and all of you who have been adding one or two things to this forum. I passed my ccnp route(640-902) 2day with a score of 918. the labs are all the same(ciscopassguide 642.902.v.2010-11-27.by.goggy.272q) except that some hosts are locked.but if u know wat u’r doing, u will find ur way around it.

For the Load balance Q with V= 2 on ALL the router,
the asnwer from my point of view is 2.
since the rule says that : Any FD that is LESS THAN the (BEST FD)*2 , can enter the load balancing list.
From A it will load to D and F.
From D it will continue E then H.
From F, G will fail the rule, and E will win and continue to H

So only the BEST Path : ADEH and AFEH.

Please correct me if am wrong.

The correct answer is 3. Think every one has received full marks for it. explanation in the previous comments.

If the correct answer was 3 router B wouldn’t load balance traffic across C and E. Try to think about this from router’s B perspective.

Correct answer is 4!

And show me any reason why router B will not forward traffic through router C? It will forward packets through C and E!

Correct answer is 4!

The Correct Answer is 2.
Since, the Unequel Cost Load Balancing Rule says that “The Feasible Successor’s FD must be less than the Product of (Successor’s FD * Variance)”

FS < Successor FD * Variance
NOT
FS D–>E–>H, FD= 30) and
Two Feasible Successor
(A–>B–>E–>H, FD= 60) and (A–>F–>E–>H, FD = 40)

NOW,
(Successor’s FD * Variance)
30*2= 60

So, Any Feasible Successor with a Feasible Distance Less than 60 will be copied into the Routing Table for Unequel Load Balancing.

There is only one Feasible Successor with a FD Less than 60, Which is (A–>F–>E–>H, FD = 40).
The other Feasible Successor (A–>B–>E–>H, FD= 60) is Equel to 60 not Less than 60.

So, The Traffic from the Internet to the Data Centre will be LOAD BALANCED over 2 ROUTES NOT 3 ROUTES.

The Correct Answer is 2.
Since, the Unequel Cost Load Balancing Rule says that “The Feasible Successor’s FD must be less than the Product of (Successor’s FD * Variance)”

FS < Successor FD * Variance
NOT
FS E–>H, FD= 30) and
Two Feasible Successor
(A–>B–>E–>H, FD= 60) and (A–>F–>E–>H, FD = 40)

NOW,
(Successor’s FD * Variance)
30*2= 60

So, Any Feasible Successor with a Feasible Distance Less than 60 will be copied into the Routing Table for Unequel Load Balancing.

There is only one Feasible Successor with a FD Less than 60, Which is (A–>F–>E–>H, FD= 40).
The other Feasible Successor (A–>B–>E–>H, FD= 60) is Equel to 60 not Less than 60.

So, The Traffic from the Internet to the Data Centre will be LOAD BALANCED over 2 ROUTES NOT 3 ROUTES.

“The Feasible Successor’s FD must be less than the Product of (Successor’s FD * Variance)” SHOULD BE:

“The Feasible Successor’s FD must be LESS THAN OR EQUAL the Product of (Successor’s FD * Variance)”

so correct is 4 ;]

No It’s NOT…

How on earth did you come up with 4 paths.

The Correct Answer is 2.
Since, the Unequel Cost Load Balancing Rule says that “The Feasible Successor’s FD must be less than the Product of (Successor’s FD * Variance)”

FS < Successor FD * Variance NOT FS H, FD= 30) and
Two Feasible Successor
(A–>B–>E–>H, FD= 60) and (A–>F–>E–>H, FD = 40)

NOW,
(Successor’s FD * Variance)
30*2= 60
So, Any Feasible Successor with a Feasible Distance Less than 60 will be copied into the Routing Table for Unequel Load Balancing.

There is only one Feasible Successor with a FD Less than 60, Which is (A–>F–>E–>H, FD= 40).
The other Feasible Successor (A–>B–>E–>H, FD= 60) is Equel to 60 not Less than 60.

So, The Traffic from the Internet to the Data Centre will be LOAD BALANCED over 2 ROUTES NOT 3 ROUTES

The Correct Answer is 2.
Since, the Unequel Cost Load Balancing Rule says that “The Feasible Successor’s FD must be less than the Product of (Successor’s FD * Variance)”

FS H, FD= 30) and
Two Feasible Successor
(A–>B–>E–>H, FD= 60) and (A–>F–>E–>H, FD = 40)

NOW,
(Successor’s FD * Variance)
30*2= 60
So, Any Feasible Successor with a Feasible Distance Less than 60 will be copied into the Routing Table for Unequel Load Balancing.

There is only one Feasible Successor with a FD Less than 60, Which is (A–>F–>E–>H, FD= 40).
The other Feasible Successor (A–>B–>E–>H, FD= 60) is Equel to 60 not Less than 60.

So, The Traffic from the Internet to the Data Centre will be LOAD BALANCED over 2 ROUTES NOT 3 ROUTES

“EIGRP includes all routes that have a metric of less than or equal to 40 and satisfy the feasibility condition.”[1]

source:
http://www.cisco.com/en/US/tech/tk365/technologies_tech_note09186a008009437d.shtml

correct is 4 :d

Not It’s Not…
And Where did you get the Number 40???
“40″ From the Source Document is Just an Example!!!!

Just do a LAB see by yourself what’s what???

Did you do a LAB?

This example proves that the Feasible Successor’s FD must be LESS THAN OR EQUAL the Product of (Successor’s FD * Variance) to get into routing table..

There is a lot’s of confusion around variance… The “CCNP Route Official Certification Guide” says that “Any FS routes whose metric is less than the product of the variance times the FD are considered to be equal routes and may b e placed into the routing table, ”
And in the same book there is Practice Question from the Chapter 3 Q.no. 9. which says it’s LESS THAN OR EQUEL.

I did many LAB’s before but never gave attention towards it.
To make myself clear I just did a LAB and I found It’s LESS THAN OR EQUEL TO.
It clearly shows 3 paths.
I don’t know whatever the book says BUT IOS won’t lie..

SO, THE RIGHT ANSWER IS 3..
AND Cisco Press Correct the Mistake.
and Thanks rewrew

Did you check Router’s B routing table?

I am sure that router B will load balance and that means the correct answer is 4

ok guys I’ve just finished the lab to prove that the correct answer is that the traffic is load balanced across 4 paths

as I wrote before router A balanced between B, D and F and router B balanced between C and E, altogether traffic is balanced across 4 path :)

here are the routing tables from each of router (10.11.0.0/24 is DC’s network):

A#sh ip route
Codes: C – connected, S – static, I – IGRP, R – RIP, M – mobile, B – BGP
D – EIGRP, EX – EIGRP external, O – OSPF, IA – OSPF inter area
N1 – OSPF NSSA external type 1, N2 – OSPF NSSA external type 2
E1 – OSPF external type 1, E2 – OSPF external type 2, E – EGP
i – IS-IS, L1 – IS-IS level-1, L2 – IS-IS level-2, ia – IS-IS inter area
* – candidate default, U – per-user static route, o – ODR
P – periodic downloaded static route

Gateway of last resort is not set

10.0.0.0/24 is subnetted, 12 subnets
C 10.0.0.0 is directly connected, FastEthernet0/2
D 10.0.1.0 [90/37120] via 10.0.6.2, 00:13:38, FastEthernet0/1
[90/39680] via 10.0.0.2, 00:13:35, FastEthernet0/2
[90/39680] via 10.0.5.2, 00:13:18, FastEthernet0/3
D 10.0.2.0 [90/37120] via 10.0.6.2, 00:14:47, FastEthernet0/1
[90/43520] via 10.0.0.2, 00:13:35, FastEthernet0/2
[90/39680] via 10.0.5.2, 00:13:18, FastEthernet0/3
D 10.0.3.0 [90/38400] via 10.0.6.2, 00:12:25, FastEthernet0/1
[90/46080] via 10.0.0.2, 00:12:25, FastEthernet0/2
[90/40960] via 10.0.5.2, 00:11:58, FastEthernet0/3
D 10.0.4.0 [90/35840] via 10.0.5.2, 00:12:24, FastEthernet0/3
C 10.0.5.0 is directly connected, FastEthernet0/3
C 10.0.6.0 is directly connected, FastEthernet0/1
D 10.0.7.0 [90/30720] via 10.0.6.2, 00:25:02, FastEthernet0/1
D 10.0.8.0 [90/33280] via 10.0.6.2, 00:22:34, FastEthernet0/1
[90/40960] via 10.0.0.2, 00:13:35, FastEthernet0/2
[90/35840] via 10.0.5.2, 00:13:18, FastEthernet0/3
D 10.0.9.0 [90/33280] via 10.0.6.2, 00:13:38, FastEthernet0/1
[90/38400] via 10.0.0.2, 00:13:35, FastEthernet0/2
[90/35840] via 10.0.5.2, 00:13:18, FastEthernet0/3
D 10.0.10.0 [90/33280] via 10.0.6.2, 00:13:20, FastEthernet0/1
[90/40960] via 10.0.0.2, 00:13:20, FastEthernet0/2
[90/33280] via 10.0.5.2, 00:13:18, FastEthernet0/3
D 10.0.11.0 [90/161280] via 10.0.6.2, 00:20:56, FastEthernet0/1
[90/168960] via 10.0.0.2, 00:13:35, FastEthernet0/2
[90/163840] via 10.0.5.2, 00:13:18, FastEthernet0/3
—————————————————————————————–
B#sh ip route
Codes: C – connected, S – static, I – IGRP, R – RIP, M – mobile, B – BGP
D – EIGRP, EX – EIGRP external, O – OSPF, IA – OSPF inter area
N1 – OSPF NSSA external type 1, N2 – OSPF NSSA external type 2
E1 – OSPF external type 1, E2 – OSPF external type 2, E – EGP
i – IS-IS, L1 – IS-IS level-1, L2 – IS-IS level-2, ia – IS-IS inter area
* – candidate default, U – per-user static route, o – ODR
P – periodic downloaded static route

Gateway of last resort is not set

10.0.0.0/24 is subnetted, 12 subnets
C 10.0.0.0 is directly connected, FastEthernet0/3
C 10.0.1.0 is directly connected, FastEthernet0/1
D 10.0.2.0 [90/33280] via 10.0.1.2, 00:16:36, FastEthernet0/1
[90/34560] via 10.0.9.2, 00:16:17, FastEthernet0/2
D 10.0.3.0 [90/35840] via 10.0.9.2, 00:15:51, FastEthernet0/2
[90/38400] via 10.0.1.2, 00:15:51, FastEthernet0/1
D 10.0.2.0 [90/32000] via 10.0.1.2, 00:19:17, FastEthernet0/1
[90/35840] via 10.0.9.2, 00:13:54, FastEthernet0/2
D 10.0.5.0 [90/35840] via 10.0.9.2, 00:14:01, FastEthernet0/2
[90/40960] via 10.0.0.1, 00:14:50, FastEthernet0/3
D 10.0.6.0 [90/33280] via 10.0.9.2, 00:18:55, FastEthernet0/2
[90/38400] via 10.0.0.1, 00:19:33, FastEthernet0/3
D 10.0.7.0 [90/30720] via 10.0.9.2, 00:18:55, FastEthernet0/2
D 10.0.8.0 [90/30720] via 10.0.9.2, 00:18:55, FastEthernet0/2
C 10.0.9.0 is directly connected, FastEthernet0/2
D 10.0.10.0 [90/30720] via 10.0.9.2, 00:18:55, FastEthernet0/2
D 10.0.11.0 [90/158720] via 10.0.9.2, 00:18:55, FastEthernet0/2
[90/161280] via 10.0.1.2, 00:16:34, FastEthernet0/1
——————————————————————————————-
C#sh ip route
Codes: C – connected, S – static, I – IGRP, R – RIP, M – mobile, B – BGP
D – EIGRP, EX – EIGRP external, O – OSPF, IA – OSPF inter area
N1 – OSPF NSSA external type 1, N2 – OSPF NSSA external type 2
E1 – OSPF external type 1, E2 – OSPF external type 2, E – EGP
i – IS-IS, L1 – IS-IS level-1, L2 – IS-IS level-2, ia – IS-IS inter area
* – candidate default, U – per-user static route, o – ODR
P – periodic downloaded static route

Gateway of last resort is not set

10.0.0.0/24 is subnetted, 12 subnets
D 10.0.0.0 [90/39680] via 10.0.1.1, 00:17:36, FastEthernet0/2
C 10.0.1.0 is directly connected, FastEthernet0/2
C 10.0.2.0 is directly connected, FastEthernet0/1
D 10.0.3.0 [90/34560] via 10.0.2.2, 00:16:43, FastEthernet0/1
D 10.0.4.0 [90/39680] via 10.0.1.1, 00:14:46, FastEthernet0/2
[90/39680] via 10.0.2.2, 00:14:24, FastEthernet0/1
D 10.0.5.0 [90/39680] via 10.0.2.2, 00:14:53, FastEthernet0/1
[90/39680] via 10.0.1.1, 00:14:53, FastEthernet0/2
D 10.0.6.0 [90/37120] via 10.0.1.1, 00:17:36, FastEthernet0/2
[90/37120] via 10.0.2.2, 00:17:26, FastEthernet0/1
D 10.0.7.0 [90/34560] via 10.0.1.1, 00:17:36, FastEthernet0/2
[90/34560] via 10.0.2.2, 00:17:26, FastEthernet0/1
D 10.0.8.0 [90/32000] via 10.0.2.2, 00:17:26, FastEthernet0/1
[90/34560] via 10.0.1.1, 00:17:28, FastEthernet0/2
D 10.0.9.0 [90/32000] via 10.0.1.1, 00:17:36, FastEthernet0/2
[90/34560] via 10.0.2.2, 00:17:26, FastEthernet0/1
D 10.0.10.0 [90/34560] via 10.0.1.1, 00:17:36, FastEthernet0/2
[90/34560] via 10.0.2.2, 00:17:26, FastEthernet0/1
D 10.0.11.0 [90/157440] via 10.0.2.2, 00:17:26, FastEthernet0/1
——————————————————————————————–
D#sh ip route
Codes: C – connected, S – static, I – IGRP, R – RIP, M – mobile, B – BGP
D – EIGRP, EX – EIGRP external, O – OSPF, IA – OSPF inter area
N1 – OSPF NSSA external type 1, N2 – OSPF NSSA external type 2
E1 – OSPF external type 1, E2 – OSPF external type 2, E – EGP
i – IS-IS, L1 – IS-IS level-1, L2 – IS-IS level-2, ia – IS-IS inter area
* – candidate default, U – per-user static route, o – ODR
P – periodic downloaded static route

Gateway of last resort is not set

10.0.0.0/24 is subnetted, 12 subnets
D 10.0.0.0 [90/38400] via 10.0.6.1, 00:16:27, FastEthernet0/1
D 10.0.1.0 [90/34560] via 10.0.7.2, 00:20:38, FastEthernet0/2
D 10.0.2.0 [90/34560] via 10.0.7.2, 00:17:36, FastEthernet0/2
D 10.0.3.0 [90/35840] via 10.0.7.2, 00:17:10, FastEthernet0/2
D 10.0.4.0 [90/35840] via 10.0.7.2, 00:15:13, FastEthernet0/2
D 10.0.5.0 [90/33280] via 10.0.6.1, 00:16:09, FastEthernet0/1
C 10.0.6.0 is directly connected, FastEthernet0/1
C 10.0.7.0 is directly connected, FastEthernet0/2
D 10.0.8.0 [90/30720] via 10.0.7.2, 00:25:23, FastEthernet0/2
D 10.0.9.0 [90/30720] via 10.0.7.2, 00:25:50, FastEthernet0/2
D 10.0.10.0 [90/30720] via 10.0.7.2, 00:25:40, FastEthernet0/2
D 10.0.11.0 [90/158720] via 10.0.7.2, 00:23:44, FastEthernet0/2
——————————————————————————————–
E#sh ip route
Codes: C – connected, S – static, I – IGRP, R – RIP, M – mobile, B – BGP
D – EIGRP, EX – EIGRP external, O – OSPF, IA – OSPF inter area
N1 – OSPF NSSA external type 1, N2 – OSPF NSSA external type 2
E1 – OSPF external type 1, E2 – OSPF external type 2, E – EGP
i – IS-IS, L1 – IS-IS level-1, L2 – IS-IS level-2, ia – IS-IS inter area
* – candidate default, U – per-user static route, o – ODR
P – periodic downloaded static route

Gateway of last resort is not set

10.0.0.0/24 is subnetted, 12 subnets
D 10.0.0.0 [90/38400] via 10.0.9.1, 00:17:48, FastEthernet0/4
D 10.0.1.0 [90/32000] via 10.0.9.1, 00:21:59, FastEthernet0/4
D 10.0.2.0 [90/32000] via 10.0.8.2, 00:18:57, FastEthernet0/2
D 10.0.3.0 [90/33280] via 10.0.8.2, 00:18:31, FastEthernet0/2
D 10.0.4.0 [90/33280] via 10.0.10.1, 00:16:34, FastEthernet0/3
D 10.0.5.0 [90/33280] via 10.0.10.1, 00:16:41, FastEthernet0/3
D 10.0.6.0 [90/30720] via 10.0.7.1, 00:27:11, FastEthernet0/1
C 10.0.7.0 is directly connected, FastEthernet0/1
C 10.0.8.0 is directly connected, FastEthernet0/2
C 10.0.9.0 is directly connected, FastEthernet0/4
C 10.0.10.0 is directly connected, FastEthernet0/3
D 10.0.11.0 [90/156160] via 10.0.8.2, 00:25:05, FastEthernet0/2
——————————————————————————————–
F#sho ip route
Codes: C – connected, S – static, I – IGRP, R – RIP, M – mobile, B – BGP
D – EIGRP, EX – EIGRP external, O – OSPF, IA – OSPF inter area
N1 – OSPF NSSA external type 1, N2 – OSPF NSSA external type 2
E1 – OSPF external type 1, E2 – OSPF external type 2, E – EGP
i – IS-IS, L1 – IS-IS level-1, L2 – IS-IS level-2, ia – IS-IS inter area
* – candidate default, U – per-user static route, o – ODR
P – periodic downloaded static route

Gateway of last resort is not set

10.0.0.0/24 is subnetted, 12 subnets
D 10.0.0.0 [90/40960] via 10.0.10.2, 00:17:41, FastEthernet0/1
[90/40960] via 10.0.5.1, 00:17:38, FastEthernet0/3
D 10.0.1.0 [90/34560] via 10.0.10.2, 00:22:58, FastEthernet0/1
D 10.0.2.0 [90/34560] via 10.0.10.2, 00:19:56, FastEthernet0/1
D 10.0.3.0 [90/35840] via 10.0.10.2, 00:17:07, FastEthernet0/1
[90/35840] via 10.0.4.2, 00:17:07, FastEthernet0/2
C 10.0.4.0 is directly connected, FastEthernet0/2
C 10.0.5.0 is directly connected, FastEthernet0/3
D 10.0.6.0 [90/33280] via 10.0.10.2, 00:17:41, FastEthernet0/1
[90/33280] via 10.0.5.1, 00:17:38, FastEthernet0/3
D 10.0.7.0 [90/30720] via 10.0.10.2, 00:28:00, FastEthernet0/1
D 10.0.8.0 [90/30720] via 10.0.10.2, 00:27:43, FastEthernet0/1
D 10.0.9.0 [90/30720] via 10.0.10.2, 00:28:00, FastEthernet0/1
C 10.0.10.0 is directly connected, FastEthernet0/1
D 10.0.11.0 [90/158720] via 10.0.10.2, 00:26:05, FastEthernet0/1
———————————————————————————————–
G#sh ip route
Codes: C – connected, S – static, I – IGRP, R – RIP, M – mobile, B – BGP
D – EIGRP, EX – EIGRP external, O – OSPF, IA – OSPF inter area
N1 – OSPF NSSA external type 1, N2 – OSPF NSSA external type 2
E1 – OSPF external type 1, E2 – OSPF external type 2, E – EGP
i – IS-IS, L1 – IS-IS level-1, L2 – IS-IS level-2, ia – IS-IS inter area
* – candidate default, U – per-user static route, o – ODR
P – periodic downloaded static route

Gateway of last resort is not set

10.0.0.0/24 is subnetted, 12 subnets
D 10.0.0.0 [90/46080] via 10.0.4.1, 00:17:39, FastEthernet0/1
[90/46080] via 10.0.3.2, 00:17:35, FastEthernet0/2
D 10.0.1.0 [90/38400] via 10.0.3.2, 00:17:35, FastEthernet0/2
[90/39680] via 10.0.4.1, 00:17:39, FastEthernet0/1
D 10.0.2.0 [90/34560] via 10.0.3.2, 00:17:35, FastEthernet0/2
C 10.0.3.0 is directly connected, FastEthernet0/2
C 10.0.4.0 is directly connected, FastEthernet0/1
D 10.0.5.0 [90/35840] via 10.0.4.1, 00:17:39, FastEthernet0/1
D 10.0.6.0 [90/38400] via 10.0.4.1, 00:17:39, FastEthernet0/1
[90/38400] via 10.0.3.2, 00:17:35, FastEthernet0/2
D 10.0.7.0 [90/35840] via 10.0.4.1, 00:17:39, FastEthernet0/1
[90/35840] via 10.0.3.2, 00:17:35, FastEthernet0/2
D 10.0.8.0 [90/33280] via 10.0.3.2, 00:17:35, FastEthernet0/2
[90/35840] via 10.0.4.1, 00:17:36, FastEthernet0/1
D 10.0.9.0 [90/35840] via 10.0.4.1, 00:17:39, FastEthernet0/1
[90/35840] via 10.0.3.2, 00:17:35, FastEthernet0/2
D 10.0.10.0 [90/33280] via 10.0.4.1, 00:17:39, FastEthernet0/1
[90/35840] via 10.0.3.2, 00:17:35, FastEthernet0/2
D 10.0.11.0 [90/158720] via 10.0.3.2, 00:17:35, FastEthernet0/2

THE RIGHT ANSWER IS 4 — proved by LAB !!!

*here are the routing tables from each of router (10.0.11.0/24 is DC’s network):

Hello guys..

does anyone have ccnp route 642-902 cbt, pass4sure or any dumps. kindly send me links on shariq2009@live.com or u can send here as well.

Thanksss

Would you agree with the following definitions?

1. A Neighbour with (RD)<(FD of Successor) is a FS.
2. Any (and only) FS that have a RD<(variance*(FD of successor)) are added to the routing table.

guys

do u have Route GNS3 LAB?
Kindly give a link.. if u got

hi guys, does anyone have the ccnp route dumps? email them at desmo@live.be

i have the following error .. Anyone has any idea what is opcode=1 ?

R2
==
03:58:47: AS 1, Flags 0×0, Seq 0/0 idbQ 0/0 iidbQ un/rely 0/0
03:58:48: EIGRP: FastEthernet0/0: ignored packet from 10.1.23.1, opcode = 1 (missing authentication)
03:58:49: EIGRP: FastEthernet0/0: ignored packet from 10.1.23.1, opcode = 5 (missing authentication)

R2 Running-config
=========
key chain test
key 2
key-string cisco
accept-lifetime 16:50:00 Jul 19 2011 17:00:00 Jul 19 2011
send-lifetime 16:50:00 Jul 19 2011 17:00:00 Jul 19 2011

interface FastEthernet0/0
ip address 10.1.23.2 255.255.255.252
no ip redirects
ip authentication mode eigrp 1 md5
ip authentication key-chain eigrp 1 cisco
no ip mroute-cache
speed 100
full-duplex
no cdp enable
!

R3 running-config
==========
key chain 2
key 2
key-string cisco
accept-lifetime 15:00:00 Jul 21 2011 17:00:00 Jul 21 2011
send-lifetime 15:00:00 Jul 21 2011 17:00:00 Jul 21 2011
!

interface FastEthernet0/1
ip address 10.1.23.1 255.255.255.252
ip authentication mode eigrp 1 md5
ip authentication key-chain eigrp 1 cisco
speed 100
full-duplex
!

asldavid,

just change the following command:

ip authentication key-chain eigrp 1 cisco

to

ip authentication key-chain eigrp 1 1

after eigrp 1 the name of key chain should be written not the key-string, as well in both routers try to give same key-chain name not 1 and 2.

try these it will work.

Thanks,

I think the confusion regarding the unequal cost load balancing question stems from these two issues:

1. A feasible successor is a path whose reported distance is less than the feasible distance (current best path)

2. For Variance, EIGRP includes all routes that have a metric of less than or equal to (FD x Variance) and satisfy the feasibility condition

http://www.cisco.com/en/US/tech/tk365/technologies_tech_note09186a008009437d.shtml#var

http://www.cisco.com/en/US/tech/tk365/technologies_white_paper09186a0080094cb7.shtml#feasibleandreported

So, best path is A – D – E – H with a FD of 30 ( x 2 for Variance) = 60

Router B is reporting an AD of 20 for path B – E – H to Data Centre
Router F is reporting an AD of 20 for path F – E – H to Data Centre

So, Router A has 1 Succesor route (A-D-E-H) and 2 Feasible Successor routes (A-B-E-H, A-F-E-H)

Both Feasible Successors become active for load balancing when Variance is increased to 2 based on rule #2 above.
FD for path A – D – E – H = 30 x 2 = 60
FD for path A – B – C – H = 60
FD for path A – F – E – H = 40

The other issue with this question is that they should have specified from which router’s perspective this is happening, because as a previous poster mentioned, router B will load balance as well. The question should state “traffic from router A destined to DC is load balanced over how many paths?” I think taking into account the load balancing happening on router B is “over thinking” the question. I’m going with 3 as my answer when I do the exam tomorrow.

…To configure EIGRP on a network use the following command ….R1enable .R1 configure terminal .R1 config router eigrp 100 .R1 config-router network 192.168.3.0 ..R1 config-router network 172.16.1.0.R1 config-router end….From the command above we used the command router eigrp 100 to turn on the protocol all other routers on the network must be configured the same way each router must declare its directly connected network to be seen by other routers on the network and most importantly all routers on the same network running EIGRP must be configured with the same process number 100…. …….You want to redistribute routes that were learned by the router into the EIGRP routing process..using the redistribute command ..Redistributing Routes into EIGRP.HQ configure terminal.HQ config ip route 192.168.2.0 255.255.255.0 172.16.1.5 .HQ config router eigrp 100.HQ config-router redistribute static.HQ config-router end.HQ ……Setting the properties of the routes that are redistributed from another routing protocol with the default-metric command …HQ configure terminal .HQ config router eigrp 100.HQ config-router redistribute rip.HQ config-router default-metric 1000 100 250 100 1500.HQ config-router end…Use the show ip protocols command to view your route redistribution…..

Please update EIGRP New Questions

I really liked the article, and the very cool blog

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i’ll give route exam on 15th of july plz send me the latest dumps on bozo.tiger@gmail.com

Pls help!
Item 4 under EIGRP on P4S v278Q rev. 6
Q. refer to the exhibit. EIGRP is configured on all routers. Autosummarization is enabled on routers R2 and Rbut it is disabled on router R1. Which two EIGRP routes will be seen in the routing table of router R3.
Ans. 10.10.10.0/24 and 172.16.0.0/16

I cant figure out why 172.16.0.0/16 is one of the answer instead of 172.16.0.0/24?
Can anybody explain it? Thanks!

Please tell me anyone that question on this site and pass4sure are same or not?

@ spark_rod
as the question said, autosummarization is enabled on R2, so the router autosummarized 172.16.0.0/24 to 172.16.0.0/16 which is a class B network. Although it is disabled on R1 but still before it receive the network address is has been autosummarized by R2.

Can you guys tell me, how can I find dump CCNP route questions with explanation answers?

@umka: google the exact question statement and search in the results.

Hello there, I believe your internet site may be getting browser compatibility problems. When I seem at your web site in Safari, it appears very good but when opening in Web Explorer, it’s some overlapping. I just needed to provide you with a fast heads up! Other then that, amazing weblog!

Hi all, i know the argument re: whether variance rule is “Any FS routes whose calculated metric is less than or equal to the product of variance times FD are added to the IP routing table, assuming the maximum-paths setting allows more routes”. I stumbled upon a post while i was searching for the answer –> Wendon Odom has replied to this himself to iron out any doubts.
See https://learningnetwork.cisco.com/message/102605#102605

The answer is <= and NOT <

@9tut,

Which statement about a non-zero value for the load metric (k2) for EIGRP is true?
Select the best response.
A. A change in the load on an interface will cause EIGRP to recalculate the routing metrics
and send a corresponding update out to each of its neighbors.
B. EIGRP calculates interface load as a 5-minute exponentially weighted average that is
updated every 5 minutes.
C. EIGRP considers the load of an interface only when sending an update for some other
reason.
D. A change in the load on an interface will cause EIGRP to recalculate and update the
administrative distance for all routes learned on that interface.

Tell me please, where you can find the official documentation on this issue:
What happens when the EIGRP topology changes, DUAL notified–>neighbor adjacency is deleted and etc…?