EIGRP Questions 3

Is question 7 A ok and shouldn’t it be D

Could you pls explane

@blackcisco
for Q7 A is correct because to use “ip default-network” it should be defined in the network command but here that is not present as network 20.20.20.20.

so for EIGRP default n/w can only be propaged using network and using ip default-network or by redistributing the static default n/w.

Not true

first condition must be met: FD(Successor)>AD(Feasible Successor) to avoid routing loops!
+ A-D-E-H with a metric of 30 (10+10+10)
+ A-F-E-H with an AD of 20
+ A-F-E-H with an AD of 20

then second condition is variance2 FD(S) > FD(Feasible Successor )

variance 2 of (S)=2*30=60
+ A-B-E-H with a metric of 60 (40+10+10)
+ A-F-E-H with a metric of 40 (20+10+10)

Result is + A-F-E-H with a metric of 40 (20+10+10)

Correct answer is B

from:
https://learningnetwork.cisco.com/message/102605#102605

“On the technical point, it looks like there was somewhat general agreement by the end of this thread that the logic is <=, not simply <. I tested again in lab, and agree – a tie on variance*bestmetric compared to feasible-successors-metric means it's a candidate for inclusion.

My apologies that the confusion caused in part by the error in my ROUTE book. I listed it as simply "less than" in one place, and with the correct "less than or equal to" in another. The error was on the bottom of page 89. It'll be in the errata in a few days; this was the first notice I'd seen on the problem.

I also see your comment about not knowing how to contact Cisco Press, so in case it helps for the general audience here… there's two ways to get in touch with them.The first is listed in the front matter of their books, under heading "Feedback Information" – it just says to email feedback at ciscopress.com. The other and better option (in my opinion), is to go to ciscopress.com, and go to the bottom "Contact Us". The next page shows a form. If you select "Submit Errata" from the list in the form that follows, it goes to the admin for Cisco Press, and she distributes it to the correct author if it's a technical issue. That's true for any Cisco Press books.

Again, my apologies for my part in causing the confusion.
Wendell "

IN QUESTION 1, CORRECT ANSWER IS "B" (2PATHS)!

Ya I agree that B is the correct answer. But what is the correct answer on the Cert Exam? Will they still have us choose C?

for Q7

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

Can any one verify the answer for Q1, as per me ans B is correct the traffic will be load balanced across two paths coz for A route to be included in unequal cost load sharing “The metric of the lowest cost route A-D-E-H (30) when multiplied by variance(2) must be greater than the metric of the route to be added”

so A-D-E-H 30*2=60 > A-F-E-H 20+10+10=40 None of the other path satisfies this condition so as per me there should only be two paths across which load balance will happen.

Answer B is correct any anyone verify this.

Please Correct me if i am wrong:
Q1: With a variance of “1″, routes must be exactly equal to be also added into the routing table. (next to the Succesor route)
There is only one router with a metric of 30 eg.ADEH, so this is the only route in the RT.
However there are 2 FS routes ABEH (AD 20) and AFEH (AD 20).
Now with a variance of “2″ the metric may be =< 60 for a FS to be viable for Load balancing.
There are 5 routes that are =< 60 BUT only routes that are FS are eligible to be considered for LoadBalancing which brings it back to three routes ADEH, ABEH and AFEH.
Therefor i stil think we have 3paths on which packets will be loadbalanced.

Q1. Ok, from the A’s perspective there are 3 routes. But when traffic gets to router B, it will load-balance it across B-C-H and B-E-H because from B’s persective B-E-H = 20 and B-C-H = 30. So we have 4 paths: A-B-C-H, A-B-E-H, A-D-E-H, A-F-E-H, which means the answer is D?

About Q1: why DataCenter network has no metric? It’s not possible (as far as it’s a real network). It has to have (at least =1)!
So, (10+10+10+1)*2 > (40+10+10+1).

Q1 the key word here is load balancing. In order to load balance the routes must be a successor route which places it in the routing table. Whats getting mixed in the picture are feasible successors that show up in the topology output but not the routing table.

Also, I’m thinking a link can not be duplicated between paths, they have to be unique. Is the load really being balanced if two paths are using the same link? So the only routes that will show up in the routing table are paths less than or equal to twice the Successor and unique.

The format in the topology table is feasible distance/advertised distance
A-B-C-H = 75/30

The feasible distance of A-B-C-H is 75 which is why it is not a successor but a feasible successor with a advertised distance of 30 which less than the variance of 2.

variance of two = 60
two paths in the routing table: A-D-E-H (30) and A-F-G-H(60)
feasible successor based on advertised distance in the topology table : A-B-C-H(30)

Seriously, about Q1, does anyone consider answer D? Please, correct me if I’m wrong.

@kiva

So you’re saying that A-B-C-H and A-B-E-H are no “unique” paths? I mean that I understand that they use the same link, but still: they are no the same! Besides, why whould they put 4 options to chose from? From what you’re saying it is obvious that there cannot be more than 3 “unique” paths, so the options really are A, B or C

@sagittarius Lets use a different terminology of overlapping links. Again this strictly my thinking. If we presume the paths can not overlap (or use the same link) in a path then A-B-C-H and A-B-E-H have overlapping links. There are three possible paths but we have to distinguish between a successor and a feasible successor. Two paths are successors in the routing table for load balancing. The third is a feasible successor not in the routing table but waiting in the topology database as backup.

The question is not how many paths are available but how paths are available for load balancing. Load balancing will be performed across the paths in the routing table – two paths.

@kiva

wait, what is wrong with A-B-E-H? The total metric for it is 60 (with AD of 20) which is ok both for feasible condition and for variance: FD is 30, 30*2 = 60, 30 (FD) > 20 (AD). If you have doubts about variance read cisco.com: http://www.cisco.com/en/US/docs/ios/12_2/iproute/command/reference/1rfeigrp.html#wp1024988

@Sagittarius

A-B-E-H overlaps with the primary path A-D-E-H.

Dear Kiva and dear all,

So, it mean the C is correct ( it have 3 way to reach), I check from Cisco.
Tanveer

I was wrong, we need load balance not how many path, so it have only two path, Answer 2 is correct.
Thanks.

Dear all, as per cisco
http://www.cisco.com/en/US/docs/ios/12_2/iproute/command/reference/1rfeigrp.html#wp1024988

first condition must be met: FD(Successor)>AD(Feasible Successor)
+ A-D-E-H with a metric of 30 (10+10+10) ****This is FD****
+ A-F-E-H with an AD(10+10) of 20
+ A-B-E-H with an AD(10+10) of 20

then second condition is variance2 FD(S) = > FD(Feasible Successor ) most of us know, that only second condtion must be also less, but no, The first condition must be less, not the second.

variance 2 of (S)=2*30=60
+ A-B-E-H with a metric of 60 (40+10+10)
+ A-F-E-H with a metric of 40 (20+10+10)

That is equle or less then 60. so we have Three path for loadbalance. 100%.

Correct answer is C
If some one have prove,please share it.

@Salem

ok, I completely agree with you, but you are thinking from router A’s perspective. On router A we will see THREE routes: via B, via D and via F (not A-B-E-H, A-D-E-H, A-F-E-H), This means that when traffic goes to router B it is completely up to router B across what paths will it load-balance packets. There is no way router A can make router B to use on B-E-H !!!

I agree with the last comment. Router B will load-balance across two links!

Then it same with router F , when trafic will go to F it have two link ( FEH and FGH ), so where Router F will send the trafic.
In my point , When trafic will reched to Router B then will chose only BEH by low metric (10+10). Router A have no concer what is next after B, Router A will just calculate only metric to decid trafic.

@Salem

With F it’s not the same, because F-E-H = 20, which is FD, F-G-H = 40 with AD = 20, so in this case the feasible condition is not met: AD should be LESS then FD (and we have AD=FD).
With B, the feasible condition IS met and, as variance = 2, router B WILL install both B-E-H and B-C-H in route table.

Please, prove if it’s wrong!

I’m really intrigued with this topology. I’ve actually recreated it in GNS, mainly to see that the variance multiplier also includes the “equal” in the <= condition:

path <= best path * variance (and yes, there is a "=" in that formula).

Anyway, I also believe 4 paths is the answer, but I can't seem to get the much needed proof, because it is not behaving as I would expect: I opened the http service on router H's loopback, to simulate a server in the Enterprise, then connected my Windows Loopback adapter to router A and opened web browser and pointed it to Router H's loopback. I also set some route-maps with access-lists and ip policy route-map on interfaces in the incoming paths from router A, to check the "10 permit tcp any any eq www 432 matches" on both router C and router E. However, although the best path on router B is through router E, it still uses router C exclusively for web traffic. If I shut down router C, it uses router E. When I turn router C back on, it shifts to router C. I don't know what to believe :( I really want to know the right answer.

My logical guess would be 4 paths: ADEH, AFEH, ABEH and ABCH , but I can't seem o get the much needed proof. Maybe I'm doing something wrong :)

It’s that you simulated it, cause to be honest I didn’t have time to do that (so far). The behavior that you see is because the router using CEF as a switching mechanism actually does per session load-sharing (per source-destination), so traffic from the same source-destination pair will go through only one path. The goal here is to configure “variance” value on all router and than just examine the routing table.

Also, as far as i remember, process-switching by default performs per packet load-sharing, so you can try to disable CEF and try again.

Q1 OK this is weird. I simulated the network using GNS3. The two paths in the routing table are not the paths I thought. It appears overlapping links is acceptable. They are:
A-D-E-H and A-F-E-H. The third branch from A-B is not considered at all. It doesn’t meet the variance 2 condition. It’s 4x the Successors FD value so we throw that branch out of consideration.

Because variance 2 is configured on all routers, I’m thinking the FD and the AD distance has to be evaluated at each router. This is what it’s looking like, 1) the FD condition is evaluated as <= the successor but 2) the AD is evaluated as < the successor.

1) From CCNP Route 642-902
2) http://www.cisco.com/en/US/tech/tk365/technologies_white_paper09186a0080094cb7.shtml#feasibleandreported see the section Feasible Distance, Reported Distance, and Feasible Successor

Ok, it’s definitely weird. Maybe you have an error in link metrics assignment? Anyway, I’ll simulate it too, later.

PS. I’ve passed my 642-902 exam today (scored 965) and I didn’t get this Q1 :)

In my logic.

A-D-E-H with a metric of 30 (10+10+10) ****This is FD****

First condition must be met: FD(Successor)>AD(Feasible Successor)
Here we will not add variance of 2.

B-E-H with an AD(10+10) of 20
F-E-H with an AD(10+10) of 20
So we have only two way as per first condition.

Other two are will met the first condition. We need less then 30 ( no variance add here)
B-C-H With AD ( 15+15) of 30 it is = 30 so never concider we need less.
F-G-H with AD ( 20 +20) of 40> 30 it is also bigger os never concider, we need less.

The second condition is variance2 FD(S) = > FD(Feasible Successor)

variance 2 of (S)=2*30=60
+ A-B-E-H with a metric of 60 (40+10+10)
+ A-F-E-H with a metric of 40 (20+10+10)
so we have three path.

Note: if A-B is more then 40 like 50 then A-B-E-H ( 50+10+10) 70 then only we have two path, it is bigger then FD 30×2=60 so it will not add, we need = or <.

@Salem

Do you agree that if you do a show ip route on Router B you will see 2 routes to the enterprise network (with B-E-H = 20 as the best path and B-C-H = 30 which is added to the routing table because of the configured variance) ??

So A has 3 routes to the “Data Center Network” in the routing table, and B has 2 routes. When A sends multiple packets to B, do you agree that B, because it has the 2 paths, will in turn load-balance across those 2 paths, giving a total of 4 paths overall ?

Question 1 is about pathes not just routes on the edge (A) router.
“A” will have 3 pathes + “B” splits one path.
so, I agree about Q1 – answer should be D, as Bogdan & Aninymous explained.

EIGRP is a link vector protocol, unlike OSPF it does not have a database of the area topology. EIGRP references the feasible distance (FD) and the advertised distance(AD) of directly connected routers.

Two tests are performed. At router A the successor path is A-D=10. Visually A-D-E-H is the successor. The first test compares the successor FD x 2 = 20 to potential feasible successor (FS) path. A-F=20 is 20 and does not pass the first test and not an option. The first test is discussed in Wendell Odoms book CCNP Router 642-902 page 90.

The second test is the AD < 20? ADs from F are F-E=10 and F-G=20. F-E passes and F-G fails. See the link for the < than test.
http://www.cisco.com/en/US/tech/tk365/technologies_white_paper09186a0080094cb7.shtml#feasibleandreported. The actual algorithm is probably way more involved than this.

This gives us A-F-E. For simplicity, E takes us straight to the data center through H. Thus we have two paths A-D-E-H and A-F-E-H.

I used the interface bandwidth command to weigh the costs in the diagram; where 10 is the preferred bandwidth and 40 the least preferred.

10 10000
15 7500
20 5000
40 2500

router A – topology
——————-
P 192.168.1.100/30, 1 successors, FD is 266240, serno 73 data center subnet
via 192.168.1.18 (266240/263680), FastEthernet0/1 router A-D
via 192.168.1.34 (522240/263680), FastEthernet1/0 router A-F
via 192.168.1.2 (1034240/263680), FastEthernet2/0 router A-B

router F – topology
——————-
P 192.168.1.100/30, 1 successors, FD is 263680, serno 34 data center subnet
via 192.168.1.30 (263680/261120), FastEthernet1/0 router F-E
via 192.168.1.33 (524800/266240), FastEthernet2/0 router F-A
via 192.168.1.38 (519680/517120), FastEthernet0/1 router F-G

router A – route
—————-
D 192.168.1.100 [90/522240] via 192.168.1.34, 00:24:16, FastEthernet1/0
[90/266240] via 192.168.1.18, 00:24:16, FastEthernet0/1

@ sagittarius , would you please share what P4s you have used ?

thanks

Please go this to check EIGRP.

http://bestitshop.blogspot.com/2011/12/load-balancing-in-eigrp-configuration.html

This problem is a pain the butt! So I changed the tos values to use delay metric only. Now I’m getting three routes in the routing table.

@Salem thank you for the clarification of when the variance is used.

route table contains paths A-D-E-H, A-F-E-H, and A-B-E-H in that order.

@kiva @Salem

You keep thinking only from A’s perspective! Consider routing table on B!

Daer Kiva

when Variance is used. it will used in first condition or second condition.

I think variance is using in second condition, if first condtion will pass without adding variance.
variance will add when we calculate the metric, (the second condtion). Total FD of feasible successor must be = < then metric ( FD x 2 of successor).
So it will take 3 path, that i wrote before.
see here
http://bestitshop.blogspot.com/2011/12/load-balancing-in-eigrp-configuration.html

For question 1:
I think this question is not worderd specific enough. It will load balance over 3 links for only one hop. Once router B and F get the packets it will be a different story as they have their own respective S and FS.

Today i pass with 100% score ( 1000)
I got four LABs , Saving cammand is not working in three Labs, so i was very upset , so do not worry go to next, but you have to chack saving cammand in all labs, most of qusetion came from this site but
4 new qusetion that i review before this site for all new qusetion, So please chech all site.
http://bestitshop.blogspot.com/2011/12/latest-question-ccnp-642-902-route-2012.html

can any one shae new dump. i have exam next week

Hi, guys

regarding Question 3 you stated “EIGRP routing updates are triggered only by a change in network topology (like links, interfaces go up/down, router added/removed), and not by change in interface load or reliability -> A & D are not correct.”.

But on the other hand, my CCNP Route Certification Guide says: “Cisco recommends against using k-values k2, k4, and k5, because a nonzero value for these parameters causes the metric calculation to include interface load and reliability. The load and reliability change over time, which causes EIGRP to reflood topology data, and may cause routers to continually choose different routes (route flapping).”

Wouldn’t be A) option the correct one then?

Thanks :)

Should not Q2 instead be “A” to be correct? If the FD is equal to, it is still acceptable, no?

Bruno, I think it might be because EIGRP will update the interface when sending an update for a DIFFERENT interface/router/path altogether. It is true that “C: EIGRP considers the load of an interface only when sending an update for some other reason.” But there are MANY reasons to be sending an update – not all of which have to do with that particular interface…so it still slows down and may cause the routes to flap.

#7, 9, 10, 11 were on my test

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!

kindly explain Q.7

Question 1
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

By CCNP ROUTE 642-902 Official Certification Guide

@kannan
The question asked how R1 can generate default route to downstream routers. A will do this but B will not. A will generate an external EIGRP default route. If the destination attempted to be reached at the router redistributing the default route does not know the destination then it will send an unreachable message to the receiver
B only generates a default route downstream if the 20.0.0.0 network or 20.20.20.0 network is advertised by the EIGRP process(so A is right for external EIGRP)

@xpman

Ok, but network 20.0.0.0 or 20.20.20.0 could be advertised by EIGRP, You will not see this in the routing table, because direct connection has lower AD. In answer A, the default route points to Null0 interface, so unknown addresses will be dropped by R1. I still believe there is a typo in that question, because A could work in case of ‘ip route 0.0.0.0 0.0.0.0 E0/0′ command or B could work as well if there is a def route to E0/0 on router R1 and 20.0.0.0 is advertised by EIGRP.

about question 7 : i found this to be very clear info :
http://blog.ipexpert.com/2010/03/29/eigrp-ip-default-network-command/

@ปลาวาฬทราย

Pretty sure the rule says it will load balance only over links with LESS than the variance x lowest metric.

“Use the variance n command in order to instruct the router to include routes with a metric of LESS than n times the minimum metric route for that destination.”

http://www.cisco.com/en/US/tech/tk365/technologies_tech_note09186a008009437d.shtml#topic1

in question 7 why not b or d???