CCNA – EIGRP Questions
Here you will find answers to EIGRP Questions
Note: If you are not sure about EIGRP, please read my EIGRP tutorial
Question 1
Refer to the exhibit, when running EIGRP what is required for R1 to exchange routing updates with R3?
A – AS numbers must be changed to match on all the routers
B – Loopback interfaces must be configured so a DR is elected
C – The no auto-summary command is needed on R1 and R3
D – R2 needs to have two network statements, one for each connected network
Answer: A
Question 2:
As a Cisco technician, you need to know EIGRP protocol very well. Which of the following is true about EIGRP successor routes? (Choose two)
A – A successor route is used by EIGRP to forward traffic to a destination
B – Successor routes are stored in the neighbor table following the discovery process
C – Successor routes are flagged as “active” in the routing table
D – A successor route may be backed up by a feasible successor route
E – Successor routes are stored in the neighbor table following the discovery process.
Answer: A D
Explanation:
B is not correct because neighbor table only contains a list of directly connected EIGRP routers that have an adjacency with this router, it doesn’t contain successor routes.
C is not correct because successor routes are not flagged as “active”, they are always the best route to reach remote networks and are always used to send packets.
A and D are correct because successor route is the best and primary route to a remote network. It is stored in the routing table and topology table. If this route fails, a backup route (called feasible successor route) in the topology table will be used to route traffic to a destination.
Question 3:
Which two statements are true regarding EIGRP? (Choose two)
A – Passive routes are in the process of being calculated by DUAL
B – EIGRP supports VLSM, route summarization, and routing update authentication
C – EIGRP exchanges full routing table information with neighboring routers with every update
D – If the feasible successor has a higher advertised distance than the successor route, it becomes the primary route
E – A query process is used to discover a replacement for a failed route if a feasible successor is not identified from the current routing information
Answer: B E
Explanation:
Diffusing Update Algorithm (DUAL) is the algorithm for selecting and maintaining the best path to each remote network. DUAL tracks all the routes advertised by neighbors and selects routes based on feasible successors. It inserts lowest cost paths into the routing table (these routes are known as primary routes or successor routes) -> A is not correct.
EIGRP is still a distance-vector protocol, but has certain features that belong to link-state algorithms (like OSPF) than distance-vector algorithms. For example, EIGRP sends a partial routing table update, which includes just routes that have been changed, not the full routing table like distance-vector algorithms -> C is not correct.
The feasible successor route will become the primary route when its advertised distance is lower than the feasible distance of the successor route. The feasible successor route can be used in the event that the successor route goes down. Notice that the feasible successor route does not get installed in the routing table but is kept in the topology table as a backup route -> D is not correct.
“Support VLSM, route summarization, and routing update authentication” are the features of EIGRP -> B is correct.
When a route fails and has no feasible successor, EIGRP uses a distributed algorithm called Diffusing Update Algorithm (DUAL) to discover a replacement for a failed route. When a new route is found, DUAL adds it to the routing table -> E is correct.
Question 4
Which type of EIGRP route entry describes a feasible successor?
A. a primary route,stored in the routing table
B. a backup route,stored in the routing table
C. a backup route,stored in the topology table
D. a primary route,stored in the topology table
Answer: C
Explanation
Feasible successor is a route whose Advertised Distance is less than the Feasible Distance of the current best path. A feasible successor is a backup route, which is not stored in the routing table but stored in the topology table.
Question 5
Refer to the exhibit. Given the output from the show ip eigrp topology command, which router is the feasible successor?
router# show ip eigrp topology 10.0.0.5 255.255.255.255 IP-EIGRP topology entry for 10.0.0.5/32 State is Passive, Query origin flag is 1, 1 Successor(s), FD is 41152000 |
A.
10.1.0.1 (Serial0), from 10.1.0.1, Send flag is 0×0 Composite metric is (46152000/41640000), Route is Internal Vector metric: Minimum bandwidth is 64 Kbit Total delay is 45000 Microseconds Reliability is 255/255 Load is 1/255 Minimum MTU is 1500 Hop count is 2 |
B.
10.0.0.2 (Serial0.1), from 10.0.0.2, Send flag is 0×0 Composite metric is (53973248/128256), Route is Internal Vector Metric: Minimum bandwidth is 48 Kbit Total delay is 25000 Microseconds Reliability is 255/255 Load is 1/255 Minimum MTU is 1500 Hop count is 1 |
C.
10.1.0.3 (Serial0), from 10.1.0.3, Send flag is 0×0 Composite metric is (46866176/46354176), Route is Internal Vector metric: Minimum bandwidth is 56 Kbit Total delay is 45000 microseconds Reliability is 255/255 Load is 1/255 Minimum MTU is 1500 Hop count is 2 |
D.
10.1.1.1 (Serial0.1), from 10.1.1.1, Send flag is 0×0 Composite metric is (46763776/46251776), Route is External Vector metric: Minimum bandwidth is 56 Kbit Total delay is 41000 microseconds Reliability is 255/255 Load is 1/255 Minimum MTU is 1500 Hop count is 2 |
Answer: B
Explanation
To be the feasible successor, the Advertised Distance (AD) of that route must be less than the Feasible Distance (FD) of the successor. From the output of the “show ip eigrp topology 10.0.0.5 255.255.255.255″ we learn that the FD of the successor is 41152000.
Now we will mention about the answers, in the “Composite metric is (…/…)” statement the first parameter is the FD while the second parameter is the AD of that route. So we need to find out which route has the second parameter (AD) less than 41152000 -> only answer B satisfies this requirement with an AD of 128256.
Question 6
A network administrator is troubleshooting an EIGRP problem on a router and needs to confirm the IP addresses of the devices with which the router has established adjacency. The retransmit interval and the queue counts for the adjacent routers also need to be checked. What command will display the required information?
A. Router# show ip eigrp adjacency
B. Router# show ip eigrp topology
C. Router#show ip eigrp interfaces
D. Router#show ip eigrp neighbors
Answer: D
Explanation
Below is an example of the show ip eigrp neighbors command. The retransmit interval (Smooth Round Trip Timer – SRTT) and the queue counts (Q count, which shows the number of queued EIGRP packets) for the adjacent routers are listed:
Question 7
Refer to the exhibit. How many paths can the EIGRP routing process use to forward packets from HQ_Router to a neighbor router?
HQ_Router# show ip protocols Routing Protocol is “eigrp 109″ Outgoing update filter list for all interfaces is not set Incoming update filter list for all interfaces is not set Default networks flagged in outgoing updates Default networks accepted from incoming updates EIGRP metric weight K1=1, K2=0, K3=1, K4=0, K5=0 EIGRP maximum hopcount 100 EIGRP maximum metric variance 3 Redistributing: eigrp 109 EIGRP NSF-aware route hold timer is 240s Automatic network summarization is not in effect Maximum path: 4 Routing for Networks: 20.10.10.0/24 172.30.10.0/24 192.168.1.0 Routing Information Sources: Gateway Distance Last Update 20.10.10.2 90 00:13:12 172.30.10.2 90 01:13:06 Distance: internal 90 external 170 HQ_Router# |
A. two equal-cost paths
B. two unequal-cost paths
C. three equal-cost paths
D. three unequal-cost paths
E. four equal-cost paths
F. four unequal-cost paths
Answer: F
Explanation
The “Maximum path: 4″ means EIGRP can use up to 4 equal-cost paths to forward packets from HQ_Router to a neighbor router. But here the variance is set to 3 which allows unequal-cost paths. Therefore in this case EIGRP can use up to four unequal-cost paths.
Question 8
IP address and routing for the network are configured as shown in the exhibit. The network administrator issues the show ip eigrp neighbors command from Router1 and receives the output shown below the topology. Which statement is true?
A. It is normal for Router1 to show one active neighbor at a time to prevent routing loops.
B. Routing is not completely configured on Router3.
C. The IP addresses are not configured properly on the Router1 and Router3 interfaces.
D. The no auto-summary command configured on the routers prevents Router1 and Router2 from forming a neighbor relationship.
Answer: B
Explanation
From the output of Router1, we learn that Router1 has not established neighborship with R3 yet. Also from the “show running-config” on Router3 we notice that the “network 192.168.3.0″ statement is missing -> the configuration on Router3 is not complete.
I PASSED CCNA EXAM TODAY THANKS TO ALL MIGHTY ALLAH
960/1000
which labs doing in exzam
Can anyone guide me what kind of labs would i expect to come, and the network diagram is over there and we have to configure that and tick the right answer. Is that so?
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48 ques for exams including 3 simulation, I had EIGRP, Acesslist2 and VTP . Make sure you practice the simulation, use packet tracer or gns3. Best wishes to all!
Alhamdulillah! Thanks to Great Allah! I have passed CCNA with score 933. This site is very useful!
hiiiiiii……i will pass ccna today al hamdo llah. thanks for god ..thanks for my family.. thanks 9tut. i got 920.
This was a great help, but you do not explain the variance variable in the EIGRP tutorial. I am worried that 9tut may be omitting details of certain subjects that may appear on the actual exam since the questions you offer use those details…
@Mbwana
@geedub
Variance defines a range of metrics that allowed for balancing: metrics of this routes should be less then (FD of successor)*variance. (Feasible condition should be satisfied too).
So if variance =1, only equal-cost paths are allowed.
If variance >1 then unequal-cost paths are possible, but their number is no more then Maximum Path.
Correct me if I’m wrong.
q7
hi peeps..how many questions does this exam contains?
thanks
If the Variance is = 1, then there will be 4 equal paths. But since the variance is = 3, which means that the (3xAD)<FD principle converted the unequals to equals, now it can be up to 4 unequal paths. if the variance was set to 2 still there were 4 unequal paths, with diferente (VxAD)<AD. AD=Advertized Distance = Reported Distance.
If the maximum path is changed to 5, and having a variance higher that 1, then there will be 5 unequal paths. If variance is changed back to 1, then there will be 5 equal paths.
i failed last 3weeks ago with 683
i passed today with 920
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@this is messed up ,what was in your exam?
@ this is messed up , my infomation had 48 questions (including the 3 typical Labs(simulations) : ACL2, EIGRP and VTP). now a days this qustns r cums na..?
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Hi Can anyone explain me why is option C correct?
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@9tut… Is it just me or is the network diagram above in Q8 different than that in Q8 of the premium member quiz?
@LR: It’s a mistake. Thanks for your detection. I updated it!
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a friend of mine wrote and failed yesterday. He got stuck at an EIGRP lab of which the first question was: Change the EIGRP AS number without using the “no” command. To be honest, I can’t remember this either
it’s impossible to my knowledge anyway
@9tut: thanks for your effort on this site, I am writing my exam very soon, pls I will appreciate more study materials like dumps from u, my email is asholet98@hotmail.com
Thanks in a million!
Hi 9tut
Question 8 if host 10.0.8.2/24 sent packet to host 10.0.4.0/24 then packet should go R1 >R3 , R1 > R2 >R3 or both . if router config with eigrp , rip or ospf .
normally we using show ip route to get detail , so how identify packet travel from R1 to R3 or R1 to R2 to R3
Helllllloooooo guys. I passed my exam last Wednesday with a hit of 907. I was really happy. The questions were all the same from the dumps(SHEKAR). Only a very very few were new. Nonetheless the new qns had the same concept rounding off. The sims were the usual EIGRP, ACL & VTP. Only in ACL there was just alittle play around, this time it was host B. The EIGRP a new router was added. And well there is something to note on the VTP sim. The show mac-address-table is a bug to the sim. Therefore, you can either use show mac or rather show mac address-table.
My thanks to the 9tut online community, to my colleagues Denis & Rogers, my teachers, and to God almighty. I wish you all the best guys. I’m going for the CCNA security soon and am planning to take my exam this coming October. And I hope to eat Christmas with my CCNP. cheers and all the best!
q3 Explanation:
“The feasible successor route will become the primary route when its advertised distance is lower than the feasible distance of the successor route.”
Is this correct or should it read
“The feasible successor route will become the primary route when its feasible distance is lower than the feasible distance of the successor route”‘
hi all
@9tut,xallax,anyone
while configuring EIGRP r v supposed to configure the “no passive-interface” cmd on the interface compulsarily everytime…….????
plz help me i hv my xm on 29th…….tnx in advance
@ankushk
np is interface is configured wid “passive-interface…..” command displayed as in running-config thn we hav to remove followed by no
not everytym
@ankushk–> :D typo sorry !
no if interface is configured wid “passive-interface…..” command displayed as in running-config thn we hav to remove followed by no
not everytym
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9tut, i really appreciate your work on this site. thanks so much.
Anyone need help or question, asholet98@hotmail.com is my email
hi all,
thanks to 9tut, this site is really helpful, i passed my exams with 935/1000.
all sims are same, only slight different in exam no more 48 questions, now its 50 questions.
@ 9tut/ All
1st thank you for this great site.
Q7….the answer why not be eather C or D?.
I know EIGRP do 4 equal cost links by default (same rip and ospf) or 6 unequal cost links ( if manualy configured ).. but in this exhibit there are only 3 active interfaces for EIGRP so why we can’t answer 3 here? thanks
@Ruben, a broadcast domain = a subnet, and there three subnets in the show ip eigrp topology. there are three subnets in the topology, the 192.168.90.0, 192.168.80.0 ip is .28 and the 192.168.81.0 ip is .28 also, all in 255.255.255.0 subnet mask. so the answer is three
i hope will help.
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Q. 8, since the 192.168.3.0 is directly connected to R1 and R3, does it need to be advertised on these routers again? Somebody pls help me
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