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.
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question 7 please. i can’t get it .
passed CCNA Cert 24 july I dont usually bother with dumps but had a go and was suprised at finding questions from dump in 640-802 i used Spike dump and Seqar also the acl simlet from 9 tut was there though a little bit different Vtp was there and the same as 9tut eigrp also thanks 9tut
rishabh,
We can use varience for load balance as follow. if there are 4 path and put varience 3
so there are 4 unequal path having. if we put 2, have 2 same paths per each.
(confg- touter)#varience 3
erangadpm@gmail.com
In Questions 7 , if the variance is 2 then there is two path if the variance is 3 then 4 unequal path how ….? if the variance is 5 then…….?
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@rishabh, @chanaka, @paul…
chanaka, I think you got the variance concept totally wrong. *Please read EIGRP chapter on Wendell Odom’s, ICND2 Official Cert Guide, 3rd edition (pg 436 – 437 to be exact)*
The above question lists a variance that is set to some number. As soon as the variance set to a value above 1, we can be confident that it’s DEFINITELY going to be un-equal cost load balancing.
The output also shown that the maximum paths is 4. So that’s how we get the answer “maximum 4 un-equal cost load balancing paths are possible with this configuration”
i’ll try to briefly explain how the ‘variance’ works. You should actually refer to the text for proper explanation.
Remember you can load balance only with the routes that are in your routing table. And you also know that the router only puts the EIGRP successor route in the routing table. For multiple successor routes to exist in the routing table, their eigrp cost must be exactly the same! Then only the router can use these paths for “equal-cost load balancing” up-to the number of paths defined by the max paths setting (default to 4 max, can go upto 16 as i can remember)
if you want to use a feasible successor for “un-equal-cost load balancing”, you must first get that route into the routing table. Remember that FS(Feasible Successor) only lives in the topology table while the successor exists in the routing table. So the condition that has to be met in order to get the FS to the routing table is this:
eigrp cost of the successor route(Feasible Distance) <= ( variance * feasible distance of the feasible successor)
Then only it will be added to the routing table for un-equal cost load balancing. but remember this choice is only available for Feasible Successors. The condition for feasible successors is that:
The Advertised Distance(aka Reported Distance) of the Feasible Successor < The feasible distance of the successor
hth someone. it's not difficult guys.. just needs to take sometime to get your head around the concept and it's pretty clear after that.
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@Nimal
Isn’t the correct formula
eigrp cost of the successor route(Feasible Distance)*variance >= ( feasible distance of the feasible successor)?
In this case the variance of 3 says- for the feasible successor route with the FD 3 times worse than the successor’s Cost is Ok to become the successor as well (no need to be exactly equal, be 3 times worse only – and you are the successor as well).
people, what do you think?
Great explanation Nimal on 8/23 Glad you pointed out that once VAR > 1 then expect un-equal cost. Yep guys you need to read your book. Also good example on Page 452 of Sybex CCNA 6th Ed.
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@eric_c, you are correct! I’ve made a mistake typing that formula there. thanks for pointing it out :) unfortunately this site doesn’t allow editing comments. So what I should have said is:
If you want to use a feasible successor for “un-equal-cost load balancing”, you must first get that route into the routing table. Remember that FS(Feasible Successor) only lives in the topology table while the successor exists in the routing table. So the condition that has to be met in order to get the FS to the routing table is this:
eigrp cost of the successor route(Feasible Distance) * variance >= ( feasible distance of the feasible successor)
And yes, if a variance of ’3′ satisfies the above formula for a Feasible Successor route, it means that FS route is 3 times worse like you said.
Basically this means: “sure we can use both routes to send packets to the same destination, and we can use the two routes to do unequal cost load balancing.. but packets going through the FS route is most likely to reach the destination slower than the successor route(hence the name unequal cost).
Whereas with equal cost load balancing, all equal cost routes are equally good to send packets to the destination..”
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qs 2 and 5 in my exam
Hi all,
I appeared for CCNA 640-802 yesterday .. I passed it with 854/1000 marks.
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I could not attempt last 5 questions due to time-out… (Don’t know how 2 hrs were not enough for me for 50 questions. Actually, I spent more time on Eigrp Lab though I already practised it on 9Tut but swapping the screens between questions,scenario and router access etc.. took my time)
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