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2013 Latest Cisco 350-001 Exam Section 3: EIGRP (19 Questions)

QUESTION NO: 1
An EIGRP multicast flow timer is defined as which of the following?
A. The timeout timer after which EIGRP retransmits to the neighbor in non CR mode, through unicasts.
B. The time interval that EIGRP hello packets are sent.
C. The timer after which EIGRP will not forward multicast data traffic.
D. The timer interval between consecutive transmitted EIGRP hello intervals.
E. The timeout timer after which EIGRP retransmits to the neighbor in CR mode, through unicasts.
F. None of the above.
Answer: E
Explanation:
After pair of routers become neighbors, they will send routing updates (and other packets) to one another using a reliable multicast scheme. For example, if router one has a series of packets which must be transmitted to routers two, three, and four such as a routing table update, it will send the first packet to the EIGRP multicast address, 224.0.0.10, and wait for an acknowledgment from each of its neighbors on its Ethernet interface (in this case, routers two, three and four). Let’s assume that routers two and four do answer, but router three does not.
Router one will wait until the multicast flow timer expires on the Ethernet interface, then send out a special packet, a sequence TLV, telling router three not to listen to any further multicast packets from router one, then it will continue transmitting the remainder of the update packets as multicast to all other routers on the network. The sequence TLV indicates an out-of-sequence multicast packet. Those routers not listed in the packet enter Conditional Receive (CR) mode, and continue listening to multicast. While there are some routers in this mode, the Conditional Receive bit will be set in multicast packets. In this case, router one will send out a sequence TLV with router three listed, so routers two and four will continue listening to further multicast updates.

QUESTION NO: 2
Which components are factored in by default when an EIGRP metric is calculated? (Choose all that apply)
A. MTU
B. Delay
C. Load
D. Bandwidth
E. Reliability
Answer: B, D
Explanation:
By default, EIGRP uses only bandwidth and Delay when calculating the metric. EIGRP uses these scaled values to determine the total metric to the network:
1. metric = [K1 * bandwidth + (K2 * bandwidth) / (256 – load) + K3 * delay] * [K5 /
(reliability + K4)] The default values for K are:
1.
K1 = 1
2.
K2 = 0
3.
K3 = 1
4.
K4 = 0
5.
K5 = 0 For default behavior, you can simplify the formula as: Metric = Bandwidth + Delay
Incorrect Answers:
A. The MTU is tracked but never used in calculating the metric for IGRP or EIGRP at
any time.
C, E. Although Load and Reliability are K values that can indeed be factored into the
metric, by default their K value is 0 so they are not used.
Reference:

http://www.cisco.com/warp/public/103/eigrp-toc.html#eigrpmetrics

QUESTION NO: 3 The topology of a network changes causing an EIGRP router to go into the active state. The DUAL process shows a new route that meets the EIGRP Feasibility Condition. In regards to this specific route, which of the following is true?
A. The Feasible Distance of the new route must be equal to one.
B. The Feasible Distance of the new route must be higher than one.
C. The Reported Distance of the new route must be equal to Feasible Distance.
D. The Reported Distance of the new route must be higher than Feasible Distance.
E. The Reported Distance of the new route must be lower than Feasible Distance.
Answer: E
Explanation:
The following are some terms relating to EIGRP:
1.
Feasible Distance: The lowest calculated metric to each destination
2.
Feasibility Condition: A condition that is met if a neighbor’s advertised distance to a destination is lower that the router’s Feasible Distance to that same destination.
3.
Successor: The neighbor that has been selected as the next hop for a given destination based on the Feasibility Condition.
Reference:
Jeff Doyle, Routing TCP/IP, Volume I, Chapter 8: Enhanced Interior Gateway Routing Protocol (EIGRP), p.336-337, Cisco Press, (ISBN 1-57870-041-8)
Incorrect Answers:
A: The metric of the new route needs only to be less than the current metric to the
destination (feasible distance), and does not necessarily need to equal one.

B: It is feasible that the new metric to the destination could equal one, and also be lower
than the current metric.
C, D: The reported distance must be lower than the feasible distance.

Additional info:
The Feasible Condition is met when the receiving router has a Feasible Distance (FD) to a particular network and it receives an update from a neighbor with a lower advertised or Reported Distance (RD) to that network. The neighbor then becomes a Feasible Successor (FS) for that route because it is one hop closer to the destination network. There may be a number of Feasible Successors in a meshed network environment.
The RD for a neighbor to reach a particular network must always be less than the FD for the local router to reach that same network. In this way EIGRP avoids routing loops. This is why routes that have RD larger than the FD are not entered into the Topology table.
Reference:
Ravi Malhotra, IP Routing, Chapter 4: Enhanced Interior Gateway Routing Protocol (EIGRP), O’Reilly Press, January 2002 (ISBN 0-596-00275-0)

QUESTION NO: 4
Which of the following EIGRP packets require an acknowledgement? (Choose all that apply)
A. Hello
B. Query
C. Reply
D. Update
E. Ack
F. None of the above
Answer: B, C, D
Explanation:
Updates are used to convey reachability of destinations. When a new neighbor is discovered, update packets are sent so the neighbor can build up its topology table. In this case, update packets are unicast. In other cases, such as a link cost change, updates are multicast. Updates are always transmitted reliably. Queries and replies are sent when destinations go into Active state. Queries are always multicast unless they are sent in response to a received query. In this case, it is unicast back to the successor that originated the query. Replies are always sent in response to queries to indicate to the originator that it does not need to go into Active state because it has feasible successors. Replies are unicast to the originator of the query. Both queries and replies are transmitted reliably. EIGRP reliable packets are: Update, Query and Reply. EIGRP unreliable packets are: Hello and Ack.
Incorrect Answers:
A, E. Hellos are multicast for neighbor discovery/recovery. They do not require acknowledgment. A hello with no data is also used as an acknowledgment (ack). Acks are always sent using a unicast address and contain a non-zero acknowledgment number. Reference: Cisco BSCN version 1.0 study guide, pages 6-18.

QUESTION NO: 5
Which of the following types of EIGRP packets contain the Init flag?

A. Hello/Ack
B. Query
C. Reply
D. Update
E. None of the above
Answer: D
Explanation:
In EIGRP header there is an 8-bit flag value. The rightmost bit is init.
Which when set to 0x00000001 indicates that the enclosed route entries are the first in a
new neighbor relationship.
Also the route entries are carried in update packet not hello packet.

Additional Info:
The following debug output displays the Init Sequence increasing only with the update packet. Router# debug eigrp packetEIGRP: Sending HELLO on Ethernet0/1 AS 109, Flags 0x0, Seq 0, Ack 0EIGRP: Sending HELLO on Ethernet0/1 AS 109, Flags 0x0, Seq 0, Ack 0EIGRP: Sending HELLO on Ethernet0/1 AS 109, Flags 0x0, Seq 0, Ack 0EIGRP: Received UPDATE on Ethernet0/1 from 192.195.78.24, AS 109, Flags 0x1, Seq 1, Ack 0EIGRP: Sending HELLO/ACK on Ethernet0/1 to 192.195.78.24, AS 109, Flags 0x0, Seq 0, Ack 1EIGRP: Sending HELLO/ACK on Ethernet0/1 to 192.195.78.24, AS 109, Flags 0x0, Seq 0, Ack 1EIGRP: Received UPDATE on Ethernet0/1 from 192.195.78.24, AS 109, Flags 0x0, Seq 2, Ack 0Incorrect Answers:
A. Hellos are multicast for neighbor discovery/recovery. They do not require acknowledgment. A hello with no data is also used as an acknowledgment (ack). Acks are always sent using a unicast address and contain a non-zero acknowledgment number. B, C. Queries and replies are sent when destinations go into Active state. Replies are always sent in response to queries to indicate to the originator that it does not need to go into Active state because it has feasible successors. Replies are unicast to the originator of the query. Both queries and replies are transmitted reliably. Reference: “Routing TCP/IP” Jeff Doyle Pg364

QUESTION NO: 6 In your EIGRP network you notice that the neighbor relationship between two of your routers was recently restarted. Which of the following could have occurred to have caused this? (Choose all that apply)
A. The clear ip route command was issued.
B. The ARP cache was cleared.
C. The IP cache was cleared.
D. An update packet with Init flag set from a known, already established neighbor relationship was received by one of the routers.
E. The IP EIGRP neighbor relationship was cleared manually.
Answer: D, E
Explanation:
D as well as E will result in EIGRP relationship to be restarted.
The reason for D: If a router receives an update packet with the init flag set it clearly
implies that this packet is the first after a new neighbor relationship has been established.
The reason for E: If we clear the IP EIGRP neighbor relationship it will automatically
result in EIGRP neighbor relationship to be restarted.

Incorrect Answers:
A. This will clear the IP routing table, but will not have any affect on the EIGRP neighbor relationship.
B. This will only clear the MAC address learned ARP cache.
C. This also will not have any affect on the EIGRP neighbor relationship.

QUESTION NO: 7
The TestKing EIGRP network has a router named Router TK2. Router TK2 is connected to an EIGRP neighbor, TK1. TK1 is defined as a stub. With regard to this network, which of the following are true?
A. Router TK1 will not advertise any network routes to TK2.
B. Router TK2 will send only summary routes to TK1.
C. Router TK2 will not query TK1 about any internal route.
D. Router TK2 will not query TK1 about any external route.
E. Router TK2 will not query TK1 about any route.
F. None of the above.
Answer: E
Explanation:
E is the best choice, as an EIGRP router will not query a stub neighbor about any route.
Incorrect Answers:
A. TK1 will still be required to advertise its network routes to the neighbor, even though
it is configured as a stub.

B. TK2 still sends all routes to TK1.
C, D. Although both of these are true, since TK2 will not query TK1 about any route, E is
a better choice.

Reference:
http://www.cisco.com/univercd/cc/td/doc/product/software/ios120/120newft/120limit/120s/120s15/eigrpst

QUESTION NO: 8
The TestKing network uses EIGRP as the routing protocol and has an ISDN connection that is used as a backup to their frame relay network. The ISDN link successfully comes up when the frame relay network fails, but no routing traffic will pass over the ISDN link. What could be the cause of this problem?
A. The dialer-list is blocking EIGRP.
B. The EIGRP configuration is incorrect.
C. The encapsulation is different on the opposite ends of the link.
D. There is a network type mismatch.
E. The broadcast keyword is missing from the dialer-map.
Answer: E
Explanation:
For routing protocol traffic to pass over the ISDN link, the broadcast keyword must be present in the dialer map.
Incorrect Answers:
A. Although the dialer list may indeed be blocking EIGRP updates, so that these updates do not initiate ISDN calls, one the ISDN link is up, all traffic will be able to traverse this link, and not just the traffic that is defined as interesting. C, D. If this were the case, there would be a problem with the ISDN link connecting in the first place.

QUESTION NO: 9
How is the metric for a summarized route derived when the interface summary command for EIGRP is used?
A. It is derived from the route that has the biggest metric.
B. It is derived from the route that has the smallest metric.
C. It is derived from the interface that has the summary command configured on it.
D. It is derived from the route that has the shortest matching mask.
E. It is derived from the default-metric.
Answer: B
Explanation:
According to Cisco’s EIGRP design guide, “The metric is the best metric from among the
summarized routes.”
Reference:

http://www.cisco.com/univercd/cc/td/doc/product/software/ios122/122cgcr/fipr_c/ipcprt2/1cfeigrp.htm#1001078

“…EIGRP will advertise the summary address out the interface with a metric equal to the
minimum of all more specific routes.”

QUESTION NO: 10
Routers Testking1, Testking2, and Testking3 are all configured for EIGRP as shown below:
Testking1 has the following configuration:
router eigrp 1 network 192.168.10.0 redistribute connected Which routes would show up in the routing table of Testking3 as EIGRP routes? (Choose all that apply)
A. 10.1.0.0/16
B. 10.0.0.0/24
C. 10.0.0.0/8
D. 10.1.1.0/24
E. 192.168.10.0/24
Answer: C, E
Explanation: EIGRP will perform auto-summarization of External Routes. Since the 10.1.1.0 network was redistributed into EIGRP via a connected network, this will automatically make this route external to EIGRP. The 192.168.10.0 network will also show up in the routing table as an EIGRP route through the normal EIGRP process. Additional Info:
Auto-Summarization
EIGRP performs an auto-summarization
each time it crosses a border between two different major networks.

For example, in Figure 13, Router Two advertises only the 10.0.0.0/8 network to Router
One, because the interface Router Two uses to reach Router One is in a different major
network.

On Router One, this looks like the following:
one#show ip eigrp topology 10.0.0.0
IP-EIGRP topology entry for 10.0.0.0/8
State is Passive, Query origin flag is 1, 1 Successor(s), FD is 11023872
Routing Descriptor Blocks:

172.16.1.1 (Serial0), from 172.16.1.2, Send flag is 0x0
Composite metric is (11023872/10511872), Route is Internal
Vector metric:
Minimum bandwidth is 256 Kbit
Total delay is 40000 microseconds
Reliability is 255/255
Load is 1/255
Minimum MTU is 1500
Hop count is 1

metric is the best metric from among the summarized routes. Note that the minimum bandwidth

56k.
Leading the way in IT testing and certification tools, www.testking.com
– 236 –
On the router doing the summarization, a route is built to null0 for the summarized address:
two#show ip route 10.0.0.0 Routing entry for 10.0.0.0/8, 4 known subnets Attached (2 connections) Variably subnetted with 2 masks Redistributing via eigrp 2000 C 10.1.3.0/24 is directly connected, Serial2 D 10.1.2.0/24 [90/10537472] via 10.1.1.2, 00:23:24, Serial1 D 10.0.0.0/8 is a summary, 00:23:20, Null0 C 10.1.1.0/24 is directly connected, Serial1
The route to 10.0.0.0/8 is marked as a summary through Null0. The topology table entry for this summary route looks like the following:
two#show ip eigrp topology 10.0.0.0 IP-EIGRP topology entry for 10.0.0.0/8 State is Passive, Query origin flag is 1, 1 Successor(s), FD is 10511872 Routing Descriptor Blocks:
0.0.0.0 (Null0), from 0.0.0.0, Send flag is 0x0
(note: the 0.0.0.0 here means this route is originated by this router)
Composite metric is (10511872/0), Route is Internal
Vector metric:
Minimum bandwidth is 256 Kbit
Total delay is 20000 microseconds
Reliability is 255/255
Load is 1/255
Minimum MTU is 1500
Hop count is 0

Incorrect Answers:
A, B, D. Any more specific routes in the 10.0.0.0 network will be summarized into one 10.0.0.0/8 network. Again, since the 10.0.0.0 network was learned by EIGRP only via redistribution, it is external as far as EIGRP is concerned.

QUESTION NO: 11 The Testking EIGRP network topology is displayed below, along with the EIGRP metric values for each link: From the perspective of router A shown above, which of the following routers would be considered the successor and the feasible successors to the 172.16.9.0/24 network? (Select two choices below)
A. B is the successor
B. C is the successor
C. D is the successor
D. B is a feasible successor
E. C is a feasible successor
F. D is a feasible successor.
G. E is a feasible successor
Answer: A, E
Explanation:
The following are some terms relating to EIGRP:
1.
Feasible Distance: The lowest calculated metric to each destination
2.
Feasibility Condition: A condition that is met if a neighbor’s advertised distance to a destination is lower that the router’s Feasible Distance to that same destination.
3.
Successor: The neighbor that has been selected as the next hop for a given destination based on the Feasibility Condition.
4.
Feasible Successor: A neighbor whose Reported Distance (RD) is less than the Feasible Distance (FD). The Feasible Condition is met when the receiving router has a Feasible Distance (FD) to a particular network and it receives an update from a neighbor with a lower advertised or Reported Distance (RD) to that network. The neighbor then becomes a Feasible Successor (FS) for that route because it is one hop closer to the destination network. There may be a number of Feasible Successors in a meshed network environment. The RD for a neighbor to reach a particular network must always be less than the FD for the local router to reach that same network. In this way EIGRP avoids routing loops. This is why routes that have RD larger than the FD are not entered into the Topology table. In this example, Router B would be the successor, with a feasible distance of 7680 (2560+2560+2560). Therefore, only routers with an AD of less than 7680 will become successors. In this case, router C will have an Advertised Distance of 6400 so it is a FS. Router D has a RD of 10240, and since it must be less than the current FD, it will not become a FS.
Reference:
Jeff Doyle, Routing TCP/IP, Volume I, Chapter 8: Enhanced Interior Gateway Routing Protocol (EIGRP), p.336-337, Cisco Press, (ISBN 1-57870-041-8)
The following display further describes an example of a Feasible Successor:

QUESTION NO: 12
The Testking EIGRP network is displayed in the following diagram:

The associated EIGRP metric is listed as shown above for each of the links. Based on this information, what is the reported distance (advertised distance) to network 172.16.9.0/24 from router C to router A?
A. 5120
B. 6400
C. 17,920
D. 10,240
E. 11,520
F. 2560
G. None of the above
Answer: B
Explanation:
The following are some terms relating to EIGRP:
1.
Feasible Distance: The lowest calculated metric to each destination

2.
Feasibility Condition: A condition that is met if a neighbor’s advertised distance to a
destination is lower that the router’s Feasible Distance to that same destination.

3.
Successor: The neighbor that has been selected as the next hop for a given destination
based on the Feasibility Condition.

4.
Feasible Successor: A neighbor whose Reported Distance (RD) is less than the
Feasible Distance (FD).
In the example above, the Feasible Distance to network 172.16.9.0/24 from C to A would
be the distance that this network is from router C. In this case, the distance is
2560+3840=6400, so Choice B is correct.
The following diagram provides another example for calculating the RD:

QUESTION NO: 13
The Testking EIGRP network is displayed in the exhibit below:

The “Show IP EIGRP neighbor” command is issued on the router A. Router A is configured with the default EIGRP settings. After issuing this command, which of the following answer choices correctly describe the expected output?
A. routerA#show ip eigrp neighbor IP-EIGRP neighbors for process 1 H Address Interface Hold Uptime SRTT RTO Q Seq
(Sec) (ms) Cnt Num
1 10.1.1.2 Et1 13 12:00:53 12 300 0 620
0 10.1.2.2 S0 174 12:00:56 17 200 0 645
B.
routerA#show ip eigrp neighbor IP-EIGRP neighbors for process 1 H Address Interface Hold Uptime SRTT RTO Q Seq
(Sec) (ms) Cnt Num
1 10.1.1.2 Et1 20 12:00:53 12 300 0 620
0 10.1.2.2 S0 190 12:00:56 17 200 0 645
C.
routerA#show ip eigrp neighbor IP-EIGRP neighbors for process 1 H Address Interface Hold Uptime SRTT RTO Q Seq
(Sec) (ms) Cnt Num
1 10.1.1.2 Et1 174 12:00:53 12 300 0 620
0 10.1.2.2 S0 13 12:00:56 17 200 0 645

D.
routerA#show ip eigrp neighbor IP-EIGRP neighbors for process 1 H Address Interface Hold Uptime SRTT RTO Q Seq
(Sec) (ms) Cnt Num
1 10.1.1.2 Et1 185 12:00:53 12 300 0 620
0 10.1.2.2 S0 19 12:00:56 17 200 0 645
Answer: A
Explanation:
The value in the Hold column of the command output should never exceed the hold time,
and should never be less than the hold time minus the hello interval (unless, of course,
you are losing hello packets). If the Hold column usually ranges between 10 and 15
seconds, the hello interval is 5 seconds and the hold time is 15 seconds. If the Hold
column usually has a wider range – between 120 and 180 seconds – the hello interval is 60
seconds and the hold time is 180 seconds.
The EIGRP default timer settings are:
Hello Interval: 5 seconds for all high speed links

60 seconds for low speed links (T1 or less)

The default hold timer is less 3 times the hello interval. Since this question tells us that
the default values are used, the Router A would have a value of not more than 15 seconds
for the Ethernet peer and 180 seconds for the serial peer, so choice A is correct.

Incorrect Answers:
B, D. The timers for both the Ethernet and serial peers are above the maximum theoretical values for a working EIGRP network, assuming that the default timers are being used.
C. The timer values in this choice is wrong. High speed links such as ethernet use a shorter hello interval than low speed T1 links.

QUESTION NO: 14 The Testking EIGRP network, along with the configured bandwidth statements of the routers and the delay of each link, is shown below: Assuming that all EIGRP routers are all using default configurations, what path would router A choose to route packets to network A?
A. Router A takes the path through router B.
B. Router A takes the path through router C.
C. Router A would load balance to both router B and router C.
D. Neither path would be chosen as there is a loop in the network.
E. The metrics shown are too large, and the route to network A would be considered unreachable.
Answer: B
Explanation:
When all 5 of the K values are set to the default values, the EIGRP metric calculation for
each link is found by the default formula of (Bandwidth + Delay) x 256. The metric
calculation is the same as IGRP, but the result is multiplied by 256 for finer granularity.
In this case, the bandwidth component is found in the same was as the OSPF metric,
which is 10,000,000/bandwidth. It is important to note that only the minimum outgoing
bandwidth is used, so along any path from A to Z, the slowest link among all the hops is
used as the chosen metric for the bandwidth portion. This is true for both IGRP and
EIGRP (See Routing TCP/IP by Jeff Doyle, page 243-244). The delay metric is found by
adding the total delay of the path (in microseconds) and dividing by 10.

For this question, the shortest path can be found by comparing the two different choices
that we really have (through router B or router C). For the path through router B the
bandwidth metric is:
(10 million/56) + (24000/10) x 256 = (178571 + 2400) x 256 = 46328683.

For the path going through router C:
(10 million/128) + (70000/10) x 256 = (78125 + 7000) x 256 = 21792000.

Note that only the lowest bandwidth metric was used along the entire path, where as the delay was added at each hop. Also note that the calculation for the path from router D to network A was omitted, since this value would be simply added to the metrics above would not change the answer.
Incorrect Answers:
A. The path through this router has a higher metric and so would not be used.
C. By default, EIGRP would load balance over equal cost paths. Although these paths are not equally valued, load balancing could occur despite this if the “variance” EIGRP feature was used. However, the variance command is not enabled by default.
D. Although a loop does exist, EIGRP routers maintain loop avoidance techniques, including keeping track of hop counts used. For IGRP and EIGRP, there is a maximum hop count of 100 hops.
E. Both of the metric listed above are well within the maximum limits set by EIGRP.

QUESTION NO: 15 The TestKing network is using EIGRP as the routing protocol, and the EIGRP topology information for router R4 is displayed below:
Based on the information above, which of the following statements is true?
A. The routers 116.16.34.3 and 116.16.34.9 are EIGRP neighbors to TK4.
B. The 116.16.37.0 network is reachable via the 116.16.34.9 interface.
C. A static route has been configured to summarize the 140.0.0.0 network and route it to the NULL 0 interface.
D. Interface serial 1/0 is most likely a frame relay interface with four DLCls: one to the
170.170.0.0 network, one to the 130.130.0.0 network and one to the 116.16.37.0 network.
E. All of the above
Answer: A
Explanation:
The IP address following the “via” entry is the peer that told the software about this destination. When issuing this command, the first n of these entries, where N is the number of successors, are the current successors. The remaining entries on the list are feasible successors. In the example above, the router TK4 is learning routes from both of these two peers, so they are EIGRP neighbors to TK4.
Incorrect Answers:
B. This network is reachable via the 116.16.34.3 neighbor, not 116.16.34.2.
C. The routing entry for the 140.0.0.0/8 network is known via the summary, not a static route. EIGRP uses auto-summarization by default, which has produced this route.
D. All three of these networks are known via the same IP peer. Although it is possible that 4 separate PVC’s are built to the same IP address peer, there is no reason to assume that this is the case in this example. It actually looks like there may be 3 total DLCIs on this serial interface, not 4.

QUESTION NO: 16 The Teskking WAN is displayed in the diagram below, along with the partial configuration files of routers TK1 and TK2:
Based on the above information, what would be the most likely reason that the routing tables do not contain routes for each of the remote networks?
A. The routers interfaces are not functioning properly.
B. IP routing is not enabled on the routers.
C. The routers are not members of the same autonomous system.
D. The routers only pass locally significant routing information.
E. The routers are using different routing protocols.
F. Auto-summarization is not disabled on the routers.
Answer: C
Explanation:
The number following the “router eigrp” command is known as the process ID, and is used to denote the Autonomous System of the network that the router is in. The process ID can be any number between 1 and 65535 (0 is not allowed) and it can be randomly chosen, as long as it is the same for all EIGRP processes in routers that are to share the routing information. In the example above, router TK1 is configured to use EIGRP process 10, while router TK2 is using EIGRP process 11 Reference: Jeff Doyle, “Routing TCP/IP volume 1” page 377.

QUESTION NO: 17
Two TestKing routers are configured for EIGRP as shown in the following exhibit:

Router TestKing1 and router TestKing2 are unable to form an EIGRP neighbor relationship. What are two reasons for this problem? (Select two).
A. The bandwidth settings on the interfaces do not match.
B. The routers belong to different autonomous systems.
C. EIGRP can not form a neighbor relationship using secondary addresses.
D. The network statement under router EIGRP does not match the subnetted network configured on the Ethernet interface.
E. Auto summarization has not been correctly configured on router TestKing1.
Answer: B, C
Explanation:
EIGRP, unlike OSPF, checks for the Autonomous System number on neighboring routers before becoming neighbors. EIGRP will only form a neighbor relationship with other routers in the same AS. Since EIGRP always sources data packets from the primary address, Cisco recommends that you configure all routers on a particular subnet with primary addresses that belong to the same subnet. Routers do not form EIGRP neighbors over secondary networks. Therefore, if all routers’ primary IP addresses do not agree, problems can arise with neighbor adjacencies.
Incorrect Answers:

A. Manually setting the bandwidth will affect the overall metric of the individual EIGRP routes, but will not affect the state of the neighbor relationship.
D. Although they do not match, EIGRP will still work as long as the EIGRP interface is included within the subnet mask used for the EIGRP process.
E. In EIGRP, automatic summarization is on by default. Whether this is enabled or disabled will have no effect on the neighbor relationship.

QUESTION NO: 18
TestKing.com is designing a large network with core, distribution, and access layers. EIGRP is the routing protocol that will be used throughout the network. Each distribution router has WAN connectivity to at least 20 access routers. Every router in the network has an explicit route to every possible subnet. All hosts in the network should be able to reach any other host, anywhere within the network. What should be done to optimize the routing configuration?
A. Ensure IP address space is allocated so that routes can be summarized at the core routers.
B. Filter routes in the distribution layer so that every access router doesn’t have an explicit route to every subnet.
C. Filter routes in the access layer so that every access router doesn’t have an explicit route to every subnet.
D. Ensure IP address space is allocated so that routes can be summarized at each distribution router.
Answer: D
Explanation:
The best way to reduce the number of routes within the routing table is via route summarization. In order to optimize the network using this approach, summarization should take place on the distribution layer of a three tiered network design. When doing this, it is important to ensure that proper planning takes place to ensure that enough IP address space is allocated at each distribution router, in order to summarize all of the remote locations into one single network route. Incorrect Answers:
A. Core routers should focus solely on routing data packets as quickly as possible. The use of any ancillary technologies such as access lists, packet classification, and route filtering. These technologies are best suited to be placed on the distribution layer network devices.
B, C: Either of these choices could result in some hosts becoming unreachable from other hosts within the TestKing network.

QUESTION NO: 19
The TestKing frame relay network is displayed below:

Assume that no subinterfaces are used on router TestKing1 and EIGRP is the routing protocol in use. What is the effect on routing updates if router TestKing1 learns about network A from router TestKing2, assuming default configurations are being used?
A. Router TestKing1 will advertise the route to network A to router TestKing3.
B. Router TestKing1 will advertise the route to network A to router TestKing2 and TestKing3.
C. Router TestKing1 will load balance between router TestKing2 and router TestKing3.
D. Router TestKing1 will not advertise the route to network A to router TestKing3.
Answer: D
Explanation:
Split horizon controls the sending of IP Enhanced IGRP update and query packets. When split horizon is enabled on an interface, these packets are not sent for destinations for which this interface is the next hop. This reduces the possibility of routing loops. By default, split horizon is enabled on all interfaces.
Split horizon blocks route information from being advertised by a router out of any interface from which that information originated. This behavior usually optimizes communications among multiple routing devices, particularly when links are broken. However, with nonbroadcast networks (such as Frame Relay and SMDS), situations can arise for which this behavior is less than ideal. In the example above, when router TestKing1 sees the route advertisement from TestKing2, it will not advertise this route out the same interface, which is also the connection to TestKing3. If we were to use sub-interfaces or disable split horizons on TestKing1, then it would indeed advertise the route.
Incorrect Answers:
A, B: The Split Horizon rule, which is enabled by default, prevents this.
C: TestKing1 will only load balance over equal cost routes by default. In this case,
TestKing2 will always be used to reach network A since the route via TestKing3 will be
seen with a higher metric.
Note: We could configure TestKing1 to load balance in this situation, using the
“variance” router process configuration command.

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