Answer: B 

Explanation: You can use the passive-interface command in order to control the advertisement of routing information. The command enables the suppression of routing updates over some interfaces while it allows updates to be exchanged normally over other interfaces. With most routing protocols, the passive-interface command restricts outgoing advertisements only. But, when used with Enhanced Interior Gateway Routing Protocol (EIGRP), the effect is slightly different. This document demonstrates that use of the passive-interface command in EIGRP suppresses the exchange of hello packets between two routers, which results in the loss of their neighbor relationship. This stops not only routing updates from being advertised, but it also suppresses incoming routing updates. This document also discusses the configuration required in order to allow the suppression of outgoing routing updates, while it also allows incoming routing updates to be learned normally from the neighbor. 

Answer: C 

Explanation: 

Here we see that Rtr3 is configured to use PPP encapsulation, but Rtr1 has not been configured for any kind of encapsulation. The default on Cisco router serial interfaces is HDLC, not PPP, so there is an encapsulation mismatch. 

Answer: A 

Explanation: 

Router_E learns two subnets subnets 208.149.23.64 and 208.149.23.96 via Router_A 

through FastEthernet interface. The interface cost is calculated with the formula 108 / 

Bandwidth. For FastEthernet it is 108 / 100 Mbps = 108 / 100,000,000 = 1. Therefore the 

cost is 12 (learned from Router_A) + 1 = 13 for both subnets -> 

The cost through T1 link is much higher than through T3 link (T1 cost = 108 / 1.544 Mbps = 

64; T3 cost = 108 / 45 Mbps = 2) so surely OSPF will choose the path through T3 link -> 

Router_E will choose the path from Router_A through FastEthernet0/0, not Serial1/0. 

In fact, we can quickly eliminate answers B, C and D because they contain at least one 

subnet learned from Serial1/0 -> they are surely incorrect. 

Answer: A,C 

Answer: B 

Explanation: 

A broadcast storm can consume sufficient network resources so as to render the network unable to transport normal traffic. 

Topic 6, Simulation 

Answer: B,C,D 

Explanation: 

OSPF implements a two-tier hierarchical routing model that uses a core or backbone tier known as area zero (0). Attached to that backbone via area border routers (ABRs) are a number of secondary tier areas. 

The hierarchical approach is used to achieve the following: 

. Rapid convergence because of link and/or switch failures 

. Deterministic traffic recovery 

. Scalable and manageable routing hierarchy, reduced routing overhead. 

Answer: A,C 

Explanation: 

Reference 1: http://www.ciscopress.com/articles/article.asp?p=24090&seqNum=4 

Link state protocols, sometimes called shortest path first or distributed database protocols, are built around a well-known algorithm from graph theory, E. W. Dijkstra'a shortest path algorithm. Examples of link state routing protocols are: Open Shortest Path First (OSPF) for IP The ISO's Intermediate System to Intermediate System (IS-IS) for CLNS and IP DEC's DNA Phase V Novell's NetWare Link Services Protocol (NLSP) Although link state protocols are rightly considered more complex than distance vector protocols, the basic functionality is not complex at all: 

1.

 Each router establishes a relationship—an adjacency—with each of its neighbors. 

2.

 Each router sends link state advertisements (LSAs), some 

3.

 Each router stores a copy of all the LSAs it has seen in a database. If all works well, the databases in all routers should be identical. 

4.

 The completed topological database, also called the link state database, describes a graph of the internetwork. Using the Dijkstra algorithm, each router calculates the shortest path to each network and enters this information into the route table. OSPF Tutorial 

Answer: A 

Explanation: 

Serial 0/0 interface is administratively down therefore, you will have to run the "no shutdown" command to enable the interface for data. 

Answer: A 

Explanation: 

Serial Connection 

cl_3_dte_male 

cl_2_dce 

The picture on the left shows a V.35 DTE cable with a male DB60 connector and a male standard 34-pin Winchester-type connector. The right picture shows a V.35 DCE serial cable with a male DB60 connector and a female 34-pin Winchester-type connector. As you probably guessed already, the male connector of the DTE cable is attached to the DCE cable's female connector, this is depicted in the picture below. This is known as a back-to-back connection, and 'simulates' a WAN link. In a real world setup, the DTE cable's male connector typically connects to a port on a CSU/DSU provided by a service provider (i.e. telco), which in turn connects to a CSU/DSU at another location, thru a T1 link for example. 

The DB60 connector connects to a Serial interface on a router. 

cl_4_malefemale Reference: http://www.techexams.net/techlabs/ccna/lab_hardware.shtml 

Answer: B,C 

Explanation: 

http://www.cisco.com/en/US/tech/tk365/technologies_white_paper09186a0080094e9e.sht ml#t6 

The cost (also called metric) of an interface in OSPF is an indication of the overhead required to send packets across a certain interface. The cost of an interface is inversely proportional to the bandwidth of that interface. A higher bandwidth indicates a lower cost. There is more overhead (higher cost) and time delays involved in crossing a 56k serial line than crossing a 10M Ethernet line. The formula used to calculate the cost is: Cost = 10000 0000/bandwidth in bps For example, it will cost 10 EXP8/10 EXP7 = 10 to cross a 10M Ethernet line and will cost 10 EXP8/1544000 =64 to cross a T1 line. By default, the cost of an interface is calculated based on the bandwidth; you can force the cost of an interface with the ip ospf cost <value> interface subconfiguration mode command.