Correct Answers: D

Explanation:
This issue would NOT be caused by executing the passive interface command on the Gi0/4 interface of R1. This command prevents the advertisement of RIP routes on that interface. If that command had been issued, the traffic would not be forwarded to R1 because R3 would not know about the route to the LAN off of R2. This command would also prevent R3 from knowing about the default route to ISP1. Since the traffic is being routed to ISP1, this command must not have been executed.

All of the other options could potentially because traffic destined for R2 to be forwarded from R1 to ISP1, rather than to R2.

It is true that a default route exists on R1 that leads to ISP1. If this default route did not exist, the traffic destined for R2 would simply be dropped at R1 instead of being forwarded to ISP1.

If the network command has not been executed on the interface leading to the LAN off of R2, the network leading to the LAN off R2 would not advertised by R2.
That would make R1 unaware of this destination. In that case, R1 would use the default route to send traffic destined for R2 to ISP1. We know such a default route must exist, or the traffic would simply be dropped at R1.

If RIPv2 has not been enabled on R2, R2 would not be receiving or advertising any RIP routes. When the packets destined for the network off of R2 arrive at R1, R1 will have not have a route to that network. In that case R1 will forward the traffic to ISP1 using the default route.

Correct Answers: A

Explanation:
The best route to the 192.168.5.0/24 network from the perspective of router F will have an OSPF assigned cost of 2. There are three possible loop-free paths to get from router F to the 192.168.5.0/24 network. The default OSPF costs for a 100 MB link, a T1 link, and a T3 link are 1, 64, and 2, respectively.

The three paths and the calculation of their costs are shown:

Router F to Router E to Router A: 1 + 1 = 2 Router F to Router C to Router A: 2 + 1 = 3
Router F to Router B to Router D to Router C to Router A: 64 + 64 + 64 + 1 = 193

Each OSPF route calculates the cost of its path to a network, and passes that value on to the next router, which will then add to it the cost to reach that neighbor. For example, the routing table of Router E would look like this for the route to 192.168.5.0/24:

O 192.168.5.0 [110/1] via

Router F would add its own cost to reach Router E to the cost of reaching 192.168.5.0/24, resulting in the following output:

O 192.168.5.0 [110/2] via

110 is the administrative distance of OSPF.

Correct Answers: B

Explanation:
Host A is not receiving a DHCP configuration because its initial DHCP Discover frame is a broadcast, and routers do not forward broadcast frames by default.

A DHCP client sends out a DHCP Discover packet when booting up, enveloped within an Ethernet broadcast frame. The broadcast frame will be flooded by switches, but filtered by routers. There must either be a DHCP server on the local subnet or a DHCP Relay Agent, which will forward the request from the local subnet to the DHCP server.

The DHCP server is not on the wrong subnet. A DHCP server can be centrally located and configured to support multiple remote subnets, as long as those subnets have DHCP Relay Agents configured to forward the DHCP Discover requests.

No information is provided on the DHCP server configuration. The router is the most obvious cause of the problem, so this option is incorrect.

Port security can be configured to restrict hosts based on the MAC address, but the scenario does not provide information on any port security configurations. The router is the most obvious cause of the problem as shown in the network exhibit.

Correct Answers: A, F, G

Explanation:
The given exhibit has three problems:
– The 172.16.1.0/30 segment requires more user address space.
– Interface Fa0/4 has an IP address that belongs to the 172.16.2.0/26 segment. Interface –
– Fa0/3 has an IP address outside the 172.16.3.0/25 segment.

The 172.16.1.0/30 segment, as configured, will only support two hosts. This segment needs to support three hosts, the two servers, and the Fa0/1 interface. The number of hosts that a subnet is capable of supporting is a function of the number of host bits in the subnet mask. When that has been determined, the following formula can be used to determine the number of hosts yielded by the mask:
2n – 2 = X
(where n = the number of host bits in the mask and X = the number of hosts supported)

In this example with a 30-bit mask, 2 host bits are left in the mask. When that is plugged into the formula, it yields only two usable addresses. The -2 in the formula represents the two addresses in each subnet that cannot be assigned to hosts, the network ID and the broadcast address. Therefore, the segment should be configured with the 172.16.1.0/29 address range, which supports up to six hosts.

Interface fa0/4, as configured, has an IP address that belongs to the 172.16.2.0/26 segment. With a 26-bit mask and the chosen class B address, the following network IDs are created:

172.16.0.0
172.16.0.64
172.16.1.128
172.16.1.192
172.16.2.0
172.16.2.64
172.16.2.128
172.16.2.192
172.16.2.0
172.16.2.64
172.16.2.128
172.16.2.192
…and so on, incrementing each time by 64 in the last octet

The 172.16.2.0/26 segment is allocated host addresses in the 172.16.2.1 through 172.16.2.62 range (the last address, 172.16.2.63, is the broadcast address and cannot be assigned). Interface fa0/4 should be assigned an IP address in the 172.16.2.64/26 range, which includes host addresses in the 172.16.2.65 through 172.16.2.126 range.

Interface Fa0/3, as configured, has an IP address outside the 172.16.3.0/25 segment. With a 25-bit mask and the chosen class B address, the following network IDs are created:

172.16.0.0
172.16.0.128
172.16.1.0
172.16.1.128
172.16.2.0
172.16.2.128
172.16.3.0
172.16.3.128
…and so on, incrementing each time by 128 in the last octet

Interface Fa0/3 should be allocated an IP address in the 172.16.3.1 through 172.16.3.126 range.

The 172.16.2.0/26 segment does not require more user address space. With a 26-bit mask, 6 bits are left for hosts, and by using the above formula it can be determined that it will yield 62 hosts. It requires 32.

The 172.16.2.64/26 segment does not require more user address space. With a 26-bit mask, 6 bits are left for hosts, and by using the above formula it can be determined that it will yield 62 hosts. It requires 32.

Interface Fa0/2 does not have an IP address that belongs to the 172.16.2.64/26 segment. The 172.16.2.64/26 segment includes addresses 172.16.2.65- 172.16.5.126. Because its address is 172.16.2.1, it belongs in the 172.16.2.0/26 network (from 172.16.2.1-172.16.2.62), so it is correctly configured.

Correct Answers: D

Explanation:
Router C is not displayed in the neighbor table of Router A, which indicates that Router C and Router A are not forming a neighbor relationship or exchanging information. This is because Router C does not have EIGRP configured for its S1 interface. You can see this is missing from its configuration in the output of the show run command for RouterC. To solve the issue, you should execute the network 192.168.9.0 command under the EIGRP 56 configuration on Router C. Then Router C will start sending hellos on that interface and the two routers will become neighbors.

The show ip eigrp neighbors command displays the following information for each EIGRP neighbor. In parentheses is the value of each found in the output of router A for Router B:

IP address (192.168.10.2)
Local interface (S1)
Retransmit interval (13)
Queue count (100)

There is no need to execute the auto-summary command on Router A. It will not affect the establishment of a neighbor relationship between Routers A and C.

There is no need to correct the IP address on the S1 interface of Router C. The address 192.168.9.1 is correctly located in the same subnet as the address on S0 of Router A.

Finally, changing the EIGRP configuration on Router C to EIGRP 55 will not help. Router C will not start sending hellos on its S1 interface until EIGRP is enabled on the S1 interface. Until then, the Routers A and C will not form a neighbor relationship and will not share information.

Correct Answers: B, D

Explanation:
If both RIPv2 and EIGRP are in use, the EIGRP route will be placed in the routing table. If RIPv2 is in use, the path taken will be R2 – R3.

EIGRP has a default administrative distance (AD) of 90, while RIPv2 has a default administrative distance (AD) of 120. The route learned by the routing protocol with the lowest AD will be placed in the routing table.

If you wanted to force R2 to use the RIPv2 route instead of the EIGRP route, this could be accomplished by changing the administrative distance of RIPv2 to a value less than 90, such as 80. The commands that would accomplish this are:

R2(config)# router rip
R2(config-router)# distance 80

If either of the versions of RIP is in use, hop count is used to determine the route. The path with the least number of hops is R2 – R3. If RIPv1 is in use, the path taken would be R2 – R3, not R2 – R4 – R3, because R2 – R3 has a lower hop count.

If EIGRP is in use, the path R2 – R4 – R3 will not be the only path taken. EIGRP load-balances two equal cost paths when they exist, and R2 – R4 – R3 and R2 – R1 – R3 are of equal cost so would both be used.

Correct Answers: A

Explanation:
A hub should be placed at the highlighted box to produce a total of two broadcast domains and seven collision domains in the network. Network devices segment collision domains and broadcast domains in the following manner:

– Hub: A Layer 1 device with all ports in same collision domain and broadcast domain.

– Bridge/Switch: Layer 2 devices on which all ports are in different collision domains, but in the same broadcast domain (assuming that all ports are in the same VLAN or no VLAN is configured).

– Routers: A Layer 3 device on which every port is a separate collision as well as broadcast domain.

The bridge shown in the graphic has three ports populated by active links, resulting in three collision domains. The switch shown in the exhibit has four ports populated with the links, resulting in four collision domains. Together these two devices create seven collision domains.
Because the scenario requires that there be no more than seven collision domains, the device in the highlighted box must not create any further collision domains. A hub is a device that has all its ports in the same collision domain and will not create any further collision domains in the topology.

A bridge or switch cannot be the correct option because these will also add collision domains.

In the exhibit, the router has two ports with active links, which will result into two broadcast domains. Because the scenario states there are no more than two broadcast domains, the device in the highlighted box must not be a router. Routers are used to segment broadcast domains.

Correct Answers: D

Explanation:
The command show standby brief displays the output in the exhibit. It is used to display a summary of the HSRP groups of which the switch is a member. The summary information it provides includes the group number, priority, state, active device address, standby address, and group address. In the exhibit, the interface VLAN 64 is a member of HSRP group 2. Its priority in the group is 100 and it is currently the standby switch. Since preemption is configured (as indicated by the P following the priority), we know that the priority of this switch must be lower than the priority of the active device. The active device has an IP address of 192.168.64.10 and the group IP address is 192.168.64.1.

The command show standby can be used to display detailed information about HSRP groups of which a switch is a member. It does not provide the quick summary display of the exhibit. This command displays information about HSRP on all configured interfaces and for all HSRP groups. It also displays hello timer information and the expiration timer for the standby switch. The command syntax is show standby [type number [group]].

Below is an example of this command’s output:

In the above output, Router A is load-sharing traffic for VLAN 5. It is active for group 1 and standby for group 2. The router at address 192.168.23.3 is active for group 2 and standby for group 1. This allows traffic to be sent to both routers while still allowing for redundancy. Router A was also configured with the standby 1 preempt command (results seen in line 1), which allows it to resume its role as active for group 1 if it comes back up from an outage.

The command show interface vlan is not a complete command. A VLAN number must follow the command. When provided with a VLAN number, the output would display the status of the SVI, but no HSRP information.
The command show hsrp is not a valid command due to incorrect syntax.

Correct Answers: D

Explanation:
The firewall has as its primary responsibility the regulation of network traffic flow based on different trust levels for different computers or networks. In the network diagram shown in the exhibit, a firewall protects the network from unauthorized access attempts. A firewall can be implemented in hardware or software. Firewalls permit, deny, or filter data packets coming into and going out of the network. This helps prevent unauthorized access attempts from outside the network.

The primary function of a router is to perform routing between two subnets or between dissimilar network technologies. Routers can provide limited firewall functionality, but a firewall is a dedicated hardware or software solution with the primary responsibility of securing the network. A router does not have as its primary responsibility the regulation of network traffic flow based on different trust levels.

Switches work at Layer 2 in the Open System Interconnection (OSI) model and perform the function of separating collision domains. A switch does not have as its primary responsibility the regulation of network traffic flow based on different trust levels.

A hub is a device that provides a common connection point for network devices. The primary responsibility of a hub is not to regulate network traffic flow based on different trust levels.

Correct Answers: A

Explanation:
SwitchA will become the root bridge. The bridge ID, also known as the switch ID, is used to elect the root bridge in a redundant network topology. The bridge ID has two components:
– Switch’s priority number: Configured as 32768 on Cisco switches by default
– Switch’s Media Access Control (MAC) address: The burnt-in hardware address of the network interface card

The switch with the lowest bridge ID is selected as the root bridge. If the same priority number is configured on two or more switches in the network, the switch with the lowest MAC address will become the root. Bridge Protocol Data Units (BPDUs) communicate the details of the switch with the lowest bridge ID in the network. The election process for the root bridge takes place every time there is a topology change in the network. A topology change may occur due to the failure of a root bridge or the addition of a new switch in the network. The root bridge originates BPDUs every two seconds, which are propagated by other switches throughout the network. BPDUs are used as keepalives between switches, and if a switch stops receiving BPDUs from a neighboring switch for ten intervals (20 seconds), it will assume a designated role for the network segment.

Neither SwitchB nor SwitchC will become the root bridge. Although both have an equal priority value to SwitchA (32768), the MAC addresses of SwitchB and SwitchC are higher than that of SwitchA.

The root bridge can be determined with the information given. If the diagram did not indicate MAC addresses, then the root bridge would not be able to be determined, since the priorities are equal.