1. D and F. Of the remaining answers, Ethernet defines both physical and data-link protocols, PPP is a data-link protocol, IP is a network layer protocol, and SMTP and HTTP are application layer protocols.
2. A and G. Of the remaining answers, IP is a network layer protocol, TCP and UDP are transport layer protocols, and SMTP and HTTP are application layer protocols.
3. B. Adjacent-layer interaction occurs on one computer, with two adjacent layers in the model. The higher layer requests services from the next lower layer, and the lower layer provides the services to the next higher layer.
4. B. Same-layer interaction occurs on multiple computers. The functions defined by that layer typically need to be accomplished by multiple computers—for example, the sender setting a sequence number for a segment and the receiver acknowledging receipt of that segment. A single layer defines that process, but the implementation of that layer on multiple devices is required to accomplish the function.
5. A. Encapsulation is defined as the process of adding a header in front of data supplied by a higher layer (and possibly adding a trailer as well).
6. D. By convention, the term frame refers to the part of a network message that includes the data-link header and trailer, with encapsulated data. The term packet omits the data-link header and trailer, leaving the network layer header with its encapsulated data. The term segment omits the network layer header, leaving the transport layer header and its encapsulated data.
7. B. The term frame refers to the data-link (that is, Layer 2) data structure created by a Layer 2 protocol. As a result, the matching OSI term for protocol data units (PDUs) mentions that same layer, that is, Layer 2 PDU, or L2PDU.
1. A. The IEEE defines Ethernet LAN standards, with standard names that begin with 802.3, all of which happen to use cabling. The IEEE also defines wireless LAN standards, with standard names that begin with 802.11, which are separate standards from Ethernet.
2. C. The number before the word BASE defines the speed, in megabits per second (Mbps). 1000 Mbps equals 1 gigabit per second (1 Gbps). The T in the suffix implies twisted-pair or UTP cabling, so 1000BASE-T is the UTP-based Gigabit Ethernet standard name.
3. B. Crossover cables cross the wire at one node’s transmit pin pair to the different pins used as the receive pins on the other device. For 10- and 100-Mbps Ethernet, the specific crossover cable wiring connects the pair at pins 1 and 2 on each end of the cable to pins 3 and 6 on the other end of the cable, respectively.
4. B, D, and E. Routers, wireless access point Ethernet ports, and PC NICs all send using pins 1 and 2, whereas hubs and LAN switches transmit on pins 3 and 6. Straight-through cables connect devices that use opposite pin pairs for sending, because the cable does not need to cross the pairs.
5. B. Multimode fiber works with LED-based transmitters rather than laser-based transmitters. Two answers mention the type of transmitters, making one of those answers correct and one incorrect.
Two answers mention distance. The answer that mentions the longest distance possible is incorrect because single-mode cables, not multimode cables, provide the longest distances. The other (correct) answer mentions the tradeoff of multimode being used for distances just longer than UTP’s 100 meter limit, while happening to use less expensive hardware than single mode.
6. B. NICs (and switch ports) use the carrier sense multiple access with collision detection (CSMA/CD) algorithm to implement half-duplex logic. CSMA/CD attempts to avoid collisions, but it also notices when collisions do occur, with rules about how the Ethernet nodes should stop sending, wait, and try again later.
7. C. The 4-byte Ethernet FCS field, found in the Ethernet trailer, allows the receiving node to see what the sending node computed with a math formula that is a key part of the error-detection process. Note that Ethernet defines the process of detecting errors (error detection), but not error recovery.
8. B, C, and E. The pre-assigned universal MAC address, given to each Ethernet port when manufactured, breaks the address into two 3-byte halves. The first half is called the organizationally unique identifier (OUI), which the IEEE assigns to the company that builds the product as a unique hex number to be used only by that company.
9. C and D. Ethernet supports unicast addresses, which identify a single Ethernet node, and group addresses, which can be used to send one frame to multiple Ethernet nodes. The two types of group addresses are the broadcast address and multicast address.
1. B. The standard HDLC header does not include a Type field, which identifies the type of packet encapsulated inside the HDLC frame.
2. B and D. The physical installation uses a model in which each router uses a physical Ethernet link to connect to some SP device in an SP facility called a point of presence (PoP). The Ethernet link does not span from each customer device to the other. From a data-link perspective, both routers use the same Ethernet standard header and trailer used on LANs; HDLC does not matter on these Ethernet WAN links.
3. A. PC1 will send an Ethernet frame to Router 1, with PC1’s MAC address as the source address and Router 1’s MAC address as the destination address. Router 1 will remove the encapsulated IP packet from that Ethernet frame, discarding the frame header and trailer. Router 1 will forward the IP packet by first encapsulating it inside an HDLC frame, but Router 1 will not encapsulate the Ethernet frame in the HDLC frame but rather the IP packet. Router 2 will de-encapsulate the IP packet from the HDLC frame and forward it onto the Ethernet LAN, adding a new Ethernet header and trailer, but this header will differ. It will list Router 2’s MAC address as the source address and PC2’s MAC address as the destination address.
4. C. Routers compare the packet’s destination IP address to the router’s IP routing table, making a match and using the forwarding instructions in the matched route to forward the IP packet.
5. C. IPv4 hosts generally use basic two-branch logic. To send an IP packet to another host on the same IP network or subnet that is on the same LAN, the sender sends the IP packet directly to that host. Otherwise, the sender sends the packet to its default router (also called the default gateway).
6. A and C. Routers do all the actions listed in all four answers; however, the routing protocol does the functions in the two listed answers. Independent of the routing protocol, a router learns routes for IP subnets and IP networks directly connected to its interfaces. Routers also forward (route) IP packets, but that process is called IP routing, or IP forwarding, and is an independent process compared to the work of a routing protocol.
7. C. Address Resolution Protocol (ARP) does allow PC1 to learn information, but the information is not stored on a server. The ping command does let the user at PC1 learn whether packets can flow in the network, but it again does not use a server. With the Domain Name System (DNS), PC1 acts as a DNS client, relying on a DNS server to respond with information about the IP addresses that match a given hostname.
1. A and B. The command in the question is an EXEC command that happens to require only user mode access. As such, you can use this command in both user mode and enable mode. Because it is an EXEC command, you cannot use the command (as shown in the question) in configuration mode. Note that you can put the word do in front of the EXEC command while in configuration mode (for example, do show mac address-table) to issue the command from inside any configuration mode.
2. B. The command referenced in the question, the reload command, is an EXEC command that happens to require privileged mode, also known as enable mode. This command is not available in user mode. Note that you can put the word do in front of the EXEC command while in configuration mode (for example, do reload) to issue the command from inside any configuration mode.
3. B. SSH provides a secure remote login option, encrypting all data flows, including password exchanges. Telnet sends all data (including passwords) as clear text.
4. A. Switches (and routers) keep the currently used configuration in RAM, using NVRAM to store the configuration file that is loaded when the switch (or router) next loads the IOS.
5. F. The startup-config file is in NVRAM, and the running-config file is in RAM.
6. B and C. The exit command moves the user one config mode backward, toward global configuration mode, or if already in global configuration mode, it moves the user back to enable mode. From console mode, it moves the user back to global configuration mode. The end command and the Ctrl+Z key sequence both move the user back to enable mode regardless of the current configuration submode.
1. A. A switch compares the destination MAC address to the MAC address table. If a matching entry is found, the switch forwards the frame out the appropriate interface. If no matching entry is found, the switch floods the frame.
2. C. A switch floods broadcast frames, multicast frames (if no multicast optimizations are enabled), and unknown unicast destination frames (frames whose destination MAC address is not in the MAC address table).
3. A. A switch floods broadcast frames, multicast frames (if no multicast optimizations are enabled), and unknown unicast destination frames (frames whose destination MAC address is not in the MAC address table).
4. B. Switches need to learn the location of each MAC address used in the LAN relative to that local switch. When a switch receives a frame, the source MAC identifies the sender. The interface in which the frame arrives identifies the local switch interface closest to that node in the LAN topology.
5. C. The show interfaces status command lists one line of output per interface. Cisco Catalyst switches name the type of interface based on the fastest speed of the interface, so 10/100 interfaces would be Fast Ethernet. With a working connection, ports from FastEthernet 0/1 through 0/10 would be listed in a connected state, while the rest would be listed in a notconnected state.
6. D. For the correct answer, each entry lists the learned MAC address. By definition, dynamically learned MAC addresses are learned by looking at the source MAC address of received frames. (That fact rules out one of the incorrect answers as well.)
The show mac address-table dynamic command lists the current list of MAC table entries, with three known entries at the point at which the command output was gathered. The counter in the last line of output lists the number of current entries, not the total number of learned MAC addresses since the last reboot. For instance, the switch could have learned other MAC addresses whose entries timed out from the MAC address table.
Finally, the answer that claims that port Gi0/2 connects directly to a device with a particular MAC address may or may not be true. That port could connect to another switch, and another, and so on, with one of those switches connecting to the device that uses the listed MAC address.
1. B. If both commands are configured, IOS accepts only the password as configured in the enable secret command.
2. A. To answer this question, it might be best to first think of the complete configuration and then find any answers that match the configuration. The commands, in vty line configuration mode, would be password password and login. Only one answer lists a vty subcommand that is one of these two commands.
Of note in the incorrect answers:
One answer mentions console subcommands. The console does not define what happens when remote users log in; those details sit in the vty line configuration.
One answer mentions the login local command; this command means that the switch should use the local list of configured usernames/passwords. The question stated that the engineer wanted to use passwords only, with no usernames.
One answer mentions the transport input ssh command, which, by omitting the telnet keyword, disables Telnet. While that command can be useful, SSH does not work when using passwords only; SSH requires both a username and a password. So, by disabling Telnet (and allowing SSH only), the configuration would allow no one to remotely log in to the switch.
3. B and C. SSH requires the use of usernames in addition to a password. Using the username global command would be one way to define usernames (and matching passwords) to support SSH. The vty lines would also need to be configured to require the use of usernames, with the login local vty subcommand being one such option.
The transport input ssh command could be part of a meaningful configuration, but it is not a global configuration command (as claimed in one wrong answer). Likewise, one answer refers to the username command as a command in vty config mode, which is also the wrong mode.
4. A, D, and F. To allow access through Telnet, the switch must have password security enabled, at a minimum using the password vty line configuration subcommand. In addition, the switch needs an IP address (configured under one VLAN interface) and a default gateway when the switch needs to communicate with hosts in a different subnet.
5. B and C. To allow SSH or Telnet access, a switch must have a correct IP configuration. That includes the configuration of a correct IP address and mask on a VLAN interface. That VLAN interface then must have a path out of the switch via ports assigned to that VLAN. In this case, with all ports assigned to VLAN 2, the switch must use interface VLAN 2 (using the interface vlan 2 configuration command).
To meet the requirement to support login from hosts outside the local subnet, the switch must configure a correct default gateway setting with the ip default-gateway 172.16.2.254 global command in this case.
6. A. The logging synchronous line subcommand synchronizes the log message display with other command output so the log message does not interrupt a show command’s output. The no ip domain-lookup command is not a line subcommand. The other two incorrect answers are line subcommands but do not configure the function listed in the question.
1. F. Cisco switches do not have a command to disable autonegotiation of speed and duplex. Instead, a switch port that has both speed and duplex configured disables autonegotiation.
2. E. Cisco switches can be configured for speed (with the speed command) and duplex (with the duplex command) in interface configuration mode.
3. A and D. The IEEE autonegotiation rules dictate that if a device attempts autonegotiation but the other side does not participate, use the slowest speed it supports. However, Cisco switches override that logic, instead sampling the electrical signal to detect the speed used by the connected device, so the switch will operate at 1000 Mbps. The switch uses the IEEE default setting for duplex based on the speed, and the IEEE default for duplex when using 1000 Mbps is to use full duplex. So in this case, the switch will match both the speed and the duplex setting made on the PC.
4. A, B, and D. The disabled state in the show interfaces status command is the same as an “administratively down and down” state shown in the show interfaces command. The interface must be in a connected state (per the show interfaces status command) before the switch can send frames out the interface.
5. A and D. SW2 has effectively disabled IEEE standard autonegotiation by configuring both speed and duplex. However, Cisco switches can detect the speed used by the other device, even with autonegotiation turned off. Also, at 1 Gbps, the IEEE autonegotiation standard says to use full duplex. If the duplex setting cannot be negotiated, both ends use 1 Gbps, full duplex.
6. D. For the two answers about a duplex mismatch, that condition does cause collisions, particularly late collisions, but only the side using CSMA/CD logic (the half-duplex side) has any concept of collisions. So, if switch SW1 was using half duplex, and switch SW2 using full duplex, SW1 would likely see late collisions and see that counter increment over time.
If switch SW2 had shut down its interface, switch SW1’s interface would be in a down/down state, and none of the counters would increment. Also, if both switch ports had been configured with different speeds, again the ports would be in a down/down state, and none of the interface counters would increment.
1. B. A VLAN is a set of devices in the same Layer 2 broadcast domain. A subnet often includes the exact same set of devices, but it is a Layer 3 concept. A collision domain refers to a set of Ethernet devices, but with different rules than VLAN rules for determining which devices are in the same collision domain.
2. D. Although a subnet and a VLAN are not equivalent concepts, the devices in one VLAN are typically in the same IP subnet and vice versa.
3. B. 802.1Q defines a 4-byte header, inserted after the original frame’s destination and source MAC address fields. The insertion of this header does not change the original frame’s source or destination address. The header itself holds a 12-bit VLAN ID field, which identifies the VLAN associated with the frame.
4. A and C. The dynamic auto setting means that the switch can negotiate trunking, but it can only respond to negotiation messages, and it cannot initiate the negotiation process. So, the other switch must be configured to trunk or to initiate the negotiation process (based on being configured with the dynamic desirable option).
5. A and B. The configured VTP setting of VTP transparent mode means that the switch can configure VLANs, so the VLAN is configured. In addition, the VLAN configuration details, including the VLAN name, show up as part of the running-config file.
6. B and C. The show interfaces switchport command lists both the administrative and operational status of each port. When a switch considers a port to be trunking, this command lists an operational trunking state of “trunk.” The show interfaces trunk command lists a set of interfaces—the interfaces that are currently operating as trunks. So, both of these commands identify interfaces that are operational trunks.
7. A and B. On switches that do not use VTP (by using VTP modes off or transparent), the switch lists all VLAN configuration in the configuration file (making one answer correct). Also, the show vlan brief command lists all defined VLANs, regardless of VTP mode and regardless of shutdown state. As a result, the two answers that mention commands are correct.
The two incorrect answers are incorrect because VLAN 30 has been shut down, which means the switch will not forward frames in that VLAN, regardless of whether they arrive on access or trunk ports.
8. B. The first list of VLAN IDs includes all VLANs (1–4094) except those overtly removed per the details in any switchport trunk allowed vlan interface subcommands on the trunk interface. If no such commands are configured, the first list in the output will include 1–4094. The two incorrect answers that mention VLAN 30 both list conditions that change the second of two lists of VLANs in the command output, while STP’s choice to block an interface would impact the third list.
1. A and B. Listening and learning are transitory port states, used only when moving from the blocking to the forwarding state. Discarding is not an STP port state.
2. C. The smallest numeric bridge ID wins the election.
3. C and D. Listening and learning are transitory port states used only when moving from the blocking to the forwarding state. Discarding is not an STP port state. Forwarding and blocking are stable states.
4. B. Nonroot switches forward Hellos received from the root; the root sends these Hellos based on the root’s configured Hello timer.
5. B and D. RSTP uses port state forwarding, learning, and discarding. Forwarding and learning perform the same functions as the port states used by traditional STP.
6. A and D. With RSTP, an alternate port is an alternate to the root port when a switch’s root port fails. A backup port takes over for a designated port if the designated port fails.
7. D. The PortFast feature allows STP to move a port from blocking to forwarding without going through the interim listening and learning states. STP allows this exception when the link is known to have no switch on the other end of the link, removing the risk of a switching loop. BPDU Guard is a common feature to use at the same time as PortFast because it watches for incoming bridge protocol data units (BPDU), which should not happen on an access port, and prevents the loops from a rogue switch by disabling the port.
1. A. Of the four answers, only pvst and rapid-pvst are valid options on the command. Of those, the rapid-pvst option enables Rapid Per VLAN Spanning Tree (RPVST+), which uses RSTP. The pvst option enables Per VLAN Spanning Tree (PVST) which uses STP, not RSTP. The other two options, if attempted, would cause the command to be rejected because the option does not exist.
2. A and C. The system ID extension (or extended system ID) part of a bridge ID contains 12 bits and sits after the 4-bit priority field and before the 48-bit system ID. Switches use this field to store the VLAN ID when using STP or RSTP to build spanning trees per VLAN. So of the two answers that mention the system ID extension, the one that lists the VLAN ID, in this case 5, is correct.
The output also lists a priority of 32773. However, that output lists the decimal equivalent of the 16-bit priority value. In reality, this decimal value is the sum of the configured decimal priority plus the VLAN ID: 32768 + 5 = 32773. So in this case, the root’s configured priority is 32,768.
3. A, B, and D. The Cisco Rapid Per VLAN Spanning Tree (RPVST+) creates one spanning tree instance per VLAN. To do so, it sends BPDUs per-VLAN. Each switch identifies itself with a unique Bridge ID (BID) per VLAN, made unique per-VLAN by adding the VLAN ID to the system ID extension 12-bit field of the BID. RVPST also adds a new Type-Length Value (TLV) to the BPDU itself, which includes a place to list the VLAN ID. Finally, when transmitting the BPDUs over VLAN trunks, the switch uses a trunking header that lists the VLAN ID (a practice sometimes called tunneling in 802.1Q.) The receiving switch can check all three locations that list the VLAN ID to ensure that they all agree about what VLAN the BPDU is describing. Of the four answers, the three correct answers describe the three actual locations in which RPVST+ lists the VLAN ID.
4. D. IOS uses the channel-group configuration command to create an EtherChannel. Then the term etherchannel is used in the show etherchannel command, which displays the status of the channel. The output of this show command then names the channel a PortChannel. The only answer that is not used somewhere in IOS to describe this multilink channel is Ethernet-Channel.
5. B and D. The channel-group command will direct the switch to use LACP to dynamically negotiate to add a link to an EtherChannel when the command uses the active and passive keywords, respectively. The desirable and passive keywords direct the switch to use PaGP instead of LACP. Of the four answers, the two correct answers use two LACP values, while the two incorrect answers use at least one value that would cause the switch to use PaGP, making the answer incorrect.
Of the two correct answers, both combinations result in the switches attempting to add the link to an EtherChannel using LACP as the negotiation protocol. If both switches used the passive keyword, they would both sit and wait for the other switch to begin sending LACP messages and therefore never attempt to add the link to the channel.
6. C. EtherChannel load distribution, or load balancing, on Cisco Catalyst switches uses an algorithm. The algorithm examines some fields in the various headers, so messages that have the same values in those fields always flow over the same link in a particular EtherChannel. Note that it does not break the frames into smaller fragments nor use a round-robin approach that ignores the header values, and it does not examine link utilization when making the choice.
1. B and D. The general rule to determine whether two devices’ interfaces should be in the same subnet is whether the two interfaces are separated from each other by a router. To provide a way for hosts in one VLAN to send data to hosts outside that VLAN, a local router must connect its LAN interface to the same VLAN as the hosts and have an address in the same subnet as the hosts. All the hosts in that same VLAN on the same switch would not be separated from each other by a router, so these hosts would also be in the same subnet. However, another PC, connected to the same switch but in a different VLAN, will require its packets to flow through a router to reach Host A, so Host A’s IP address would need to be in a different subnet compared to this new host.
2. D. By definition, two address values in every IPv4 subnet cannot be used as host IPv4 addresses: the first (lowest) numeric value in the subnet for the subnet ID and the last (highest) numeric value in the subnet for the subnet broadcast address.
3. B and C. At least 7 subnet bits are needed because 26 = 64, so 6 subnet bits could not number 100 different subnets. Seven subnet bits could because 27 = 128 >= 100. Similarly, 6 host bits is not enough because 26 – 2 = 62, but 7 host bits is enough because 27 - 2 = 126 >= 100.
The number of network, subnet, and host bits must total 32 bits, making one of the answers incorrect. The answer with 8 network bits cannot be correct because the question states that a Class B network is used, so the number of network bits must always be 16. The two correct answers have 16 network bits (required because the question states the use of a Class B network) and at least 7 subnet and host bits each.
4. A and C. The private IPv4 networks, defined by RFC 1918, are Class A network 10.0.0.0, the 16 Class B networks from 172.16.0.0 to 172.31.0.0, and the 256 Class C networks that begin with 192.168.
5. A, D, and E. The private IPv4 networks, defined by RFC 1918, are Class A network 10.0.0.0, the 16 Class B networks from 172.16.0.0 to 172.31.0.0, and the 256 Class C networks that begin with 192.168. The three correct answers are from the public IP network range, and none are reserved values.
6. A and C. An unsubnetted Class A, B, or C network has two parts: the network and host parts.
7. B. An unsubnetted Class A, B, or C network has two parts: the network and host parts. To perform subnetting, the engineer creates a new subnet part by borrowing host bits, shrinking the number of host bits. The subnet part of the address structure exists only after the engineer chooses a nondefault mask. The network part remains a constant size.
1. B and C. Class A networks have a first octet in the range of 1–126, inclusive, and their network IDs have a 0 in the last three octets. 126.96.36.199 is actually a Class B network (first octet range 128–191, inclusive). All addresses that begin with 127 are reserved, so 127.0.0.0 is not a Class A network.
2. E. All Class B networks begin with values between 128 and 191, inclusive, in their first octets. The network ID has any value in the 128–191 range in the first octet, and any value from 0 to 255 inclusive in the second octet, with decimal 0s in the final two octets. Two of the answers show a 255 in the second octet, which is acceptable. Two of the answers show a 0 in the second octet, which is also acceptable.
3. B and D. The first octet (172) is in the range of values for Class B addresses (128–191). As a result, the network ID can be formed by copying the first two octets (172.16) and writing 0s for the last two octets (172.16.0.0). The default mask for all Class B networks is 255.255.0.0, and the number of host bits in all unsubnetted Class B networks is 16.
4. A and C. The first octet (192) is in the range of values for Class C addresses (192–223). As a result, the network ID can be formed by copying the first three octets (192.168.6) and writing 0 for the last octet (192.168.6.0). The default mask for all Class C networks is 255.255.255.0, and the number of host bits in all unsubnetted Class C networks is 8.
5. D. To find the network broadcast address, first determine the class, and then determine the number of host octets. At that point, convert the host octets to 255 to create the network broadcast address. In this case, 10.1.255.255 is in a Class A network, with the last three octets as host octets, for a network broadcast address of 10.255.255.255. For 192.168.255.1, it is a Class C address, with the last octet as the host part, for a network broadcast address of 192.168.255.255. Address 188.8.131.52 is a Class D address, so it is not in any unicast IP network and the question does not apply. For 172.30.255.255, it is a Class B address, with the last two octets as host octets, so the network broadcast address is 172.30.255.255.
1. C. If you think about the conversion one octet at a time, the first two octets each convert to 8 binary 1s. 254 converts to 8-bit binary 11111110, and decimal 0 converts to 8-bit binary 00000000. So, the total number of binary 1s (which defines the prefix length) is 8 + 8 + 7 + 0 = /23.
2. B. If you think about the conversion one octet at a time, the first three octets each convert to 8 binary 1s. 240 converts to 8-bit binary 11110000, so the total number of binary 1s (which defines the prefix length) is 8 + 8 + 8 + 4 = /28.
3. B. /30 is the equivalent of the mask that in binary has 30 binary 1s. To convert that to DDN format, write down all the binary 1s (30 in this case), followed by binary 0s for the remainder of the 32-bit mask. Then take 8 bits at a time and convert from binary to decimal (or memorize the nine possible DDN mask octet values and their binary equivalents). Using the /30 mask in this question, the binary mask is 11111111 11111111 11111111 11111100. Each of the first three octets is all binary 1s, so each converts to 255. The last octet, 11111100, converts to 252, for a DDN mask of 255.255.255.252. See Appendix A, “Numeric Reference Tables,” for a decimal/binary conversion table.
4. C. The size of the network part is always either 8, 16, or 24 bits, based on whether it is Class A, B, or C, respectively. As a Class A address, N=8. The mask 255.255.255.0, converted to prefix format, is /24. The number of subnet bits is the difference between the prefix length (24) and N, so S=16 in this case. The size of the host part is a number that, when added to the prefix length (24), gives you 32, so H=8 in this case.
5. A. The size of the network part is always either 8, 16, or 24 bits, based on whether it is Class A, B, or C, respectively. As a Class C address, N=24. The number of subnet bits is the difference between the prefix length (27) and N, so S=3 in this case. The size of the host part is a number that, when added to the prefix length (27), gives you 32, so H=5 in this case.
6. D. Classless addressing rules define a two-part IP address structure: the prefix and the host part. This logic ignores Class A, B, and C rules, and can be applied to the 32-bit IPv4 addresses from any address class. By ignoring Class A, B, and C rules, classless addressing ignores any distinction as to the network part of an IPv4 address.
7. A and B. The masks in binary define a number of binary 1s, and the number of binary 1s defines the length of the prefix (network + subnet) part. With a Class B network, the network part is 16 bits. To support 100 subnets, the subnet part must be at least 7 bits long. Six subnet bits would supply only 26 = 64 subnets, while 7 subnet bits supply 27 = 128 subnets. The /24 answer supplies 8 subnet bits, and the 255.255.255.252 answer supplies 14 subnet bits.
1. D. When using classful IP addressing concepts as described in Chapter 13, “Analyzing Subnet Masks,” addresses have three parts: network, subnet, and host. For addresses in a single classful network, the network parts must be identical for the numbers to be in the same network. For addresses in the same subnet, both the network and subnet parts must have identical values. The host part differs when comparing different addresses in the same subnet.
2. B and D. In any subnet, the subnet ID is the smallest number in the range, the subnet broadcast address is the largest number, and the usable IP addresses sit between them. All numbers in a subnet have identical binary values in the prefix part (classless view) and network + subnet part (classful view). To be the lowest number, the subnet ID must have the lowest possible binary value (all 0s) in the host part. To be the largest number, the broadcast address must have the highest possible binary value (all binary 1s) in the host part. The usable addresses do not include the subnet ID and subnet broadcast address, so the addresses in the range of usable IP addresses never have a value of all 0s or 1s in their host parts.
3. C. The mask converts to 255.255.255.0. To find the subnet ID, for each octet of the mask that is 255, you can copy the IP address’s corresponding values. For mask octets of decimal 0, you can record a 0 in that octet of the subnet ID. As such, copy the 10.7.99 and write a 0 for the fourth octet, for a subnet ID of 10.7.99.0.
4. C. First, the resident subnet (the subnet ID of the subnet in which the address resides) must be numerically smaller than the IP address, which rules out one of the answers. The mask converts to 255.255.255.252. As such, you can copy the first three octets of the IP address because of their value of 255. For the fourth octet, the subnet ID value must be a multiple of 4, because 256 – 252 (mask) = 4. Those multiples include 96 and 100, and the right choice is the multiple closest to the IP address value in that octet (97) without going over. So, the correct subnet ID is 192.168.44.96.
5. C. The resident subnet ID in this case is 172.31.77.192. You can find the subnet broadcast address based on the subnet ID and mask using several methods. Following the decimal process in the book, the mask converts to 255.255.255.224, making the interesting octet be octet 4, with magic number 256 – 224 = 32. For the three octets where the mask = 255, copy the subnet ID (172.31.77). For the interesting octet, take the subnet ID value (192), add magic (32), and subtract 1, for 223. That makes the subnet broadcast address 172.31.77.223.
6. C. To answer this question, you need to find the range of addresses in the subnet, which typically then means you need to calculate the subnet ID and subnet broadcast address. With a subnet ID/mask of 10.1.4.0/23, the mask converts to 255.255.254.0. To find the subnet broadcast address, following the decimal process described in this chapter, you can copy the subnet ID’s first two octets because the mask’s value is 255 in each octet. You write a 255 in the fourth octet because the mask has a 0 on the fourth octet. In octet 3, the interesting octet, add the magic number (2) to the subnet ID’s value (4), minus 1, for a value of 2 + 4 – 1 = 5. (The magic number in this case is calculated as 256 – 254 = 2.) That makes the broadcast address 10.1.5.255. The last usable address is 1 less: 10.1.5.254. The range that includes the last 100 addresses is 10.1.5.155 – 10.1.5.254.
1. B and E. Cisco routers have an on/off switch, but Cisco switches generally do not.
2. B. Cisco routers that do not also have any Layer 2 switch features support commands needed for Layer 3 routing as well as commands in common between Layer 2 switching and Layer 3 routing devices. In this case, the show interfaces status and show mac address-table commands happen to be commands supported on Layer 2 switches but not on routers. Both types of devices use the show running-config command. Of the answers, only the show ip interface brief command is unique to routers.
3. A and C. To route packets on an interface, the router interface configuration must include an IP address and mask. One correct command shows the correct single command used to configure both values, while one incorrect command shows those settings as two separate commands. Also, to route packets, the interface must reach an “up/up” state; that is, the show interfaces and other commands list two status values, and both must be “up.” The no shutdown command enables the interface.
4. C. If the first of the two status codes is “down,” it typically means that a Layer 1 problem exists. In this case, the question states that the router connects to a switch with a UTP straight-through cable, which is the correct cable pinout. Of the two answers that mention the shutdown command, if the router interface were shut down, the first router status code would be “administratively down,” so that answer is incorrect. However, if the neighboring device interface sits in a shutdown state, the router will sense no electrical signals over the cable, seeing that as a physical problem, and place the interface into a “down/down” state, making that answer correct.
Second, the two answers that mention interface IP addresses have no impact on the status codes of the show interfaces brief command. Both answers imply that the interface does not have an IP address configured. However, both the first and second status codes are not related to whether IP addresses have been configured or not, making both answers incorrect.
5. C and E. The show ip interface brief command lists all the interface IPv4 addresses but none of the masks. The show version command lists none of the IP addresses and none of the masks. The other three commands list both the address and mask.
6. B. A router has one IPv4 address for each interface in use, whereas a LAN switch has a single IPv4 address that is just used for accessing the switch. The rest of the answers list configuration settings that use the same conventions on both routers and switches.
1. A and C. The route defines the group of addresses represented by the route using the subnet ID and mask. The router can use those numbers to find the range of addresses that should be matched by this route. The other two answers list facts useful when forwarding packets that happen to match the route.
2. A and D. First, for the subnetting math, address 10.1.1.100, with mask /26, implies a subnet ID of 10.1.1.64. Also, mask /26 converts to a DDN mask of 255.255.255.192. For any working router interface, after adding the ip address command to configure an address and mask, the router adds a connected route for the subnet. In this case, that means the router adds a connected route for subnet 10.1.1.64 255.255.255.192. The router also adds a route called a local route, which is a route for the interface IP address with a 255.255.255.255 mask. In this case, that means the router adds a local route for address 10.1.1.100 with mask 255.255.255.255.
3. C. The ip route command can refer to the IP address of the next-hop router or to the local router’s interface. It also refers to the subnet ID and matching subnet mask, defining the range of addresses matched by the route.
4. A. The correct syntax lists a subnet number, then a subnet mask in dotted-decimal form, and then either an outgoing interface or a next-hop IP address.
5. B. The ip route command can reference an outgoing interface or a next-hop IP address, and the command lists a next-hop IP address, which rules out one answer. The command does use the correct syntax, ruling out another answer. There is no requirement for a router to have any particular interface IP addresses in relation to the configuration of an ip route command, ruling out yet another answer.
The checks that IOS uses when looking at a new ip route command include whether the outgoing interface is up/up, whether the next-hop address is reachable, and, if there is a competing route from another source, whether the other route has a better administrative distance.
6. D. Destination address 10.1.15.122 matches all the routes listed except the host route to 10.1.15.100/32. In that case, the router will choose the matching route that has the longest prefix length, that is, the prefix-style mask with the highest number. In this case, that route lists subnet 10.1.15.96 and mask /27, which lists interface G0/3/0 as the outgoing interface.
1. A and F. Of all the commands listed, only the two correct answers are syntactically correct router configuration commands. The command to enable 802.1Q trunking is encapsulation dot1q vlan_id.
2. B and C. Subinterface G0/1.1 must be in an administratively down state due to the shutdown command being issued on that subinterface. For subinterface G0/1.2, its status cannot be administratively down because of the no shutdown command. G0/1.2’s state will then track to the state of the underlying physical interface. With a physical interface state of down/down, subinterface G0/1.2 will be in a down/down state in this case.
3. C. The configuration of the Layer 3 switch’s routing feature uses VLAN interfaces. The VLAN interface numbers must match the associated VLAN ID, so with VLANs 1, 2, and 3 in use, the switch will configure interface vlan 1, interface vlan 2 (which is the correct answer), and interface vlan 3. The matching connected routes, like all connected IP routes, will list the VLAN interfaces.
As for the incorrect answers, a list of connected routes will not list any next-hop IP addresses. Each route will list an outgoing interface; the outgoing interface will not be a physical interface, but rather a VLAN interface, because the question states that the configuration uses SVIs. Finally, all the listed subnets have a /25 mask, which is 255.255.255.128, so none of the routes will list a 255.255.255.0 mask.
4. C and D. First, for the correct answers, a Layer 3 switch will not route packets on a VLAN interface unless it is in an up/up state. A VLAN interface will only be up/up if the matching VLAN (with the same VLAN number) exists on the switch. If VTP deletes the VLAN, then the VLAN interface moves to a down/down state, and routing in/out that interface stops. Also, disabling VLAN 2 with the shutdown command in VLAN configuration mode also causes the matching VLAN 2 interface to fail, which makes routing on interface VLAN 2 stop as well.
As for the incorrect answers, a Layer 3 switch needs only one access port or trunk port forwarding for a VLAN to enable routing for that VLAN, so nine of the ten access ports in VLAN 2 could fail, leaving one working port, and the switch would keep routing for VLAN 2.
A shutdown of VLAN 4 has no effect on routing for VLAN interfaces 2 and 3. Had that answer listed VLANs 2 or 3, it would definitely be a reason to make routing fail for that VLAN interface.
5. A and C. With a Layer 3 EtherChannel, the physical ports and the port-channel interface must disable the behavior of acting like a switch port, and therefore act like a routed port, through the configuration of the no switchport interface subcommand. (The routedport command is not an IOS command.) Once created, the physical interfaces should not have an IP address configured. The port-channel interface (the interface representing the EtherChannel) should be configured with the IP address.
6. B and C. With a Layer 3 EtherChannel, two configuration settings must be the same on all the physical ports, specifically the speed and duplex as set with the speed and duplex commands. Additionally, the physical ports and port-channel port must all have the no switchport command configured to make each act as a routed port. So, having a different speed setting, or being configured with switchport rather than no switchport, would prevent IOS from adding interface G0/2 to the Layer 3 EtherChannel.
As for the wrong answers, both have to do with Layer 2 configuration settings. Once Layer 2 operations have been disabled because of the no switchport command, those settings related to Layer 2 that could cause problems on Layer 2 EtherChannels do not then cause problems for the Layer 3 EtherChannel. So, Layer 2 settings about access VLANs, trunking allowed lists, and STP settings, which must match before an interface can be added to a Layer 2 EtherChannel, do not matter for a Layer 3 EtherChannel.
1. D. Both versions of RIP use distance vector logic, and EIGRP uses a different kind of logic, characterized either as advanced distance vector or a balanced hybrid.
2. C and D. Both versions of RIP use the same hop-count metric, neither of which is affected by link bandwidth. EIGRP’s metric, by default, is calculated based on bandwidth and delay. OSPF’s metric is a sum of outgoing interfaces costs, with those costs (by default) based on interface bandwidth.
3. B, C, and D. Of the listed routing protocols, only the old RIP Version 1 (RIP-1) protocol does not support variable-length subnet masks (VLSM).
4. C. LSAs contain topology information that is useful in calculating routes, but the LSAs do not directly list the route that a router should add to its routing table. In this case, R1 would run a calculation called the Shortest Path First (SPF) algorithm, against the LSAs, to determine what IP routes to add to the IP routing table.
5. B. Neighboring OSPF routers that complete the database exchange are considered fully adjacent and rest in a full neighbor state. The up/up and final states are not OSPF states at all. The 2-way state is either an interim state or a stable state between some routers on the same VLAN.
6. C. The correct answer is the one advantage of using a single-area design. The three wrong answers are advantages of using a multiarea design, with all reasons being much more important with a larger internetwork.
1. B. The network 10.0.0.0 0.255.255.255 area 0 command works because it matches all interfaces whose first octet is 10. The rest of the commands match as follows: all addresses that end with 0.0.0 (wildcard mask 255.0.0.0); all addresses that begin with 10.0.0 (wildcard mask 0.0.0.255); and all addresses that begin with 10.0 (wildcard mask 0.0.255.255).
2. A. The network 10.1.0.0 0.0.255.255 area 0 command matches all IP addresses that begin with 10.1, enabling OSPF in area 0 on all interfaces. The answer with wildcard mask 0.255.255.0 is illegal because it represents more than one string of binary 0s separated by binary 1s. The answer with x’s is syntactically incorrect. The answer with wildcard mask 255.0.0.0 means “Match all addresses whose last three octets are 0.0.0,” so none of the three interfaces are matched.
3. A and E. Of the three wrong answers, two are real commands that simply do not list the OSPF neighbors. show ip ospf interface brief lists interfaces on which OSPF is enabled but does not list neighbors. show ip interface lists IPv4 details about interfaces, but none related to OSPF. One incorrect answer, show ip neighbor, is not a valid IOS command.
4. B. With OSPFv2 interface configuration mode, the configuration looks just like the traditional configuration, with a couple of exceptions. The network router subcommand is no longer required. Instead, each interface on which OSPF should be enabled is configured with an ip ospf process-id area area-id interface subcommand. This command refers to the OSPF routing process that should be enabled on the interface and specifies the OSPFv2 area.
5. B. SPF calculates the cost of a route as the sum of the OSPF interface costs for all outgoing interfaces in the route. The interface cost can be set directly (ip ospf cost), or IOS uses a default based on the reference bandwidth and the interface bandwidth. Of the listed answers, delay is the only setting that does not influence OSPFv2 metric calculations.
6. A and D. The configuration enables OSPF and identifies the area number to use with the interface using an interface subcommand in interface mode: the ip ospf process-id area area-number command. However, to explicitly configure the router ID, the configuration must use the router-id router-id-value command, which is a command issued in OSPF router mode.
1. B and D. By default, IOS assigns Ethernet interfaces an OSPF network type of broadcast, with an OSPF interface priority of 1. As a result, both routers attempt to discover the other routers on the link (which identifies one correct answer).
The broadcast network type means that the routers also attempt to elect a DR and BDR. With a tie-in priority, the routers choose the DR based on the highest router ID (RID) values, meaning that R2 will become the DR and R1 will become the BDR. These facts combine to show why the two incorrect answers are incorrect. The other correct answer is correct because the show ip ospf neighbor command lists the local router’s neighbor relationship state (FULL) and the role filled by that neighbor (DR), which would be the output shown on R1 when R2 is acting as DR.
2. B and C. First, the OSPF point-to-point network type causes the two routers to dynamically discover neighbors, making one answer correct.
Next, IOS assigns a default OSPF interface priority of 1, so R1’s configured priority of 11 would be better in a DR/BDR election. However, the point-to-point network type causes the router to not use a DR/BDR on the interface. As a result, the answer about R1 becoming the DR is incorrect (because no DR exists at all), and the answer listing a state of “FULL/DR” is incorrect for the same reason. However, the answer that claims that R2 will be neither DR nor BDR is true because no DR or BDR is elected.
3. D. The show ip ospf interface brief command lists a pair of counters under the heading “Nbrs F/C” on the far right of the output. The first of the two numbers represents the number of fully adjacent neighbors (2 in this case), and the second number represents the total number of neighbors.
4. A and D. As worded, the correct answers list a scenario that would prevent the neighbor relationship. One correct answer mentions the use of two different OSPF areas on the potential OSPF neighbors; to become neighbors, the two routers must use the same area number. The other correct answer mentions the use of two different Hello timers, a mismatch that causes two routers to reject each other and to not become neighbors.
The two incorrect answers list scenarios that do not cause issues, making them incorrect answers. One mentions mismatched OSPF process IDs; OSPF process IDs do not need to match for two routers to become neighbors. The other incorrect answer (that is, a scenario that does not cause a problem) mentions the use of two different priority values. The priority values give OSPF a means to prefer one router over the other when electing a DR/BDR, so the setting is intended to be set to different values on different routers and does not cause a problem.
5. C. As worded, the correct answers should be a scenario that would prevent the neighbor relationship. The answers all list values that are identical or similar on the two routers. Of those, the use of an identical OSPF router ID (RID) on the two routers prevents them from becoming neighbors, making that one answer correct.
Of the incorrect answers, both routers must have the same Dead interval, so both using a Dead interval of 40 causes no issues. The two routers can use any OSPF process ID (the same or different value, it does not matter), making that answer incorrect. Finally, the two routers’ IP addresses must be in the same subnet, so again that scenario does not prevent R13 and R14 from becoming neighbors.
6. D. The OSPF shutdown command tells the OSPF process to stop operating. That process includes removing any OSPF-learned routes from the IP routing table, clearing the router’s LSDB, and closing existing OSPF neighbor relationships. In effect, it causes OSPF to stop working on the router, but it does retain the configuration so that a no shutdown command will cause the router to start using OSPF again with no changes to the configuration.
1. C. NAT, specifically the PAT feature that allows many hosts to use private IPv4 addresses while being supported by a single public IPv4 address, was one short-term solution to the IPv4 address exhaustion problem. IP version 5 existed briefly as an experimental protocol and had nothing to do with IPv4 address exhaustion. IPv6 directly addresses the IPv4 address exhaustion problem, but it is a long-term solution. ARP has no impact on the number of IPv4 addresses used.
2. A. Routers use the same process steps when routing IPv6 packets as they do when routing IPv4 packets. Routers route IPv6 packets based on the IPv6 addresses, listed inside the IPv6 header in the IPv6 packets, by comparing the destination IPv6 address to the router’s IPv6 routing table. As a result, the router discards the incoming frame’s data-link header and trailer, leaving an IPv6 packet. The router compares the destination (not source) IPv6 address in the header to the router’s IPv6 (not IPv4) routing table and then forwards the packet based on the matched route.
3. D. If you are following the steps in the book, the first step removes up to three leading 0s in each quartet, leaving FE80:0:0:100:0:0:0:123. This leaves two strings of consecutive all-0 quartets; by changing the longest string of all 0s to ::, the address is FE80:0:0:100::123.
4. B. This question has many quartets that make it easy to make a common mistake: removing trailing 0s in a quartet of hex digits. To abbreviate IPv6 addresses, only leading 0s in a quartet should be removed. Many of the quartets have trailing 0s (0s on the right side of the quartet), so make sure to not remove those 0s.
5. A. The unabbreviated version of an IPv6 address must have 32 digits, and only one answer has 32 hex digits. In this case, the original number shows four quartets and a ::. So, the :: was replaced with four quartets of 0000, making the number have eight quartets. Then, for each quartet with fewer than four digits, leading 0s were added so that each quartet has four hex digits.
6. C. The /64 prefix length means that the last 64 bits, or last 16 digits, of the address should be changed to all 0s. That process leaves the unabbreviated prefix as 2000:0000:0000:0005:0000:0000:0000:0000. The last four quartets are all 0s, making that string of all 0s be the longest and best string of 0s to replace with ::. After removing the leading 0s in other quartets, the answer is 2000:0:0:5::/64.
1. C. Unique local addresses begin with FD in the first two digits.
2. A. Global unicast addresses can begin with many different initial values, but most commonly begin with either a hex 2 or 3.
3. D. The global routing prefix is the address block, represented as a prefix value and prefix length, given to an organization by some numbering authority. All IPv6 addresses inside the company have the same value in these initial bits of their IPv6 addresses. Similarly, when a company uses a public IPv4 address block, all the addresses have the same value in the network part.
4. B. Subnetting a global unicast address block, using a single prefix length for all subnets, breaks the addresses into three parts. The parts are the global routing prefix, subnet, and interface ID.
5. D. Unique local addresses begin with a 2-hex-digit prefix of FD, followed by the 10-hex-digit global ID.
1. A. The one correct answer lists the exact same IPv6 address listed in the question, with a /64 prefix length and no spaces in the syntax of the answer. Another (incorrect) answer is identical, except that it leaves a space between the address and prefix length, which is incorrect syntax. The two answers that list the eui-64 parameter list an address and not a prefix; they should list a prefix to be correct, although neither would have resulted in the IPv6 address listed in the question.
2. B. With the eui-64 parameter, the router will calculate the interface ID portion of the IPv6 address based on its MAC address. Beginning with 5055.4444.3333, the router injects FF FE in the middle (5055.44FF.FE44.3333). Then the router inverts the seventh bit in the first byte. Mentally, this converts hex 50 to binary 01010000, changing bit 7 so that the string is 0101 0010 and converting back to hex 52. The final interface ID value is 5255:44FF:FE44:3333. The wrong answers simply list a different value.
3. A and C. Of the four answers, the two correct answers show the minimal required configuration to support IPv6 on a Cisco router: enabling IPv6 routing (ipv6 unicast-routing) and enabling IPv6 on each interface, typically by adding a unicast address to each interface (ipv6 address…). The two incorrect answers list nonexistent commands.
4. A. With an ipv6 address command configured for a global unicast address, but without a link-local address configured with an ipv6 address command, the router calculates its link-local address on the interface based on its MAC address and EUI-64 rules. The first half of the link-local address begins FE80:0000:0000:0000. The router then calculates the second half of the link-local address value by taking the MAC address (0200.0001.000A), injecting FF FE in the middle (0200.00FF.FE01.000A), and flipping the seventh bit (0000.00FF.FE01.000A).
5. B. FF02::1 is used by all IPv6 hosts on the link, FF02::5 is used by all OSPFv3 routers, and FF02::A is used by all EIGRPv6 routers. FF02::2 is used to send packets to all IPv6 routers on a link.
1. A and C. With an IPv6 address on a working interface, the router adds a connected route for the prefix (subnet) implied by the ipv6 address command. It also adds a local host route (with a /128 prefix length) based on the unicast address. The router does not add a route based on the link-local address.
2. A and C. The two correct answers show the correct subnet ID (prefix) and prefix length for the two connected subnets: 3111:1:1:1::/64 and 3222:2:2:2::/64. The answer with the /128 prefix length is shown in a local route, but those routes are not displayed by the show ipv6 route connected command. The other incorrect answer lists the entire IPv6 address with a /64 prefix length, and the entire address would not be displayed as a prefix when using a /64 prefix.
3. A. All four answers show examples of commands that use an outgoing interface. The two commands that begin with ip route define only IPv4 routes; the commands would be rejected because of the IPv6 prefixes listed in the commands. The two commands that begin with ipv6 route are syntactically correct, but the command should list the local router’s interface (an interface on the router on which the command is being configured). R5 needs to use its local S0/1/1 interface as the outgoing interface.
4. B. All four answers show examples of commands that use a next-hop router IPv6 address. Two of the answers list R5’s own IPv6 address (unicast or link-local), which is incorrect; the answer should be an address on the neighboring router, R6 in this case. For the two answers that list addresses on Router R6, the one that lists R6’s global unicast address is correct. The one that lists R6’s link-local address would also require R5’s outgoing interface, so the answer that lists FE80::FF:FE00:6 would be rejected as well.
5. C. IOS will add a new static route to the IPv6 routing table if, when using a next-hop global unicast address, the router has a working route to reach that next-hop address and there is no better (lower administrative distance) route for the exact same subnet. So, the correct answer identifies one reason why the route would not appear. The answer that mentions a better route with administrative distance of 110 is a valid reason for the static route to not appear, but the question states that no route for the subnet appears in the routing table, so clearly that competing route does not exist.
The other two answers are incorrect about the ipv6 route command. This command can use a link-local next-hop address but does not have to do so. Also, when using a global unicast address as next-hop, the command does not also require an outgoing interface parameter.
6. A and B. The output shows two static routes, as noted with the “S” code on the far left. Both were added to the IPv6 routing table because of ipv6 route commands. Both have an administrative distance of 1, which is listed as the first number in brackets.
For the two incorrect answers, note that the ipv6 address interface subcommand does cause IOS to add connected IPv6 routes to the routing table, and the phrase “directly connected” with one route might make you think this is a connected route. However, the “S” in the far left identifies the source of the route. Likewise, the answer that mentions an IPv6 routing protocol is incorrect because both routes have a code of S, meaning static.
7. B. PC1 needs to discover PC2’s MAC address. Unlike IPv4, IPv6 does not use ARP, instead using NDP. Specifically, PC1 uses the NDP Neighbor Solicitation (NS) message to request that PC2 send back an NDP Neighbor Advertisement (NA). SLAAC relates to address assignment, and not to discovering a neighbor’s MAC address.
8. A and C. The NDP RA lists the router IPv6 address, the IPv6 prefixes known on the link, and the matching prefix lengths. When using DHCPv6, the host learns the IPv6 address of the DNS server through DHCPv6 messages. For MAC addresses of on-link neighbors, hosts use NDP NS and NA messages.
1. C. The IEEE 802.3 standard defines Ethernet, while 802.11 defines Wi-Fi.
2. B. WLANs require half-duplex operation because all stations must contend for use of a channel to transmit frames.
3. C. An AP offers a basic service set (BSS). BSA is incorrect because it is a Basic Service Area, or the cell footprint of a BSS. BSD is incorrect because it does not pertain to wireless at all. IBSS is incorrect because it is an Independent BSS, or an ad hoc network, where an AP or BSS is not needed at all.
4. B. The AP at the heart of a BSS or cell identifies itself (and the BSS) with a Basic Service Set Identifier (BSSID). It also uses an SSID to identify the wireless network, but that is not unique to the AP or BSS. Finally, the radio MAC address is used as the basis for the BSSID value, but the value can be altered to form the BSSID for each SSID that the AP supports.
5. B. A workgroup bridge acts as a wireless client, but bridges traffic to and from a wired device connected to it.
6. B. In a mesh network, each mesh AP builds a standalone BSS. The APs relay client traffic to each other over wireless backhaul links, rather than wired Ethernet. Therefore, Ethernet cabling to each AP is not required.
7. D and E. Wi-Fi commonly uses the 2.5- and 5-GHz bands.
8. C and D. In the 2.4-GHz band, consecutively numbered channels are too wide to not overlap. Only channels 1, 6, and 11 are spaced far enough apart to avoid overlapping each other. In the 5-GHz band, all channels are considered to be nonoverlapping.
1. A. An autonomous AP can operate independently without the need for a centralized wireless LAN controller.
2. B. The Cisco Meraki APs are autonomous APs that are managed through a centralized platform in the Meraki cloud.
3. C. On a lightweight AP, the MAC function is divided between the AP hardware and the WLC. Therefore, the architecture is known as split-MAC.
4. B. An LAP builds a CAPWAP tunnel with a WLC.
5. A. A trunk link carrying three VLANs is not needed at all. A lightweight AP in local mode needs only an access link with a single VLAN; everything else is carried over the CAPWAP tunnel to a WLC. The WLC will need to be connected to three VLANs so that it can work with the LAP to bind them to the three SSIDs.
6. C. A unified WLC deployment model is based around locating the WLC in a central location, to support a very large number of APs.
7. A. The local mode is the default mode, where the AP provides at least one functional BSS that wireless clients can join to connect to the network. Normal and client modes are not valid modes. Monitor mode is used to turn the AP into a dedicated wireless sensor.
8. D. The SE-Connect mode is used for spectrum analysis. “SE” denotes the Cisco Spectrum Expert software. Otherwise, an AP can operate in only one mode at a time. The local mode is the default mode.
1. D. For effective security, you should leverage authentication, MIC, and encryption.
2. C. A message integrity check (MIC) is an effective way to protect against data tampering. WIPS is not correct because it provides intrusion protection functions. WEP is not correct because it does not provide data integrity along with its weak encryption. EAP is not correct because it defines the framework for authentication.
3. D. WEP is known to have a number of weaknesses and has been compromised. Therefore, it has been officially deprecated and should not be used in a wireless network. AES is not a correct answer because it is the current recommended encryption method. WPA is not correct because it defines a suite of security methods. EAP is not correct because it defines a framework for authentication.
4. C. EAP works with 802.1x to authenticate a client and enable access for it. Open authentication and WEP cannot be correct because both define a specific authentication method. WPA is not correct because it defines a suite of security methods in addition to authentication.
5. A. The TKIP method was deprecated when the 802.11 standard was updated in 2012. CCMP and GCMP are still valid methods. EAP is an authentication framework and is not related to data encryption and integrity.
6. C. WPA2 uses CCMP only. WEP has been deprecated and is not used in any of the WPA versions. TKIP has been deprecated but can be used in WPA only. WPA is not a correct answer because it is an earlier version of WPA2.
7. B. The Wi-Fi Alliance offers the WPA, WPA2, and WPA3 certifications for wireless security. WEP, AES, and 802.11 are not certifications designed and awarded by the Wi-Fi Alliance.
8. A and C. The personal mode for WPA, WPA2, and WPA3 is used to require a pre-shared key authentication. Enterprise mode uses 802.1x instead.
1. A. A lightweight AP requires connectivity to only a single VLAN, so access mode is used.
2. B. An autonomous AP must connect to each of the VLANs it will extend to wireless LANs. Therefore, its link should be configured as a trunk.
3. D. You can use HTTP and HTTPS to access the GUI of a wireless LAN controller, as well as SSH to access its CLI. While HTTP is a valid management protocol on a WLC, it is usually disabled to make the WLC more secure.
4. C. Controllers use a link aggregation group (LAG) to bundle multiple ports together.
5. D. A dynamic interface makes a logical connection between a WLAN and a VLAN, all internal to the controller.
6. C and D. A WLAN binds an SSID to a controller interface so that the controller can link the wired and wireless networks. Although the WLAN ultimately reaches a wired VLAN, it does so only through a controller interface. It is the interface that is configured with a VLAN number.
7. C. You can configure a maximum of 512 WLANs on a controller. However, a maximum of only 16 of them can be configured on an AP.
8. A and C. The SSID and controller interface are the only parameters from the list that are necessary. The VLAN number is not because it is supplied when a controller interface is configured.