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Question 1001Question →

A network technician is configuring Layer 2 neighbor discovery protocols on a Cisco Catalyst switch connected to non-Cisco IP phones and third-party switches. Which two statements regarding the default operations and characteristics of CDP and LLDP are accurate?

Select all that apply

Show answer & explanation

Answer: LLDP is an open-standard protocol defined under IEEE 802.1AB, whereas CDP is a Cisco-proprietary protocol.; CDP is enabled globally by default on Cisco IOS switches, while LLDP typically requires global enablement using the 'lldp run' command.

Answer

The two correct statements are that LLDP is an IEEE 802.1AB open standard while CDP is Cisco-proprietary, and that CDP is enabled by default on Cisco IOS switches while LLDP must be globally enabled using the 'lldp run' command.
LLDP is defined by IEEE 802.1AB as an open standard protocol, making it suitable for environments with equipment from multiple vendors, whereas CDP is a Cisco-proprietary protocol. Furthermore, Cisco switches default to having CDP enabled out-of-the-box, while LLDP must be globally enabled using the global configuration command 'lldp run'.

Step-by-Step Solution

1
Analyze standard compliance for Layer 2 discovery protocols.
LLDP is standardized under IEEE 802.1AB for multi-vendor interoperability, whereas CDP is proprietary to Cisco.
Understanding standard vs. proprietary protocol behavior is fundamental for multi-vendor network integration.
2
Evaluate default operating states on Cisco switches.
CDP is active globally out of the box on Cisco IOS, while LLDP requires explicit global activation via 'lldp run'.
Knowing default service states prevents assumptions about neighbor discovery availability.

Key Concept

Layer 2 Discovery Protocol Differences (CDP vs LLDP)
Question 1002Question →

A network administrator is assigning IP addresses to devices within the 172.16.10.16/28172.16.10.16/28 subnet. Which TWO of the following IPv4 addresses are valid host addresses that can be assigned to network interfaces on this subnet? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: 172.16.10.17; 172.16.10.28

Answer

The IPv4 addresses 172.16.10.17 and 172.16.10.28 are valid assignable host addresses.
For the prefix 172.16.10.16/28172.16.10.16/28, the subnet mask is 255.255.255.240255.255.255.240, giving a subnet block size of 1616. The subnet range spans from 172.16.10.16172.16.10.16 to 172.16.10.31172.16.10.31. Within this subnet, 172.16.10.16172.16.10.16 is the reserved network ID and 172.16.10.31172.16.10.31 is the reserved subnet broadcast address. The usable host address range is 172.16.10.17172.16.10.17 through 172.16.10.30172.16.10.30. Therefore, 172.16.10.17172.16.10.17 and 172.16.10.28172.16.10.28 are the two valid host addresses.

Step-by-Step Solution

1
Determine the subnet block size from the prefix length.
A /28 prefix corresponds to a subnet mask of 255.255.255.240. The block size in the 4th octet is 256−240=16256 - 240 = 16.
Knowing the block size helps establish the boundaries of the subnet.
2
Calculate the network address, broadcast address, and usable host range.
Network address = 172.16.10.16; Broadcast address = 172.16.10.16+16−1=172.16.10.31172.16.10.16 + 16 - 1 = 172.16.10.31; Usable host range = 172.16.10.17 to 172.16.10.30.
The first address in the subnet is reserved for the network ID, and the last address is reserved for the broadcast address.
3
Evaluate each given option against the usable host range.
172.16.10.17 and 172.16.10.28 fall within 172.16.10.17-172.16.10.30, whereas 172.16.10.16 is the network ID and 172.16.10.31 is the broadcast address.
Only IP addresses strictly between the network ID and broadcast address are assignable to host interfaces.

Key Concept

IPv4 Subnet Boundaries and Usable Host Address Calculation
Question 1003Question →

An enterprise organization is expanding its campus wireless infrastructure by deploying lightweight access points (LAPs) across several buildings. To simplify administration, a network engineer installs a central device to handle radio frequency (RF) management, client authentication policies, and control-plane processing for all LAPs. Which network component performs these centralized control functions?

Show answer & explanation

Answer: Wireless LAN Controller (WLC)

Answer

The Wireless LAN Controller (WLC) is the network component responsible for centralized control-plane management, security policy enforcement, and RF coordination in a lightweight wireless architecture.
In a centralized Cisco wireless network using Split-MAC architecture, the Wireless LAN Controller (WLC) is dedicated to handling heavy control-plane operations, including RF channel and power management, client authentication, roaming assistance, and LAP configuration management.

Step-by-Step Solution

1
Analyze the operational requirements described in the scenario.
The target component must centralize RF management, client authentication, and control-plane processing for lightweight access points.
Split-MAC architecture divides wireless responsibilities between access points and a central control entity.
2
Evaluate candidate network components against these requirements.
The Wireless LAN Controller (WLC) specifically terminates CAPWAP control tunnels from LAPs to deliver centralized management, channel assignment, power tuning, and security enforcement.
LAPs rely on the WLC for control-plane logic, whereas Layer 2 switches focus on frame switching, hypervisors manage virtual machines, and discovery protocols handle neighbor advertising.

Key Concept

Wireless LAN Controller (WLC) Role in Split-MAC Architecture
Question 1004Question →

An enterprise network administrator is configuring network isolation for virtual machines hosted on a single bare-metal server. The requirement states that traffic from different virtual machine groups must be segregated into distinct 802.1Q VLANs before traversing the server's shared physical uplink. Which component within the virtualization architecture is primarily responsible for performing Layer 2 frame switching between virtual machines and applying the required VLAN tags?

Show answer & explanation

Answer: The virtual switch (vSwitch) operating within the hypervisor

Answer

The virtual switch (vSwitch) operating within the hypervisor is responsible for Layer 2 switching and 802.1Q VLAN tagging for hosted virtual machines.
The virtual switch (vSwitch) is a software program running inside the hypervisor that functions as a Layer 2 switch. It maintains MAC tables, performs local frame forwarding between VMs on the same physical host, and tags/untags 802.1Q Ethernet frames for traffic leaving via physical uplinks.

Step-by-Step Solution

1
Identify the component that connects virtual machine virtual interfaces (vNICs) to host physical interfaces (pNICs).
The hypervisor implements a software-based Layer 2 virtual switch (vSwitch).
Virtual machines require network connectivity both locally within the host and externally.
2
Analyze which component handles Layer 2 encapsulation and forwarding features such as VLAN tagging.
The vSwitch manages port groups, maintains a virtual MAC address table, and inserts/strips 802.1Q tags on traffic.
This offloads network control logic from the individual guest operating systems while isolating VM traffic before reaching physical switches.

Key Concept

Virtual Switch (vSwitch) Operation and Functions
Question 1005Question →

A network architect is designing a two-tier spine-leaf data center fabric. Each of the 8 leaf switches connects to 48 edge servers using 10 Gbps access ports, generating 480 Gbps480\text{ Gbps} of potential access bandwidth per switch. The design policy mandates a maximum bandwidth oversubscription ratio of 3:13:1 across the spine fabric. If all spine and leaf interconnect interfaces operate at 40 Gbps, which architecture satisfies the oversubscription constraint while adhering to standard Clos topology rules?

Show answer & explanation

Answer: Deploy 4 spine switches and connect each leaf switch with a single 40 Gbps link to every spine switch.

Answer

Deploying 4 spine switches where each leaf switch has a single 40 Gbps uplink to every spine switch correctly provides 160 Gbps of uplink capacity per leaf (meeting the 3:1 oversubscription ratio) and strictly obeys Spine-Leaf bipartite graph connectivity rules.
Each leaf switch requires 480 Gbps/3=160 Gbps480\text{ Gbps} / 3 = 160\text{ Gbps} of uplink bandwidth to achieve a 3:13:1 oversubscription ratio. Using 40 Gbps links, each leaf switch needs 4 uplinks. Standard Spine-Leaf design rules dictate that every leaf switch must connect to every spine switch, and spine switches must not connect to one another. Therefore, deploying 4 spine switches with 1 link from each leaf to each spine satisfies both bandwidth and topological requirements.

Step-by-Step Solution

1
Calculate total access bandwidth per leaf switch
48×10 Gbps=480 Gbps48 \times 10\text{ Gbps} = 480\text{ Gbps} access bandwidth per leaf switch.
Determining total ingress/egress access capacity is required to calculate uplink oversubscription.
2
Calculate required uplink bandwidth based on the 3:1 oversubscription ratio limit
Required Uplink Bandwidth=480 Gbps3=160 Gbps\text{Required Uplink Bandwidth} = \frac{480\text{ Gbps}}{3} = 160\text{ Gbps}.
A 3:13:1 oversubscription ratio means uplink bandwidth must be at least one-third of the access bandwidth.
3
Determine the number of 40 Gbps uplinks required per leaf switch
160 Gbps40 Gbps=4\frac{160\text{ Gbps}}{40\text{ Gbps}} = 4 uplink connections per leaf switch.
Interconnect ports run at 40 Gbps, so 4 ports are required to achieve 160 Gbps total uplink capacity.
4
Apply Clos (Spine-Leaf) architectural connectivity rules
Deploy 4 spine switches and connect each leaf switch to every spine switch using exactly one 40 Gbps uplink.
In a 2-Tier Spine-Leaf topology, every leaf connects to every spine, spines do not connect to spines, and leaves do not connect to leaves.

Key Concept

Spine-Leaf (Clos) Topology Oversubscription and Fabric Interconnect Rules
Estimated Time:2m 0s
Question 1006Question →

Switch SW1 receives a unicast frame on ingress port GigabitEthernet0/2. The frame contains a source MAC address of 7069.5a12.1234 and a destination MAC address of 0011.2233.4455. The MAC address table of SW1 currently has no entries. Which two actions does the switch perform upon receiving this frame? (Select two.)

Select all that apply

Show answer & explanation

Answer: The switch records the source MAC address 7069.5a12.1234 and ingress port GigabitEthernet0/2 in its MAC address table.; The switch floods the frame out all active ports assigned to the same VLAN, excluding port GigabitEthernet0/2.

Answer

The switch records the source MAC address on ingress port GigabitEthernet0/2 in its MAC address table and floods the frame out all operational ports in the same VLAN except GigabitEthernet0/2.
When an Ethernet switch receives a frame, it executes two key processes: learning and forwarding. First, it reads the source MAC address (7069.5a12.1234) and maps it to the receiving interface (GigabitEthernet0/2) in its MAC address table. Second, because the destination MAC address (0011.2233.4455) is not listed in its table, the frame is treated as an unknown unicast frame. The switch floods the frame out every operational port in the VLAN except the ingress port where the frame was received.

Step-by-Step Solution

1
Examine the ingress frame's source MAC address and ingress interface.
The switch adds an entry for source MAC 7069.5a12.1234 mapped to port GigabitEthernet0/2.
Layer 2 switches perform dynamic MAC learning by recording the source address of every incoming frame.
2
Examine the ingress frame's destination MAC address against the MAC address table.
Destination MAC 0011.2233.4455 is not found in the table, identifying the frame as an unknown unicast frame.
An empty MAC table contains no matching entry for the destination.
3
Apply Layer 2 forwarding rules for an unknown unicast frame.
The frame is flooded out all active ports in the same VLAN except ingress port GigabitEthernet0/2.
Flooding ensures the unknown destination host receives the frame without retransmitting traffic out the ingress port.

Key Concept

L2 Switching Concepts and MAC Table Operation
Estimated Time:1m 0s
Question 1007Question →

A Cisco Catalyst switch receives an Ethernet frame on access port GigabitEthernet0/1 configured in VLAN 10. The frame contains a source MAC address of 0011.2233.4455 and a unicast destination MAC address of 00aa.bbcc.dde1. The switch currently maintains the following dynamic MAC address table entries:

VLANMAC AddressTypePort
100011.2233.4455DYNAMICGigabitEthernet0/2
1000aa.bbcc.dde1DYNAMICGigabitEthernet0/3
2000aa.bbcc.dde1DYNAMICGigabitEthernet0/4

Which set of actions will the switch take upon processing this ingress frame?

Show answer & explanation

Answer: Update the VLAN 10 entry for 0011.2233.4455 to GigabitEthernet0/1 and forward the frame out GigabitEthernet0/3 only.

Answer

Update the VLAN 10 entry for 0011.2233.4455 to GigabitEthernet0/1 and forward the frame out GigabitEthernet0/3 only.
When an Ethernet frame enters a switch port, the switch inspects the source MAC address to maintain its MAC address table. Since MAC 0011.2233.4455 was previously recorded on GigabitEthernet0/2 but arrived on GigabitEthernet0/1 in VLAN 10, the switch updates the table entry to point to GigabitEthernet0/1. Next, the switch inspects the destination MAC address 00aa.bbcc.dde1 within VLAN 10. Finding a matching dynamic entry for GigabitEthernet0/3 in VLAN 10, the switch forwards the frame exclusively out GigabitEthernet0/3.

Step-by-Step Solution

1
Examine the source MAC address and ingress port/VLAN context.
The source MAC address 0011.2233.4455 arrives on GigabitEthernet0/1 in VLAN 10. The current MAC table lists this address on GigabitEthernet0/2 for VLAN 10.
When a host moves or changes ports, receiving a frame on a new port triggers the switch to update (re-learn) the MAC address table entry for that specific VLAN.
2
Perform a destination MAC address lookup within the ingress VLAN.
The destination MAC address 00aa.bbcc.dde1 is present in the MAC table under VLAN 10 associated with port GigabitEthernet0/3.
Switches isolate forwarding decisions per VLAN. The matching entry in VLAN 20 on GigabitEthernet0/4 is ignored.
3
Determine the forwarding action.
The switch performs known unicast forwarding by transmitting the frame out GigabitEthernet0/3 only.
Since the destination MAC address is known within VLAN 10, no unknown unicast flooding is required.

Key Concept

Layer 2 MAC Table Learning, Station Mobility, and Per-VLAN Forwarding Decisions
Estimated Time:2m 0s
Question 1008Question →

A network administrator configures a router interface using EUI-64 address generation with the command `ipv6 address 2001:db8:100:1::/64 eui-64`. If the physical MAC address of the interface is `bc24.1122.3344`, which 64-bit interface identifier will be automatically derived?

Show answer & explanation

Answer: BE24:11FF:FE22:3344

Answer

The derived interface identifier is BE24:11FF:FE22:3344.
The 64-bit EUI-64 interface identifier is generated by taking the 48-bit MAC address (`bc24.1122.3344`), inserting the 16-bit value `FFFE` in the middle (between `11` and `22`), and flipping the 7th bit of the first byte. The binary representation of `BC` is `10111100`; flipping the 7th bit yields `10111110`, which equals `BE` in hexadecimal format. Combining these results yields `BE24:11FF:FE22:3344`.

Step-by-Step Solution

1
Split the 48-bit MAC address into two 24-bit halves.
Left half: `BC-24-11`, Right half: `22-33-44`.
EUI-64 standard requires inserting a 16-bit value between the Organizationally Unique Identifier (OUI) and the NIC-specific portion.
2
Insert the 16-bit hexadecimal value FFFE into the midpoint.
`BC-24-11-FF-FE-22-33-44`.
Expands the 48-bit MAC address to a 64-bit structure.
3
Invert the 7th bit (Universal/Local bit) of the first byte.
First byte `BC` (binary `10111100`) becomes `BE` (binary `10111110`). Combining all parts in IPv6 format yields `BE24:11FF:FE22:3344`.
In EUI-64 addresses, setting the 7th bit to 1 specifies that the interface ID is globally unique.

Key Concept

EUI-64 Interface Identifier Generation
Question 1009Question →

A network administrator needs to perform the initial out-of-band configuration of a newly unboxed Cisco Catalyst switch by connecting a laptop directly to the switch RJ-45 console port. Which cable type must be used to establish this management connection?

Show answer & explanation

Answer: Rollover cable

Answer

Rollover cable
A rollover cable is designed specifically for out-of-band management access to Cisco networking device console ports. It completely reverses the pin order from pin 1 through 8 on one connector to pins 8 through 1 on the opposing RJ-45 connector, allowing serial communication between a terminal/laptop and the device management hardware.

Step-by-Step Solution

1
Identify the purpose of the interface link
The objective is local out-of-band administrative access through the switch's RJ-45 console interface.
Console management uses serial communication signal pinouts rather than standard Ethernet network data pinouts.
2
Determine the required cabling pinout structure
A rollover cable (also known as a Yost cable or console cable) reverses the pin ordering from end to end (pin 1 to pin 8, pin 2 to pin 7, etc.), aligning the terminal adapter serial signals with the Cisco console port.
Standard Ethernet cable pinouts (T568A or T568B straight-through and crossover cables) do not map the required serial transmit and receive signals correctly.

Key Concept

Cisco Console Port Cabling and Rollover Pinouts
Question 1010Question →

An application architect is designing a lightweight heartbeat monitoring service for IoT sensors. To conserve network bandwidth over low-power links, the design requires minimizing transport layer protocol header overhead for small status payloads. Which transport layer characteristic makes UDP the most suitable choice for this application?

Show answer & explanation

Answer: UDP utilizes a fixed 8-byte header and operates without connection setup overhead.

Answer

UDP utilizes a fixed 8-byte header and operates without connection setup overhead.
UDP features a minimal, fixed 8-byte header (containing source port, destination port, length, and checksum) and operates connectionlessly without handshake or state management overhead, making it ideal for bandwidth-constrained, small-payload messaging.

Step-by-Step Solution

1
Analyze the application requirements from the scenario.
The application requires minimal transport layer header overhead for small status messages over low-bandwidth links.
Choosing between TCP and UDP depends on balancing reliability requirements against protocol overhead.
2
Compare TCP and UDP transport header sizes and connection overhead.
TCP uses a minimum 20-byte header and requires a 3-way handshake, whereas UDP uses a fixed 8-byte header and is connectionless.
UDP's 8-byte header minimizes payload ratio overhead significantly compared to TCP's 20-byte minimum header.
3
Select the option that correctly describes UDP's low-overhead characteristics.
The statement highlighting UDP's fixed 8-byte header and lack of connection setup overhead is correct.
This directly satisfies the requirement for low-power link efficiency.

Key Concept

UDP vs TCP Header Overhead and Connection Mechanics
Question 1011Question →

Two switches are connected using two physical Ethernet interfaces. A network technician configures the interfaces on Switch A with the command `channel-group 1 mode passive` to form a Link Aggregation Control Protocol (LACP) EtherChannel. Which mode must be configured on Switch B's interfaces to successfully negotiate and establish the EtherChannel bundle?

Show answer & explanation

Answer: active

Answer

active mode must be configured on Switch B.
Link Aggregation Control Protocol (LACP) requires at least one switch to actively initiate packet negotiation. Since Switch A is set to passive mode, Switch B must be configured in active mode to send LACP negotiation frames and bring up the EtherChannel.

Step-by-Step Solution

1
Identify the protocol in use.
The scenario uses LACP (IEEE 802.3ad) based on the passive mode keyword.
LACP uses active and passive modes, whereas PAgP uses desirable and auto modes.
2
Analyze LACP negotiation requirements.
Passive mode responds to LACP packets but does not send initial negotiation requests. At least one side must actively send LACP frames.
If both sides are set to passive, neither initiates negotiation and no EtherChannel is formed.
3
Determine the required remote configuration.
Switch B must be set to active mode.
An active-passive pairing successfully establishes an LACP EtherChannel bundle.

Key Concept

LACP Mode Negotiation Compatibility
Estimated Time:45s
Question 1012Question →

A network administrator executes the `show etherchannel summary` command on Switch-1 to inspect an EtherChannel connection to Switch-2:

text
Flags: D - down P - bundled in port-channel
I - stand-alone s - suspended
H - Hot-standby
SD - Layer2&Down SU - Layer2&In-use

Group Port-channel Protocol Ports
------+-------------+-----------+-----------------------------------------------
1 Po1(SD) LACP Gi0/1(I) Gi0/2(I)

Which condition on Switch-2 is the most likely cause of the `(I)` flag on the Switch-1 member interfaces?

Show answer & explanation

Answer: Switch-2 member interfaces are configured with the `channel-group 1 mode desirable` command.

Answer

Switch-2 member interfaces are configured with the `channel-group 1 mode desirable` command.
The `(I)` flag in `show etherchannel summary` output stands for stand-alone, meaning the physical interfaces are operating as separate single links rather than a bundled channel. Switch-1 is configured for LACP (Link Aggregation Control Protocol). If Switch-2 is configured with `channel-group 1 mode desirable`, it attempts to use PAgP (Port Aggregation Protocol). Because LACP and PAgP are incompatible aggregation protocols, negotiation fails completely, placing Switch-1's ports into stand-alone state.

Step-by-Step Solution

1
Analyze the CLI output flags from `show etherchannel summary`.
The port-channel Po1 is in Layer 2 Down state `(SD)`, and the physical member interfaces GigabitEthernet 0/1 and 0/2 display the `(I)` flag, indicating stand-alone mode.
The `(I)` flag signifies that the physical interfaces are operating independently as individual links because EtherChannel negotiation failed.
2
Evaluate the configured protocol and negotiation modes.
Switch-1 is attempting to form an EtherChannel using IEEE 802.3ad LACP.
LACP and Cisco proprietary PAgP protocols are mutually incompatible and cannot negotiate an aggregation bundle together.
3
Determine the configuration on Switch-2 that caused the negotiation failure.
Setting Switch-2 to `desirable` mode forces it to use PAgP. PAgP packets are ignored by Switch-1 running LACP, preventing bundle aggregation.
To form an LACP EtherChannel bundle, Switch-2 must use LACP modes (`active` or `passive`), with at least one side set to `active`.

Key Concept

LACP vs PAgP Protocol Mismatch and EtherChannel Verification
Question 1013Question →

A network administrator is designing an IPv4 addressing scheme for a local subnet that must support 6060 usable host IP addresses. The administrator assigns the smallest subnet prefix length (largest mask) that accommodates this host requirement. How many total usable host IPv4 addresses are provided by this assigned subnet?

Show answer & explanation

Answer: 62

Answer

The assigned subnet provides 62 total usable host IPv4 addresses.
To support 60 host addresses, a subnet must allocate 6 host bits because 26−2=622^6 - 2 = 62 usable addresses. 5 host bits only provide 30 usable addresses (25−22^5 - 2), which is insufficient.

Step-by-Step Solution

1
Determine the required number of host bits.
6 host bits are required (26−2=622^6 - 2 = 62).
5 host bits only yield 25−2=302^5 - 2 = 30 usable IP addresses, which is not enough for 60 hosts.
2
Calculate the usable host capacity of the subnet.
62 usable host addresses.
With 6 host bits, the total IP space is 26=642^6 = 64. Subtracting the network address and the broadcast address leaves 62 usable host IP addresses.

Key Concept

IPv4 Usable Host Calculation
Estimated Time:45s
Question 1014Question →

When a non-PortFast interface on a Cisco switch using classic IEEE 802.1D Spanning Tree Protocol (STP) comes up, it undergoes a specific sequence of port states to prevent Layer 2 loops before it begins processing user traffic. Place the following STP port states in the correct chronological order from link initialization to full data forwarding.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct operational sequence of classic 802.1D STP port states is Blocking State, followed by Listening State, then Learning State, and finally Forwarding State.
Under standard IEEE 802.1D Spanning Tree Protocol operational rules, an interface transitions sequentially from Blocking (preventing loops) to Listening (participating in BPDU role election), then to Learning (building MAC table entries), and finally reaches Forwarding (processing and forwarding user traffic).

Step-by-Step Solution

1
Identify the initial port state upon link initialization.
The switch interface starts in the Blocking state where it drops user frames and does not learn MAC addresses to prevent loops.
STP defaults to a loop-preventative state whenever a link transitions to up.
2
Determine the state focused on root bridge election and port role determination.
The interface moves to the Listening state, where BPDUs are processed and sent, but user data is still dropped and MAC learning is disabled.
The switch must determine topology roles (Root Port, Designated Port, or Alternate Port) before modifying table structures.
3
Determine the intermediate state focused on MAC table population.
After the first 15-second Forward Delay timer, the interface enters the Learning state, where it populates the MAC table from incoming frame headers but still drops user payload data.
Populating MAC addresses in advance prevents massive unicast flooding when the port starts forwarding.
4
Identify the final operational state allowing full traffic transmission.
After the second 15-second Forward Delay timer, the interface transitions to the Forwarding state.
The port is now confirmed loop-free and fully prepared to forward data frames.

Key Concept

IEEE 802.1D STP Port State Convergence Sequence
Estimated Time:45s
Question 1015Question →

An engineer wants to establish an LACP-based EtherChannel between Switch-A and Switch-B. Switch-A is configured with the `channel-group 5 mode passive` interface command. Which mode must be configured on Switch-B for the link aggregation bundle to form successfully?

Show answer & explanation

Answer: active

Answer

The interface mode on Switch-B must be set to 'active'.
Link Aggregation Control Protocol (LACP) requires at least one side of the connection to actively initiate negotiations. Because Switch-A is set to passive mode, it will only respond to LACP frames. Configuring Switch-B in active mode ensures LACP negotiation packets are transmitted, bringing up the EtherChannel.

Step-by-Step Solution

1
Identify protocol and mode characteristics
Switch-A is using LACP in passive mode.
Passive mode responds to LACP packets but does not transmit initial negotiation frames.
2
Evaluate negotiation requirements
At least one endpoint must actively initiate LACP frames.
Without an active initiator, no LACP control frames are sent.
3
Select matching mode for Switch-B
Switch-B must use 'active' mode.
Setting Switch-B to active mode initiates LACP negotiations.

Key Concept

LACP Mode Compatibility
Question 1016Question →

A network engineer is deploying IPv6 on a Cisco router interface with the burned-in MAC address `b496.910a.bcde`. The interface is enabled to automatically formulate its link-local address using standard EUI-64 formatting. What complete, compressed link-local IPv6 address will be generated on this interface?

Show answer & explanation

Answer: fe80::b696:91ff:fe0a:bcde; FE80::B696:91FF:FE0A:BCDE; fe80:0000:0000:0000:b696:91ff:fe0a:bcde; FE80:0000:0000:0000:B696:91FF:FE0A:BCDE

Answer

fe80::b696:91ff:fe0a:bcde
The MAC address `b496.910a.bcde` consists of octets `b4`, `96`, `91`, `0a`, `bc`, `de`. Inverting the 7th bit of `b4` (binary `1011 0100`) gives binary `1011 0110` (hex `b6`). Inserting `fffe` into the midpoint between `91` and `0a` creates the 64-bit interface ID `b696:91ff:fe0a:bcde`. Combining this with the IPv6 link-local prefix `fe80::/64` yields `fe80::b696:91ff:fe0a:bcde`.

Step-by-Step Solution

1
Split the 48-bit MAC address into two 24-bit halves.
Left half: `b4-96-91`, Right half: `0a-bc-de`.
The 64-bit interface identifier is constructed by inserting a 16-bit hex delimiter into the middle of the 48-bit MAC address.
2
Invert the Universal/Local (U/L) bit (the 7th bit of the first byte).
The first byte `b4` (hex `1011 0100` in binary) becomes `1011 0110` (hex `b6`).
In EUI-64 conversion, the 7th bit (U/L bit) of the first byte must be inverted (0 becomes 1 for globally unique MAC addresses).
3
Insert the 16-bit hexadecimal string `FFFE` between the two halves.
The 64-bit interface identifier becomes `b696:91ff:fe0a:bcde`.
Inserting `FFFE` expands the 48-bit MAC address into a valid 64-bit EUI-64 interface identifier.
4
Prepend the IPv6 link-local prefix (`fe80::/64`).
The full link-local address is `fe80::b696:91ff:fe0a:bcde`.
IPv6 link-local addresses use the reserved prefix `fe80::/10` with standard interface configuration defaulting to `fe80::/64` followed by the 64-bit EUI-64 interface ID.

Key Concept

IPv6 EUI-64 Link-Local Address Derivation
Estimated Time:2m 0s
Question 1017Question →

A network engineer is configuring a routed (Layer 3) EtherChannel between two Cisco Catalyst switches, Switch-1 and Switch-2, across physical interfaces GigabitEthernet 0/1 and GigabitEthernet 0/2. Switch-1 is configured with LACP in active mode (`channel-group 10 mode active`). Which TWO requirements must be met to correctly establish and operate this Layer 3 LACP EtherChannel bundle? (Select two choices.)

Select all that apply

Show answer & explanation

Answer: The `no switchport` command must be applied to both the logical interface Port-channel 10 and the physical member interfaces prior to configuring the IP address.; Switch-2 physical member interfaces can be configured with `channel-group 10 mode passive` to negotiate the LACP bundle successfully.

Answer

The correct requirements are converting both the physical member interfaces and the logical port-channel interface to Layer 3 using the 'no switchport' command before assigning the IP address, and configuring Switch-2 in LACP passive mode (or active mode) to pair with Switch-1 active mode.
For Layer 3 EtherChannels, physical member ports and the port-channel interface must be configured as routed interfaces via `no switchport`, and the IP address is assigned to the logical port-channel interface. Furthermore, LACP allows dynamic bundle establishment when one side is in active mode and the peer side is in passive mode.

Step-by-Step Solution

1
Evaluate Layer 3 EtherChannel interface configuration requirements.
Layer 3 EtherChannels require member interfaces and the port-channel interface to have switchport functionality disabled using `no switchport`. The IP address is configured on the Port-channel interface, not on individual member interfaces.
Physical member ports inherit Layer 3 configurations from the logical port-channel.
2
Analyze protocol compatibility for LACP mode negotiation.
LACP uses modes `active` and `passive`. Active-Active and Active-Passive pairings form LACP bundles. Modes `auto` and `desirable` belong to PAgP and cannot interoperate with LACP.
Mixing LACP and PAgP protocols prevents EtherChannel formation.

Key Concept

Layer 3 EtherChannel LACP Mode Negotiation and Routed Interface Setup
Question 1018Question →

A network engineer is configuring a routed (Layer 3) EtherChannel bundle (Port-channel 12) on a Cisco Catalyst switch using Link Aggregation Control Protocol (LACP). The bundle must aggregate physical interfaces GigabitEthernet1/0/1 and GigabitEthernet1/0/2, actively negotiate LACP trunking, and host the IPv4 default gateway subnet interface with IP address 192.168.100.1/30. Place the Cisco IOS configuration command steps in the correct sequential order from first to last to complete this configuration cleanly.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of commands is: 1) interface range GigabitEthernet1/0/1 - 2, 2) no switchport, 3) channel-group 12 mode active, 4) interface Port-channel 12, 5) ip address 192.168.100.1 255.255.255.252.
To create a routed (Layer 3) EtherChannel using LACP, the physical interfaces must first be selected ('interface range GigabitEthernet1/0/1 - 2') and converted to Layer 3 routed ports ('no switchport'). Next, the LACP dynamic bundle is formed using 'channel-group 12 mode active'. Afterwards, the administrator switches to the logical bundle interface context ('interface Port-channel 12') to assign the IPv4 address ('ip address 192.168.100.1 255.255.255.252').

Step-by-Step Solution

1
Select the target physical member interfaces
Enters interface range configuration mode for GigabitEthernet1/0/1 and 1/0/2.
Configuration changes must apply to both member links simultaneously.
2
Remove Layer 2 switching capabilities
Converts physical member interfaces to Layer 3 routed ports.
Issuing 'no switchport' on member interfaces prior to bundling ensures the created port-channel automatically inherits routed Layer 3 operational mode.
3
Create and bind member interfaces to the channel-group
Adds physical member interfaces to Port-channel 12 using LACP active mode.
The 'mode active' keyword specifies active LACP packet transmission to initiate bundle formation.
4
Navigate to the logical bundle interface
Enters configuration mode for interface Port-channel 12.
Logical layer configurations (such as IP addressing) must be applied directly to the port-channel interface.
5
Configure IP address on the port-channel
Assigns IP address 192.168.100.1/30 to Port-channel 12.
Layer 3 EtherChannels hold IP configuration on the logical interface, allowing traffic routing over the aggregated links.

Key Concept

Layer 3 EtherChannel LACP Configuration Sequence
Question 1019Question →

A network architect is designing a virtualized data center infrastructure utilizing bare-metal hypervisors and virtual switches. Which two statements accurately describe the roles and functions of these components? (Choose two.)

Select all that apply

Show answer & explanation

Answer: The Type 1 hypervisor runs directly on the underlying server hardware without requiring a host operating system.; The virtual switch handles Layer 2 frame forwarding between virtual machines configured on the same host.

Answer

The correct statements are that the Type 1 hypervisor runs directly on physical server hardware without requiring a host operating system, and that the virtual switch handles Layer 2 frame forwarding between virtual machines on the same host.
Type 1 hypervisors install directly onto physical hardware without requiring a host OS, providing efficient allocation of hardware resources to virtual machines. The internal virtual switch operates at Layer 2 to manage frame forwarding between virtual machines connected to the same virtual network.

Step-by-Step Solution

1
Analyze the architecture of Type 1 hypervisors
Confirm that Type 1 hypervisors (bare-metal) install directly on the hardware platform without an intermediate operating system.
Direct access to physical hardware provides higher efficiency and lower processing overhead for enterprise virtualization.
2
Examine the role of software virtual switches inside hypervisors
Confirm that virtual switches perform Layer 2 forwarding for guest virtual machine network traffic on the host.
Virtual switches mimic physical Layer 2 switches to connect virtual network interface cards (vNICs).
3
Evaluate distractor statements for conceptual misconceptions
Identify that operating on top of a host OS is characteristic of Type 2 hypervisors, and that virtual switches operate at Layer 2 rather than separating broadcast domains at Layer 3 automatically.
Hypervisor classification and OSI layer responsibilities must be strictly distinguished.

Key Concept

Hypervisor Types and Virtual Switch Operations
Question 1020Question →

A network architect is designing a Variable Length Subnet Mask (VLSM) addressing scheme for a enterprise branch using the allocated summary block 172.16.48.0/20172.16.48.0/20. Subnets must be allocated sequentially starting from the beginning of the block, prioritized from the largest host requirement to the smallest, to preserve contiguous unassigned space. The host requirements for the subnets are as follows:

- Department Alpha: 120120 hosts
- Department Beta: 6060 hosts
- Department Gamma: 2828 hosts
- Department Delta: 1212 hosts

Which IPv4 network address and subnet mask must be assigned to Department Gamma?

Show answer & explanation

Answer: 172.16.48.192172.16.48.192 with subnet mask 255.255.255.224255.255.255.224

Answer

Department Gamma should be assigned network address 172.16.48.192 with subnet mask 255.255.255.224.
The correct response identifies 172.16.48.192 255.255.255.224 as the network address and subnet mask. Calculating sequentially from the summary block 172.16.48.0/20: Department Alpha uses 172.16.48.0/25 (128 addresses, 126 usable), ending at 172.16.48.127. Department Beta uses 172.16.48.128/26 (64 addresses, 62 usable), ending at 172.16.48.191. Department Gamma requires 28 usable hosts, which fits in a /27 subnet (32 addresses, 30 usable, mask 255.255.255.224) starting at the next boundary: 172.16.48.192.

Step-by-Step Solution

1
Determine prefix length and block size for Department Alpha (120 hosts)
Requires 120 hosts. Formula 2n−2≥120  ⟹  n=72^n - 2 \ge 120 \implies n=7 host bits. Subnet mask /25 (255.255.255.128, block size 128). Range: 172.16.48.0 to 172.16.48.127.
Always allocate largest subnets first to optimize contiguous IP space.
2
Determine prefix length and block size for Department Beta (60 hosts)
Requires 60 hosts. Formula 2n−2≥60  ⟹  n=62^n - 2 \ge 60 \implies n=6 host bits. Subnet mask /26 (255.255.255.192, block size 64). Next available boundary is 172.16.48.128. Range: 172.16.48.128 to 172.16.48.191.
Department Beta starts immediately after Department Alpha's broadcast address.
3
Calculate network address and subnet mask for Department Gamma (28 hosts)
Requires 28 hosts. Formula 2n−2≥28  ⟹  n=52^n - 2 \ge 28 \implies n=5 host bits (25−2=302^5 - 2 = 30 usable hosts). Subnet mask /27 (255.255.255.224, block size 32). Next available boundary is 172.16.48.192.
Department Gamma starts at 172.16.48.192/27, providing 30 usable host IPs (172.16.48.193 - 172.16.48.222).

Key Concept

Variable Length Subnet Masking (VLSM) and Host Capacity Calculation
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