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Question 281Question →

A network technician is configuring switch interfaces for end-user PCs on an enterprise access switch. Which classic 802.1D Spanning Tree Protocol (STP) port states are bypassed when Cisco STP PortFast is enabled on an access interface?

Show answer & explanation

Answer: Listening and Learning

Answer

Listening and Learning
Enabling PortFast on an access port allows the interface to bypass the 802.1D Listening and Learning states (which total 30 seconds of forward delay) and transition directly into the Forwarding state as soon as the link comes up.

Step-by-Step Solution

1
Recall standard 802.1D STP port state transition sequence
Standard 802.1D ports transition sequentially through Blocking -> Listening (15s forward delay) -> Learning (15s forward delay) -> Forwarding.
Listening determines topology structure and processes BPDUs, while Learning populates MAC address tables before frame forwarding.
2
Determine the effect of enabling PortFast on an access interface
PortFast allows access ports directly connected to end-user hosts to transition immediately from Blocking to Forwarding.
Bypassing the 30-second combined Listening (15s) and Learning (15s) delay prevents host application and DHCP discovery timeouts upon link initialization.

Key Concept

STP PortFast operational behavior and state transition bypass
Estimated Time:45s
Question 282Question →

A network engineer is designing a Variable Length Subnet Mask (VLSM) IP addressing scheme for a newly provisioned branch office using the summary block 192.168.10.0/24192.168.10.0/24. Subnets must be allocated sequentially without wasting address space, starting from the lowest available IP address in descending order of host requirements:

- Subnet A (Staff): Requires 5555 usable host IPv4 addresses
- Subnet B (Sales): Requires 2828 usable host IPv4 addresses
- Subnet C (VoIP): Requires 1212 usable host IPv4 addresses
- Subnet D (WAN Link): Requires 22 usable host IPv4 addresses

What is the broadcast address of Subnet C?

Show answer & explanation

Answer: 192.168.10.111

Answer

The broadcast address of Subnet C is 192.168.10.111.
Subnet C requires 12 host addresses. Satisfying 2n−2≥122^n - 2 \ge 12 requires n=4n=4 host bits (1414 usable IPs), which yields a /28/28 prefix with a block size of 16. Allocating Subnet A (5555 hosts →/26\rightarrow /26, range .0–.63) and Subnet B (2828 hosts →/27\rightarrow /27, range .64–.95) places Subnet C at network address 192.168.10.96. Adding the block size of 16 gives the subnet boundary 192.168.10.96 to 192.168.10.111, making 192.168.10.111 the broadcast address.

Step-by-Step Solution

1
Determine the prefix and range for Subnet A (55 hosts)
Subnet A requires a /26 prefix (block size 64). Subnet A spans 192.168.10.0 to 192.168.10.63.
Formula 2n−2≥552^n - 2 \ge 55 yields n=6n=6 host bits (6262 usable IPs). The prefix is /26/26 (32−632-6).
2
Determine the prefix and range for Subnet B (28 hosts)
Subnet B starts at 192.168.10.64 with a /27 prefix (block size 32). Subnet B spans 192.168.10.64 to 192.168.10.95.
Formula 2n−2≥282^n - 2 \ge 28 yields n=5n=5 host bits (3030 usable IPs). The prefix is /27/27 (32−532-5).
3
Calculate the network address, block size, and broadcast address for Subnet C (12 hosts)
Subnet C starts at 192.168.10.96 with a /28 prefix (block size 16). The network range is 192.168.10.96/28, giving a broadcast address of 192.168.10.111.
Formula 2n−2≥122^n - 2 \ge 12 yields n=4n=4 host bits (1414 usable IPs). The prefix is /28/28 (32−432-4). Adding the block size of 16 to the network start 192.168.10.96 yields a range of 192.168.10.96 to 192.168.10.111.

Key Concept

VLSM block allocation order and subnet broadcast address determination
Estimated Time:2m 0s
Question 283Question →

Place the steps of the Rapid PVST+ Proposal and Agreement handshake process in the correct chronological order, starting from when a designated switch initiates fast convergence on a point-to-point link to when its port enters the Forwarding state.

Drag items to arrange them in the correct order

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Answer

The correct sequence starts with the designated switch proposing the link state, followed by the receiving switch synchronizing non-edge ports, returning an Agreement BPDU, and finally the designated switch transitioning directly to the Forwarding state.
The Rapid PVST+ (802.1w) sync process operates in a strict sequence on point-to-point links: first, a designated port proposes rapid transition; second, the neighbor switch synchronizes by blocking non-edge designated ports; third, the neighbor sends an Agreement BPDU back; fourth, the proposing port immediately moves to the Forwarding state.

Step-by-Step Solution

1
Identify the initiation of the handshake.
The designated switch transmits a Proposal BPDU to request rapid transition on the point-to-point link.
Rapid PVST+ relies on explicit point-to-point messaging rather than timer-based delays.
2
Determine the synchronization action taken by the receiver.
The receiver puts non-edge ports into the Discarding state.
This guarantees no active loops can form while the topology adapts.
3
Identify the response message.
The receiver sends an Agreement BPDU back to the proposal sender.
The agreement signals that the receiving switch has safely isolated its local ports.
4
Determine the final state transition.
The initiating designated port transitions immediately into the Forwarding state.
Upon receiving agreement, the port moves directly to forwarding without waiting for 802.1D forward delay timers.

Key Concept

Rapid PVST+ Proposal and Agreement Handshake Mechanism
Question 284Question →

A network operations center (NOC) engineer observes that workstations attached to switch SW-CORE-01 experience a 30-second initial DHCP request timeout upon link activation before successfully receiving an IP address. Interface GigabitEthernet1/0/5 is currently configured as a standard 802.1D access port. Which operational mechanism explains why enabling Cisco STP PortFast resolves this DHCP timeout issue, and what is its specific effect on STP Topology Change Notifications (TCNs)?

Show answer & explanation

Answer: PortFast allows the interface to bypass the Listening and Learning states to enter Forwarding immediately upon link up, while suppressing TCN generation when the edge port link status changes.

Answer

PortFast allows the interface to bypass the Listening and Learning states to enter Forwarding immediately upon link up, while suppressing TCN generation when the edge port link status changes.
The correct answer identifies that PortFast allows access ports connected to end hosts to transition directly to the Forwarding state without waiting through the 15-second Listening and 15-second Learning states (a total 30-second convergence delay). This eliminates initialization delays for client services such as DHCP. In addition, link state fluctuations on a PortFast-enabled port do not trigger Topology Change Notifications (TCNs), avoiding network-wide MAC address table flushing.

Step-by-Step Solution

1
Analyze standard 802.1D STP convergence behavior on access ports.
Standard 802.1D transitions through Blocking -> Listening (15s) -> Learning (15s) -> Forwarding (total 30s delay).
During the 30-second delay, frames are dropped, causing client protocols like DHCP to time out while waiting for link establishment.
2
Determine the primary operational transition bypass provided by PortFast.
PortFast moves an edge interface immediately from Blocking to Forwarding upon link bring-up.
By bypassing Listening and Learning states, connectivity is instantaneous, eliminating host initialization timeouts.
3
Analyze the secondary benefit regarding Spanning Tree Topology Change Notifications (TCNs).
PortFast suppresses TCN generation on edge link state changes.
End-user PC power cycles or disconnects should not force switches across the network to shorten their MAC table aging timers.

Key Concept

STP PortFast immediate forwarding transition and TCN suppression on access ports
Estimated Time:2m 0s
Question 285Question →

A network administrator is designing a network for a medium-sized office building. To reduce hardware costs and simplify maintenance while maintaining connectivity between access layer devices and external networks, the administrator decides to combine the responsibilities of the core layer and the distribution layer into a single set of switches. Which network topology architecture is being implemented in this enterprise design?

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Answer: Collapsed core architecture

Answer

The collapsed core architecture is being implemented because it merges the distribution layer and core layer roles into a single pair of switches.
In a collapsed core network design (also known as a two-tier campus design), the distribution layer and core layer functionality are merged into a single tier of redundant switches. This provides routing, security policy enforcement, and high-speed campus backbone connectivity on the same hardware, reducing overall cost and complexity for smaller site footprints.

Step-by-Step Solution

1
Analyze the requirements described in the scenario
The design merges core and distribution layer duties into one switch tier to save cost in a medium-sized campus.
Identifying key structural consolidations helps differentiate enterprise network models.
2
Match the requirement to the standard Cisco topology definitions
A 2-tier design where core and distribution layers are combined is specifically known as a collapsed core topology.
Collapsed core models reduce tier count while retaining access layer connections.

Key Concept

Collapsed Core Topology
Question 286Question →

A network administrator is documenting the RFC 1918 private IPv4 address space allocation for an enterprise network redesign. Match each RFC 1918 address block on the left with its corresponding CIDR prefix length and block composition description on the right.

Click a left item, then click its matching right item

Items

10.0.0.0 – 10.255.255.255
172.16.0.0 – 172.31.255.255
192.168.0.0 – 192.168.255.255

Matches

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Answer

10.0.0.0 – 10.255.255.255 matches '/8 prefix length comprising 1 Class A network block'; 172.16.0.0 – 172.31.255.255 matches '/12 prefix length comprising 16 contiguous Class B network blocks'; 192.168.0.0 – 192.168.255.255 matches '/16 prefix length comprising 256 contiguous Class C network blocks'.
Each RFC 1918 private IPv4 block aligns with a specific CIDR prefix and quantity of traditional classful networks: 10.0.0.0/8 represents one Class A block; 172.16.0.0/12 represents 16 contiguous Class B blocks (172.16.0.0 to 172.31.255.255); and 192.168.0.0/16 represents 256 contiguous Class C blocks (192.168.0.0 to 192.168.255.255).

Step-by-Step Solution

1
Identify the RFC 1918 Class A private address range
10.0.0.0 to 10.255.255.255 corresponds to 10.0.0.0/8 (1 Class A network).
RFC 1918 reserves the entire 10.0.0.0/8 network block for private IP addressing.
2
Identify the RFC 1918 Class B private address range
172.16.0.0 to 172.31.255.255 corresponds to 172.16.0.0/12, which spans 16 contiguous Class B networks (172.16.0.0/16 through 172.31.0.0/16).
A /12 prefix length allows 4 bits of subnetwork variation in the second octet (16 to 31), creating 16 Class B ranges.
3
Identify the RFC 1918 Class C private address range
192.168.0.0 to 192.168.255.255 corresponds to 192.168.0.0/16, which spans 256 contiguous Class C networks (192.168.0.0/24 through 192.168.255.0/24).
A /16 prefix length allows 8 bits of subnetwork variation in the third octet (0 to 255), creating 256 Class C ranges.

Key Concept

RFC 1918 Private IPv4 Address Space Specifications
Question 287Question →

An enterprise wireless administrator is selecting appropriate antenna types for various deployment scenarios across a corporate campus. Match each wireless antenna type on the left with its primary operational characteristic and recommended deployment scenario on the right.

Click a left item, then click its matching right item

Items

Omnidirectional dipole antenna
Patch directional antenna
Yagi directional antenna
Parabolic dish antenna

Matches

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Answer

Omnidirectional dipole antenna matches 360-degree horizontal coverage for open office environments; Patch directional antenna matches broad directional hemispherical pattern for wall mounting along corridors; Yagi directional antenna matches moderate gain focused beamwidth for short-to-medium outdoor bridges; Parabolic dish antenna matches extremely high gain tight beamwidth for long-distance point-to-point links.
Each antenna type is accurately paired with its corresponding radiation pattern, beamwidth, gain characteristic, and recommended deployment application.

Step-by-Step Solution

1
Analyze indoor open-space coverage requirements versus corridor coverage requirements.
Omnidirectional dipole antennas radiate 360 degrees horizontally for open spaces, whereas patch antennas direct RF energy forward in a hemispherical pattern ideal for hallways.
Matching antenna radiation pattern to physical space geometry prevents unnecessary RF bleed and signal degradation.
2
Evaluate outdoor point-to-point wireless bridging requirements based on distance and gain requirements.
Yagi antennas provide moderate gain for short-to-medium building links, while parabolic dish antennas provide high gain and narrow beamwidth for long-distance links.
Long-distance outdoor links require maximum directional gain to overcome free space path loss.

Key Concept

Wireless Antenna Types and RF Radiation Characteristics
Question 288Question →

A network administrator needs to configure a routed (Layer 3) LACP EtherChannel (Port-Channel 10) using physical member interfaces GigabitEthernet0/1 and GigabitEthernet0/2, and assign the IPv4 address 10.1.1.1/30 to the bundle. Place the CLI configuration steps in the correct operational sequence from first to last.

Drag items to arrange them in the correct order

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Answer

The correct configuration sequence is: First select physical interfaces with `interface range GigabitEthernet0/1 - 2`, second disable Layer 2 operation using `no switchport`, third associate member interfaces into LACP using `channel-group 10 mode active`, fourth enter logical interface mode via `interface Port-channel10`, and fifth configure the IPv4 address using `ip address 10.1.1.1 255.255.255.252`.
The correct operational sequence begins by selecting physical interfaces using `interface range GigabitEthernet0/1 - 2`, converting them to routed interfaces with `no switchport`, and binding them to the dynamic LACP channel group via `channel-group 10 mode active`. Next, configuration mode shifts to the created virtual interface using `interface Port-channel10`, where the IP address is configured with `ip address 10.1.1.1 255.255.255.252`.

Step-by-Step Solution

1
Select physical interfaces
Enters interface range configuration mode for GigabitEthernet0/1 and GigabitEthernet0/2.
Configuration options must be applied to physical interfaces simultaneously using interface range mode.
2
Convert interfaces to Layer 3
Removes switchport capabilities from member interfaces.
Physical member ports must be converted to routed ports with `no switchport` prior to creating a Layer 3 EtherChannel.
3
Configure LACP channel group
Creates Port-channel10 and enables active dynamic negotiation.
Executing `channel-group 10 mode active` dynamically creates the Port-channel interface using LACP.
4
Enter logical Port-Channel interface
Transitions CLI prompt to `config-if` mode for Port-channel10.
Network layer configuration parameters must be applied to the virtual interface rather than physical ports.
5
Assign IPv4 addressing
Applies IP subnet 10.1.1.1/30 to the EtherChannel logical interface.
Layer 3 EtherChannels route traffic based on the IP address assigned directly to the Port-Channel interface.

Key Concept

Layer 3 EtherChannel (LACP) CLI Configuration Sequence
Estimated Time:1m 30s
Question 289Question →

Match each server virtualization component or architecture on the left to its corresponding functional description on the right.

Click a left item, then click its matching right item

Items

Virtual Network Interface Card (vNIC)
Virtual Switch (vSwitch)
Type 1 Hypervisor
Type 2 Hypervisor

Matches

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Answer

Virtual Network Interface Card (vNIC) pairs with the virtualized endpoint abstraction maintaining a distinct MAC address. Virtual Switch (vSwitch) pairs with the software forwarding module routing frames between VMs and physical NICs. Type 1 Hypervisor pairs with the bare-metal virtualization layer installed directly on server hardware. Type 2 Hypervisor pairs with the hosted program running on top of an existing host operating system.
The pairings accurately reflect the core architecture of network virtualization: vNIC provides guest virtual MAC addresses and network interface emulation; vSwitch handles software-based Ethernet switching inside the hypervisor host; Type 1 hypervisors execute bare-metal on hardware for enterprise server virtualization; and Type 2 hypervisors run hosted inside an existing general-purpose host OS for desktop/testing applications.

Step-by-Step Solution

1
Differentiate hypervisor deployment models based on where the software runs relative to host hardware.
Identify Type 1 as bare-metal software operating directly on physical hardware, and Type 2 as hosted software running on top of a primary host OS.
Type 1 hypervisors interact directly with hardware hardware resources, while Type 2 hypervisors rely on host OS system calls for resource scheduling.
2
Analyze virtual networking abstractions presented to guest operating systems versus those performing intra-host traffic management.
Identify vNIC as the per-VM adapter assigned a MAC address, and vSwitch as the internal software bridge linking vNICs to each other and to host pNICs.
The vNIC presents an Ethernet interface to the guest OS, whereas the vSwitch maintains local Layer 2 MAC address tables for switching frames internally.
3
Associate each term on the left with its precise functional description on the right.
Complete all four correct pair mappings.
Ensures complete alignment between virtualization concepts and their operational definitions in enterprise network environments.

Key Concept

Virtualization components (vNIC, vSwitch) and hypervisor architecture classification (Type 1 bare-metal vs Type 2 hosted).
Estimated Time:2m 0s
Question 290Question →

An administrator enables IPv6 on a router interface using the prefix 2001:db8:feed:1::/642001:\text{db8}:\text{feed}:1::/64 with EUI-64 automatic interface identifier generation. If the interface MAC address is `1C1D.86AB.CDEF`, which IPv6 address is dynamically assigned to the interface?

Show answer & explanation

Answer: 2001:db8:feed:1:1e1d:86ff:feab:cdef

Answer

2001:db8:feed:1:1e1d:86ff:feab:cdef
Generating an IPv6 EUI-64 interface identifier from a 48-bit MAC address requires splitting the MAC address into two 24-bit halves, inserting the hexadecimal sequence FFFE in the center, and flipping the 7th bit (Universal/Local bit) of the first byte. For MAC address 1C1D.86AB.CDEF, the first byte 0x1C (binary 00011100) becomes 0x1E (binary 00011110) after flipping the 7th bit. Inserting FFFE into the center yields the 64-bit interface ID 1e1d:86ff:feab:cdef, which combines with prefix 2001:db8:feed:1::/64 to form 2001:db8:feed:1:1e1d:86ff:feab:cdef.

Step-by-Step Solution

1
Split MAC address into two 24-bit halves
First half: 1C-1D-86, Second half: AB-CD-EF
EUI-64 insertion requires placing a 16-bit payload into the middle of the 48-bit MAC address.
2
Insert the 16-bit value FFFE into the center
1C-1D-86-FF-FE-AB-CD-EF
Standard EUI-64 processing expands 48-bit MAC addresses to 64-bit interface identifiers.
3
Invert the 7th bit (Universal/Local bit) of the first octet
0x1C (0001 1100) becomes 0x1E (0001 1110), yielding 1E-1D-86-FF-FE-AB-CD-EF
RFC 4291 requires flipping the 7th bit of the IEEE MAC address when converting to EUI-64 format.
4
Combine with IPv6 prefix
2001:db8:feed:1:1e1d:86ff:feab:cdef
Appending the 64-bit interface identifier to the /64 prefix constructs the complete IPv6 address.

Key Concept

EUI-64 Interface Identifier Generation
Question 291Question →

An application developer is optimizing a custom telemetry protocol for bandwidth-constrained satellite links. To minimize transport layer encapsulation overhead, the network architecture replaces TCP with UDP. Assuming standard baseline headers with no TCP options, how many bytes of fixed header overhead are saved per packet by selecting UDP over TCP?

Show answer & explanation

Answer: 12 bytes

Answer

12 bytes of Layer 4 header overhead are saved per datagram by using UDP instead of TCP.
The correct answer is 12 bytes. A standard TCP header without optional fields requires 20 bytes to accommodate features such as sequencing, acknowledgments, dynamic windowing, and flags. Conversely, UDP utilizes a simplified 8-byte header composed of only four fields: Source Port, Destination Port, Length, and Checksum. Subtracting the 8-byte UDP header from the 20-byte baseline TCP header yields a savings of 12 bytes per packet.

Step-by-Step Solution

1
Determine the baseline size of a standard TCP header.
A standard TCP header without optional fields consists of 20 bytes (containing Source/Destination Ports, Sequence Number, Acknowledgment Number, Header Length, Flags, Window Size, Checksum, and Urgent Pointer).
Establishing the default minimum size of a connection-oriented Layer 4 protocol header.
2
Determine the fixed size of a UDP header.
A UDP header is strictly fixed at 8 bytes (consisting of 4 two-byte fields: Source Port, Destination Port, Length, and Checksum).
UDP provides connectionless delivery with minimal control overhead.
3
Calculate the difference in overhead between the two protocols.
20 bytes (TCP baseline header) - 8 bytes (UDP fixed header) = 12 bytes saved.
Subtracting UDP header size from baseline TCP header size identifies the exact per-packet byte savings.

Key Concept

Layer 4 Header Overhead and Structure (TCP 20-byte vs. UDP 8-byte)
Question 292Question →

Match each host virtualization component on the left with its corresponding operational characteristic or role on the right.

Click a left item, then click its matching right item

Items

Type 1 Hypervisor
Type 2 Hypervisor
Virtual NIC (vNIC)
Virtual Switch (vSwitch)

Matches

Show answer & explanation

Answer

Type 1 Hypervisor maps to installing directly on physical server hardware without requiring an underlying host operating system. Type 2 Hypervisor maps to running as a software application on top of a conventional host operating system. Virtual NIC maps to connecting a virtual machine's operating system to the virtual network and assigning a software-based MAC address. Virtual Switch maps to providing Layer 2 packet forwarding and VLAN segmentation between virtual machines on the same physical host.
Each component serves a specific layer in host virtualization: Type 1 hypervisors run directly on server hardware; Type 2 hypervisors run on top of an existing host OS; Virtual NICs provide network interface presentation and software MAC addresses to guest VMs; and Virtual Switches perform software-based Layer 2 forwarding and VLAN tagging within the host.

Step-by-Step Solution

1
Differentiate hypervisor architecture layers.
Type 1 hypervisors execute directly on hardware (bare-metal), whereas Type 2 hypervisors execute as an application dependent on an underlying host OS.
Classification depends on whether a primary operating system sits between the physical hardware and the hypervisor execution engine.
2
Identify the virtual network interface abstraction.
The Virtual NIC (vNIC) presents a virtualized Ethernet adapter to the guest OS, maintaining its own software-assigned MAC address.
The guest OS requires a network adapter object to generate layer 2 frames and request network resources.
3
Determine the role of host-internal switching infrastructure.
The Virtual Switch (vSwitch) handles local Layer 2 frame forwarding, MAC learning, and VLAN tagging across virtual ports on the hypervisor host.
Intra-host virtual machine switching is performed in software by the vSwitch without sending local inter-VM traffic out to physical switches unless required.

Key Concept

Host-level server virtualization components and hypervisor architecture models.
Question 293Question →

A network administrator is designing a Variable Length Subnet Mask (VLSM) scheme starting from the 172.16.128.0/18172.16.128.0/18 network block. Department A requires a subnet that supports at least 500500 usable host addresses. If the administrator allocates the first available subnet of the minimum required size to Department A, which TWO statements regarding this allocation are true?

Select all that apply

Show answer & explanation

Answer: The subnet mask required for Department A is 255.255.254.0255.255.254.0.; The broadcast address for Department A's allocated subnet is 172.16.129.255172.16.129.255.

Answer

The correct statements are that the subnet mask required for Department A is 255.255.254.0255.255.254.0 and the broadcast address for Department A's allocated subnet is 172.16.129.255172.16.129.255.
To support 500 usable host addresses, 9 host bits are necessary because 29−2=5102^9 - 2 = 510 usable addresses. This yields a /2323 prefix, which translates to a subnet mask of 255.255.254.0255.255.254.0. Allocating the first available /2323 block from 172.16.128.0/18172.16.128.0/18 results in the network address 172.16.128.0172.16.128.0 and broadcast address 172.16.129.255172.16.129.255.

Step-by-Step Solution

1
Determine the required host bits and prefix length for Department A.
Department A requires ≥500\ge 500 usable hosts. Using the formula 2h−2≥5002^h - 2 \ge 500, h=9h = 9 host bits (29−2=5102^9 - 2 = 510 usable host addresses). The prefix length is 32−9=/2332 - 9 = /23.
Host bits must satisfy the required host capacity while preserving network bits.
2
Convert prefix length /2323 to dotted-decimal subnet mask format.
A /2323 prefix corresponds to mask 255.255.254.0255.255.254.0.
The first 23 bits are set to 1 (11111111.11111111.11111110.0000000011111111.11111111.11111110.00000000).
3
Calculate the subnet boundaries for the first /2323 block within 172.16.128.0/18172.16.128.0/18.
Network address: 172.16.128.0/23172.16.128.0/23. Usable range: 172.16.128.1172.16.128.1 – 172.16.129.254172.16.129.254. Broadcast address: 172.16.129.255172.16.129.255.
The block size in the 3rd octet is 256−254=2256 - 254 = 2, covering 3rd octet values 128 and 129.
4
Verify RFC 1918 private address classification.
The range 172.16.128.0172.16.128.0–172.16.129.255172.16.129.255 is inside the 172.16.0.0/12172.16.0.0/12 private block.
RFC 1918 designates 172.16.0.0172.16.0.0 to 172.31.255.255172.31.255.255 as private IPv4 space.

Key Concept

VLSM Subnet Sizing and Boundary Calculation
Estimated Time:2m 0s
Question 294Question →

A network administrator is inspecting an IPv4 routing table on a Cisco router to understand how paths to remote networks are chosen. Match each routing table component on the left with its corresponding function or definition on the right.

Click a left item, then click its matching right item

Items

Administrative Distance (AD)
Routing Metric
Next-Hop IP Address
Destination Prefix and Subnet Mask

Matches

Show answer & explanation

Answer

Administrative Distance matches trustworthiness of the route source; Routing Metric matches path cost calculated by a specific protocol; Next-Hop IP Address matches the immediate adjacent router interface address; Destination Prefix and Subnet Mask matches the target network range against which traffic is matched.
Each routing table component serves a distinct role in packet forwarding decisions: Administrative Distance rates protocol trustworthiness, Routing Metric measures path cost for a single protocol, Next-Hop IP specifies the adjacent forwarding device, and Destination Prefix/Mask defines the destination network range.

Step-by-Step Solution

1
Identify the role of Administrative Distance (AD)
AD measures source trustworthiness (lower value = preferred route source)
Different route sources (OSPF, RIP, Static) are compared using AD first.
2
Identify the role of Routing Metric
Metric measures path cost within the same protocol
When multiple routes exist from the same routing protocol, the router uses the lowest metric.
3
Identify the role of Next-Hop IP Address
Refers to the IP address of the adjacent router
Specifies where Layer 2 framing should direct the packet next.
4
Identify the role of Destination Prefix and Subnet Mask
Specifies the target network range
The router compares destination IP addresses against the prefix length using longest prefix match.

Key Concept

Routing Table Components and Roles
Question 295Question →

A network technician displays the active Layer 2 discovery parameters on a switch port using the `show cdp interface gigabitethernet 0/1` command and observes the following output:

text
GigabitEthernet0/1 is up, line protocol is up
Encapsulation ARPA
Sending CDP packets every 60 seconds
Holdtime is 180 seconds

If the technician disables CDP on this interface and enables Link Layer Discovery Protocol (LLDP) globally and locally, which default update timer and holdtime timer values will LLDP use?

Show answer & explanation

Answer: Update timer of 30 seconds and holdtime timer of 120 seconds

Answer

Update timer of 30 seconds and holdtime timer of 120 seconds
The correct answer specifies an update timer of 30 seconds and a holdtime timer of 120 seconds. In Cisco IOS, LLDP transmits discovery advertisements every 30 seconds by default and retains neighbor information for 120 seconds (using a holdtime multiplier of 4).

Step-by-Step Solution

1
Identify the default timer parameters for Cisco Discovery Protocol (CDP)
CDP defaults to transmitting advertisement frames every 60 seconds with a holdtime value of 180 seconds.
This establishes the baseline shown in the CLI command output.
2
Recall default timer parameters for IEEE 802.1AB Link Layer Discovery Protocol (LLDP)
LLDP defaults to an update message interval of 30 seconds and a holdtime multiplier of 4 (30×4=12030 \times 4 = 120 seconds).
Standard IEEE LLDP specifications mandate shorter advertisement intervals and holdtimes than proprietary CDP.
3
Compare the option choices to select the accurate LLDP default timer values
The option specifying an update timer of 30 seconds and a holdtime timer of 120 seconds is correct.
It accurately reflects standard Cisco IOS default behavior when LLDP is enabled.

Key Concept

CDP vs LLDP Default Timers
Question 296Question →

A network administrator needs to manually configure a Catalyst switch as the primary Root Bridge for VLAN 30 using Rapid PVST+. The switch currently has default STP settings. Which configuration change correctly sets the bridge priority, and what is the resulting Bridge Priority value advertised in the Bridge Protocol Data Units (BPDUs) for VLAN 30?

Show answer & explanation

Answer: Execute 'spanning-tree vlan 30 priority 4096'; the advertised Bridge Priority in BPDUs will be 4126.

Answer

Execute 'spanning-tree vlan 30 priority 4096'; the advertised Bridge Priority in BPDUs will be 4126.
In Rapid PVST+, the Bridge ID priority consists of a 4-bit base priority and a 12-bit System ID Extension representing the VLAN ID. The command 'spanning-tree vlan 30 priority 4096' sets the base priority to a valid multiple of 4096. When combined with the System ID Extension of 30, the switch advertises a total Bridge Priority of 4126 (4096+304096 + 30) in its BPDUs.

Step-by-Step Solution

1
Determine valid STP bridge priority configuration increments.
STP bridge priority must be configured in increments of 4096 (e.g., 0, 4096, 8192, 12288, etc.).
The 16-bit Priority field in the Bridge ID (BID) is divided into a 4-bit Priority field (multiples of 4096) and a 12-bit System ID Extension field.
2
Calculate the total advertised Bridge Priority for VLAN 30.
Base Priority (4096) + System ID Extension (VLAN 30) = 4126.
Rapid PVST+ appends the 12-bit VLAN ID to the 4-bit base priority value to form the complete 16-bit priority component of the Bridge ID.

Key Concept

Rapid PVST+ System ID Extension and Bridge Priority Calculation
Question 297Question →

A network administrator executes the show ip route command on a Cisco router and observes the following routing entry:

O 10.150.1.0/24 [110/20] via 192.168.12.2, 00:10:42, GigabitEthernet0/0

Which component of this routing table entry is represented by the value 110?

Show answer & explanation

Answer: The administrative distance of the route

Answer

The value 110 represents the administrative distance of the route.
In Cisco IOS routing tables, routes learned via dynamic protocols display bracketed values in the format [Administrative Distance / Metric]. The first number, 110, is the administrative distance, which indicates the reliability of the routing source (110 is the default AD for OSPF).

Step-by-Step Solution

1
Examine the bracketed notation [110/20] in the routing table entry.
Identify the standard Cisco routing table notation format [Administrative Distance / Metric].
Cisco IOS displays the administrative distance first, followed by the metric, enclosed in square brackets.
2
Correlate the first value (110) with its corresponding routing component.
110 is identified as the Administrative Distance (AD) for an OSPF-learned route.
Administrative distance measures the trustworthiness of a routing information source.

Key Concept

Routing Table Component Interpretation (Administrative Distance vs. Metric)
Question 298Question →

A network administrator is configuring a Cisco IOS router to function as a DHCP server for a local subnet. To avoid IP address conflicts with statically configured devices such as default gateways and printers, the administrator must prevent the DHCP server from allocating addresses in the range 192.168.10.1 through 192.168.10.10. Which command should be executed in global configuration mode to accomplish this requirement?

Show answer & explanation

Answer: ip dhcp excluded-address 192.168.10.1 192.168.10.10

Answer

The command 'ip dhcp excluded-address 192.168.10.1 192.168.10.10' entered in global configuration mode correctly reserves the designated range of IP addresses so the local Cisco IOS DHCP server will not allocate them to dynamic clients.
Executing 'ip dhcp excluded-address 192.168.10.1 192.168.10.10' in global configuration mode instructs the Cisco IOS DHCP server to reserve all IP addresses from 192.168.10.1 through 192.168.10.10 inclusive, ensuring they are never offered to dynamic DHCP clients.

Step-by-Step Solution

1
Identify the configuration requirement for reserving static IP addresses on a Cisco IOS DHCP server.
Recognize that static IP addresses must be excluded before the DHCP server assigns them to pool clients to prevent duplicate IP address conflicts.
Cisco IOS DHCP servers issue any available address in a configured pool unless explicitly told not to assign specific addresses.
2
Select the appropriate Cisco IOS command mode and syntax.
Determine that global configuration mode command 'ip dhcp excluded-address <low-ip> <high-ip>' is used.
Exclusions are defined globally outside of the specific DHCP pool configuration mode context.

Key Concept

DHCP Address Exclusion Configuration
Question 299Question →

A network technician needs to verify the configured SSH version, authentication timeout, and maximum retry limits on a Cisco IOS switch named SW-Access-01. Which privileged EXEC mode command provides this specific operational summary?

Show answer & explanation

Answer: show ip ssh

Answer

The command 'show ip ssh' displays global SSH status, protocol version, authentication timeout, and retry settings on a Cisco IOS device.
The command 'show ip ssh' displays global SSH server operational parameters on Cisco IOS, including whether SSH is enabled, the active version (v1 or v2), authentication timeout, and maximum authentication retries.

Step-by-Step Solution

1
Identify the CLI verification requirement
The requirement asks for global SSH server settings including version, timeout, and retry limits.
Different verification commands display distinct aspects of remote access operations.
2
Distinguish between 'show ip ssh' and 'show ssh'
'show ip ssh' displays protocol configuration and status, while 'show ssh' displays active user connection sessions.
Understanding command output specificity is essential for Cisco IOS verification.

Key Concept

Cisco IOS SSH Verification Commands
Estimated Time:45s
Question 300Question →

A Cisco Catalyst switch running Rapid PVST+ receives BPDUs on multiple interfaces from different neighboring switches connected to the same Root Switch. Place the decision criteria in the exact sequence the switch uses to select its Root Port, starting from the first criterion evaluated to the final tie-breaker.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct evaluation sequence for Root Port selection in Rapid PVST+ is: 1. Lowest cumulative Root Path Cost received in the BPDUs, 2. Lowest Sender (Designated) Bridge ID, 3. Lowest Sender (Designated) Port ID, 4. Lowest local (Receiver) Port ID.
Rapid PVST+ follows a strict 4-step decision sequence to select a single Root Port on non-root switches: 1) Lowest cumulative Root Path Cost, 2) Lowest Sender Bridge ID, 3) Lowest Sender Port ID, and 4) Lowest local (Receiver) Port ID.

Step-by-Step Solution

1
Evaluate incoming Root Path Cost
The switch identifies the port(s) offering the shortest cumulative cost to the Root Bridge.
Path cost reflects link speed and topology length, making it the primary factor for optimal routing toward the root.
2
Break ties using upstream switch identity
If path costs are equal, the port receiving BPDUs from the upstream switch with the lowest Bridge ID (Priority + MAC) is preferred.
Bridge ID serves as the first tie-breaker when multiple paths offer equal cost through different neighboring switches.
3
Break ties using upstream port identity
If candidate ports connect to the exact same upstream switch, the port connected to the lower upstream Port ID (Priority + Port Number) is selected.
Designated Port ID identifies which specific interface on the neighboring switch is sending the BPDU.
4
Break ties using local port identity
If all preceding criteria are identical (e.g., connected via a hub), the switch selects its own local interface with the lowest Port ID.
Local Port ID guarantees a deterministic decision when upstream information offers no distinction.

Key Concept

Rapid PVST+ Root Port Selection Tie-Breaker Logic
Estimated Time:1m 30s
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