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Question 401Question →

A network technician is configuring an internal server to automatically synchronize the system clocks of all managed switches and routers across the enterprise network. Which port and transport protocol combination must be permitted through internal firewalls to allow Network Time Protocol (NTP) traffic?

Show answer & explanation

Answer: UDP port 123

Answer

UDP port 123 is the standard port and protocol combination used by Network Time Protocol (NTP) to synchronize clock times across network infrastructure components.
Network Time Protocol (NTP) utilizes UDP port 123 to send and receive time synchronization messages across IP networks. UDP is utilized because time distribution requires low delay and low protocol overhead.

Step-by-Step Solution

1
Identify the required network service from the scenario
The scenario describes synchronizing system clocks across network devices, which is handled by Network Time Protocol (NTP).
Recognizing service function is necessary to determine the corresponding standard port.
2
Determine the transport layer protocol and default port assigned to NTP
NTP operates over UDP on port 123.
UDP is chosen for time synchronization due to its lower delay and lack of handshake overhead, operating over registered port 123.

Key Concept

Network Time Protocol (NTP) Port and Transport Assignment
Question 402Question →

A network architect is assigned the IPv4 block 172.16.32.0/20172.16.32.0/20 to provision subnets for four new departments in a branch office. The requirements for usable host IP addresses are as follows: Department A requires 500500 hosts, Department B requires 250250 hosts, Department C requires 100100 hosts, and Department D requires 5050 hosts. Using Variable Length Subnet Masking (VLSM) and allocating subnets contiguously starting from the lowest available network address in the block in order of largest to smallest host requirement, what is the broadcast address of Department D's subnet, and how many total IP addresses remain unallocated within the original /20/20 block?

Show answer & explanation

Answer: Broadcast address: 172.16.35.191172.16.35.191; Unallocated IP addresses: 31363136

Answer

Broadcast address: 172.16.35.191172.16.35.191; Unallocated IP addresses: 31363136
Evaluating host requirements with VLSM yields block sizes of 512512 (/23/23), 256256 (/24/24), 128128 (/25/25), and 6464 (/26/26). Sequential assignment starting from 172.16.32.0172.16.32.0 places Department D at 172.16.35.128/26172.16.35.128/26, which spans 172.16.35.128172.16.35.128 through 172.16.35.191172.16.35.191, making 172.16.35.191172.16.35.191 the broadcast address. Subtracting total allocated addresses (960960) from the total block size of 40964096 addresses leaves 31363136 unallocated IP addresses.

Step-by-Step Solution

1
Determine total capacity of the parent block
A /20/20 network has 32−20=1232 - 20 = 12 host bits, yielding 212=40962^{12} = 4096 total IP addresses.
Calculating total initial pool size is required to find remaining unallocated space.
2
Calculate required prefix lengths and block sizes using VLSM in descending order
Department A (500500 hosts): Needs /23/23 (29−2=5102^9 - 2 = 510 usable, block size 512512). Department B (250250 hosts): Needs /24/24 (28−2=2542^8 - 2 = 254 usable, block size 256256). Department C (100100 hosts): Needs /25/25 (27−2=1262^7 - 2 = 126 usable, block size 128128). Department D (5050 hosts): Needs /26/26 (26−2=622^6 - 2 = 62 usable, block size 6464).
Subnet allocation requires accommodating host counts plus network and broadcast overhead (+2+2).
3
Assign contiguous subnet ranges from lowest available address
Dept A: 172.16.32.0/23172.16.32.0/23 (172.16.32.0−172.16.33.255172.16.32.0 - 172.16.33.255). Dept B: 172.16.34.0/24172.16.34.0/24 (172.16.34.0−172.16.34.255172.16.34.0 - 172.16.34.255). Dept C: 172.16.35.0/25172.16.35.0/25 (172.16.35.0−172.16.35.127172.16.35.0 - 172.16.35.127). Dept D: 172.16.35.128/26172.16.35.128/26 (172.16.35.128−172.16.35.191172.16.35.128 - 172.16.35.191).
Subnets must start on valid binary boundaries matching their block sizes.
4
Identify Department D broadcast address and calculate unallocated addresses
Broadcast address for Dept D is 172.16.35.191172.16.35.191. Total allocated addresses =512+256+128+64=960= 512 + 256 + 128 + 64 = 960. Remaining unallocated addresses =4096−960=3136= 4096 - 960 = 3136.
Broadcast is the last address of the subnet block, and unallocated capacity is total capacity minus used capacity.

Key Concept

Variable Length Subnet Masking (VLSM) and IPv4 Subnet Allocation
Question 403Question →

A network technician is configuring a host device and needs to verify its Media Access Control (MAC) address, which is used to uniquely identify the hardware interface on a local network segment. At which layer of the OSI model does MAC addressing function?

Show answer & explanation

Answer: Data Link Layer (Layer 2)

Answer

Data Link Layer (Layer 2)
The Data Link Layer (Layer 2) is responsible for physical hardware addressing (MAC addresses), framing, and node-to-node communication within a local network segment.

Step-by-Step Solution

1
Identify the type of addressing described in the scenario.
The scenario mentions Media Access Control (MAC) addresses, which provide hardware-level physical addressing on a local network segment.
MAC addresses are burned into network interface cards to identify devices locally.
2
Map MAC addressing to its corresponding OSI model layer.
Physical hardware addressing and framing take place at Layer 2 (Data Link Layer).
Layer 2 provides node-to-node transfer and error detection across a shared physical medium.

Key Concept

OSI Model Data Link Layer Functions
Estimated Time:45s
Question 404Question →

A network administrator is configuring an enterprise database application that requires continuous two-way communication between desktop clients and a centralized database engine. The underlying application protocol handles establishing, maintaining, and terminating the logical dialogue sessions between the endpoints, including inserting synchronization checkpoints into the data flow to recover from connection drops. At which layer of the OSI model does this session management and dialogue control operate?

Show answer & explanation

Answer: Session Layer (Layer 5)

Answer

Session Layer (Layer 5)
The Session Layer (Layer 5) of the Open Systems Interconnection (OSI) model is responsible for setting up, coordinating, maintaining, and terminating interactions and dialogues between application processes at each end of a communication path. It also manages checkpointing and recovery in the data stream so that sessions can resume smoothly after transient disruptions.

Step-by-Step Solution

1
Analyze the functional requirements in the scenario.
Identified key functions: establishing, managing, and terminating logical communication sessions between endpoints, and managing dialogue synchronization checkpoints.
Determining the core networking operations described in the stem helps map them to the correct layer of the OSI model.
2
Map the identified functions to the corresponding OSI layer.
Dialogue control, session state maintenance, and checkpoint synchronization are explicit responsibilities of Layer 5 (Session Layer).
The Session Layer organizes interaction between applications (duplex mode control, checkpointing, and session teardown).

Key Concept

OSI Session Layer (Layer 5) Functions
Question 405Question →

Match each Layer 2 switching and VLAN concept on the left with its corresponding primary function or operational characteristic on the right.

Click a left item, then click its matching right item

Items

Access Port
Trunk Port
Native VLAN
IEEE 802.1Q

Matches

Show answer & explanation

Answer

Access Port matches with carrying traffic for a single assigned VLAN to end-host devices; Trunk Port matches with carrying traffic for multiple VLANs across inter-switch connections; Native VLAN matches with carrying untagged traffic across a trunk link; IEEE 802.1Q matches with the standard protocol that inserts a 4-byte tag into Ethernet frames.
Each concept directly aligns with its fundamental Layer 2 definition: Access ports connect single-VLAN end devices, Trunk ports carry multi-VLAN traffic across switches, Native VLAN handles untagged trunk traffic, and IEEE 802.1Q is the standard 4-byte framing protocol.

Step-by-Step Solution

1
Identify the basic definitions of switch port modes.
Access ports serve individual host devices in one VLAN, whereas trunk ports aggregate multiple VLANs across switch links.
Differentiating access vs. trunk ports is essential for proper Layer 2 interface configuration.
2
Identify trunking mechanisms and standards.
IEEE 802.1Q provides frame tagging by adding 4 bytes to the header, while Native VLAN handling defines how untagged frames are managed over that trunk.
Understanding 802.1Q tagging behavior ensures correct frame forwarding and prevents trunking configuration errors.

Key Concept

Switch Port Modes, VLAN Trunking, and 802.1Q Tagging Mechanisms
Question 406Question →

A host receives an incoming data frame on its network interface card (NIC) and processes it up the protocol stack toward the destination application. Which of the following operations occur during this host de-encapsulation process? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: The Data Link layer verifies the frame checksum and removes the Layer 2 header and trailer to expose the IP packet.; The Transport layer reads the Layer 4 header port numbers to route the payload to the designated application.

Answer

During host de-encapsulation, the Data Link layer verifies the frame checksum and strips the Layer 2 header/trailer to expose the IP packet, and the Transport layer inspects the Layer 4 port numbers to deliver the payload to the correct application.
De-encapsulation on a receiving host proceeds from Layer 1 up to Layer 7. At Layer 2 (Data Link), the frame header and trailer are validated and stripped to reveal the IP packet. Once the Network layer processes and removes the IP header, the Transport layer (Layer 4) reads the TCP/UDP port numbers to deliver the raw payload to the target application.

Step-by-Step Solution

1
Analyze Layer 2 processing during inbound de-encapsulation.
The Data Link layer checks the frame checksum (FCS), verifies local MAC destination address, and strips the Ethernet header and trailer.
De-encapsulation unwraps protocol data units starting from the lowest layer up to higher layers.
2
Analyze Layer 3 processing during inbound de-encapsulation.
The Network layer verifies IP addresses, strips the IP header, and passes the upper protocol payload to Layer 4.
Layer 3 delivers the packet to the destination host and removes IP header overhead.
3
Analyze Layer 4 processing during inbound de-encapsulation.
The Transport layer evaluates destination port numbers in the header to direct the data payload to the receiving application process.
Port numbers identify specific software applications or services waiting for incoming traffic.

Key Concept

De-encapsulation host protocol unwrapping
Estimated Time:45s
Question 407Question →

Match each network topology architecture to its primary operational characteristic.

Click a left item, then click its matching right item

Items

Hub-and-Spoke Topology
Full Mesh Topology
Spine-Leaf Architecture
Point-to-Point Topology

Matches

Show answer & explanation

Answer

Hub-and-Spoke Topology matches with centralizing routing through a primary site to connect remote branch locations. Full Mesh Topology matches with interconnecting every node to every other node for maximum redundancy. Spine-Leaf Architecture matches with providing predictable, low-latency east-west data center traffic by connecting every leaf switch to every spine switch. Point-to-Point Topology matches with directly connecting exactly two dedicated endpoints.
Each topology matches its unique architectural characteristic: Hub-and-Spoke centralizes connections across sites; Full Mesh connects all nodes to all other nodes for max fault tolerance; Spine-Leaf guarantees predictable low latency in data centers by connecting each leaf to all spines; and Point-to-Point creates a direct single connection between two endpoints.

Step-by-Step Solution

1
Identify the topology designed for multi-site WAN efficiency with central control.
Hub-and-Spoke connects multiple remote spokes to a central hub site.
This structure minimizes required WAN connections compared to meshing.
2
Identify the topology providing complete physical redundancy between all nodes.
Full Mesh links every node to every other node.
It guarantees maximum redundancy so no single line failure disrupts communication.
3
Identify the modern data center architecture tailored for east-west traffic flow.
Spine-Leaf connects every leaf switch to every spine switch.
This keeps latency consistent and predictable across data center racks.
4
Identify the simplest direct link between two nodes.
Point-to-Point establishes a dedicated connection between two endpoints.
Bandwidth on a point-to-point link is reserved exclusively for those two nodes.

Key Concept

Network Topologies and Architectural Models
Question 408Question →

A network administrator is analyzing packet captures and system configurations during an enterprise IPv6 migration. Which of the following statements correctly describe IPv6 header structure features or address representation rules? (Select TWO).

Select all that apply

Show answer & explanation

Answer: The IPv6 base header has a fixed length of 40 bytes, and optional payload processing parameters are handled via chained Extension Headers.; The Hop Limit field in the IPv6 base header performs the same loop-prevention role as the Time to Live (TTL) field in IPv4, decrementing at each router hop.

Answer

The correct statements are that the IPv6 base header is a fixed 40 bytes using extension header chaining, and the Hop Limit field replaces the IPv4 Time to Live (TTL) field for loop suppression.
The IPv6 base header is standardized to a fixed 40-byte length to allow fast hardware processing; extra options are appended via Extension Headers referenced by the Next Header field. Additionally, the Hop Limit field serves the exact functional purpose as IPv4 Time to Live (TTL) to prevent routing loops.

Step-by-Step Solution

1
Evaluate IPv6 base header design characteristics
Confirm that the base IPv6 header is fixed at 40 bytes and utilizes the Next Header field to chain optional Extension Headers.
Fixed header size reduces processing overhead on routers compared to IPv4's variable header length.
2
Evaluate the function of the Hop Limit field
Confirm that Hop Limit functions identically to IPv4 TTL by decrementing at each Layer 3 device.
Packets are dropped when Hop Limit reaches zero to prevent endless loop propagation.
3
Analyze address compression rules and field functions in wrong options
Identify that using multiple double-colons in an IPv6 address breaks parsing rules, and Next Header indicates Layer 4 protocols or Extension Headers rather than Layer 2 frame sizes.
RFC 5952 forbids multiple '::' occurrences due to address ambiguity.

Key Concept

IPv6 Base Header Fields and Address Representation Rules
Estimated Time:1m 30s
Question 409Question →

A network technician is preparing a training matrix that maps network protocols and operations to their corresponding Open Systems Interconnection (OSI) model layers. Match each OSI layer on the left with its correct operational description and primary Protocol Data Unit (PDU) on the right.

Click a left item, then click its matching right item

Items

Physical Layer (Layer 1)
Data Link Layer (Layer 2)
Network Layer (Layer 3)
Transport Layer (Layer 4)

Matches

Show answer & explanation

Answer

Physical Layer matches with bit synchronization/Bits; Data Link Layer matches with MAC addressing/Frames; Network Layer matches with logical routing/Packets; Transport Layer matches with end-to-end segmentation/Segments or Datagrams.
Each OSI layer has distinct responsibilities and Protocol Data Units (PDUs): Physical Layer deals with raw Bits and signaling hardware; Data Link Layer packages data into Frames using MAC addresses; Network Layer routes Packets across subnets using IP addresses; and Transport Layer manages end-to-end communication via Segments/Datagrams using port numbers.

Step-by-Step Solution

1
Identify the primary PDU and core functions of Layer 1 (Physical).
Layer 1 processes electrical, optical, or radio signals representing raw bits.
Physical Layer operates at the lowest level without reading logical or hardware addresses.
2
Identify the primary PDU and core functions of Layer 2 (Data Link).
Layer 2 organizes bits into frames containing hardware (MAC) source and destination addresses.
Data Link handles local network segment delivery and framing.
3
Identify the primary PDU and core functions of Layer 3 (Network).
Layer 3 encapsulates data into packets with logical IP headers for inter-network routing.
Network Layer handles path selection and logical boundary crossing.
4
Identify the primary PDU and core functions of Layer 4 (Transport).
Layer 4 divides application data into segments (TCP) or datagrams (UDP) with source and destination port numbers.
Transport Layer facilitates host-to-host process communication and connection control.

Key Concept

OSI Layer Functions and PDU Mappings
Question 410Question →

A network administrator configures a single Layer 2 switch with VLAN 10 (Sales) and VLAN 20 (Marketing). Access ports are assigned to their respective VLANs, but no Layer 3 routing device or inter-VLAN routing interface is present. When a host on VLAN 10 transmits an Ethernet broadcast frame, how does the switch handle the frame?

Show answer & explanation

Answer: The switch forwards the frame exclusively out of all active ports assigned to VLAN 10, except the port on which it arrived.

Answer

The switch forwards the broadcast frame exclusively out of all active ports assigned to VLAN 10, except the port on which it arrived.
Virtual Local Area Networks (VLANs) segment a single physical switch into multiple distinct Layer 2 broadcast domains. When an ingress access port receives a broadcast frame, the switch floods that frame exclusively out of other active access ports assigned to the exact same VLAN (excluding the receiving port). Devices in different VLANs cannot receive Layer 2 broadcast frames from each other without a Layer 3 inter-VLAN routing mechanism.

Step-by-Step Solution

1
Identify the type of traffic received
The incoming traffic is an Ethernet broadcast frame (destination MAC address `FF:FF:FF:FF:FF:FF`).
Broadcast frames must be flooded out of all interfaces within the originating broadcast domain.
2
Evaluate the VLAN boundary configuration
The receiving port belongs to VLAN 10. A VLAN defines a logical Layer 2 broadcast domain.
Traffic inside one broadcast domain is logically isolated from other broadcast domains (VLAN 20) at Layer 2.
3
Determine switch forwarding behavior
The switch replicates the broadcast frame only to active member ports of VLAN 10, omitting the ingress port and ports assigned to VLAN 20.
Without a Layer 3 router or inter-VLAN routing configured, frames cannot traverse between separate VLANs.

Key Concept

VLAN Broadcast Isolation
Estimated Time:1m 0s
Question 411Question →

A network administrator is evaluating an internal real-time voice streaming application. During periods of high network congestion, some voice packets are dropped, resulting in minor audio artifacts. A technician proposes modifying the application transport protocol from UDP to TCP to eliminate packet loss. Which of the following best describes the operational consequence of implementing this proposal?

Show answer & explanation

Answer: The application will experience increased latency and audio jitter because retransmissions and flow control delay real-time packet delivery.

Answer

The application will experience increased latency and audio jitter because retransmissions and flow control delay real-time packet delivery.
Real-time media applications such as voice streaming prioritize low latency and consistent timing over guaranteed delivery. UDP is connectionless and lightweight (8-byte header), making it ideal for real-time traffic where dropping a frame is preferable to delaying the stream. Converting the application to TCP introduces connection establishment, acknowledgement overhead, sliding window flow control, and retransmissions of lost segments, which manifests as stream pause, audio jitter, and high latency.

Step-by-Step Solution

1
Analyze application requirements
Real-time voice traffic requires low latency and predictable timing over guaranteed packet delivery.
Late audio packets are useless in real-time conversations and cause disruptive delays.
2
Evaluate TCP characteristics
TCP uses a three-way handshake, sequence numbers, acknowledgements, sliding windows, and retransmissions.
When a packet is dropped, TCP holds subsequent data in buffer until the missing segment is retransmitted and acknowledged.
3
Determine impact of protocol change
Replacing UDP with TCP causes retransmitted voice frames to arrive out-of-sync, introducing jitter and noticeable delay.
UDP is preferred for voice/video because best-effort delivery without retransmission overhead maintains real-time continuity.

Key Concept

Connection-Oriented vs Connectionless Transport Protocols (TCP vs UDP)
Question 412Question →

A network administrator is provisioning IPv4 subnets within the 10.50.0.0/1610.50.0.0/16 network block. A new department requires allocation for up to 500500 usable host interfaces. The administrator assigns the smallest matching single CIDR block starting at 10.50.0.0/2310.50.0.0/23. Which of the following IP addresses represent valid usable host addresses within this assigned subnet? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: 10.50.0.25510.50.0.255; 10.50.1.110.50.1.1

Answer

The valid usable host IP addresses within the 10.50.0.0/2310.50.0.0/23 subnet are 10.50.0.25510.50.0.255 and 10.50.1.110.50.1.1.
The assigned prefix /23/23 creates a subnet spanning 10.50.0.010.50.0.0 to 10.50.1.25510.50.1.255. Within this 512-address block, 10.50.0.010.50.0.0 is the network ID and 10.50.1.25510.50.1.255 is the broadcast address. The usable host range is 10.50.0.110.50.0.1 through 10.50.1.25410.50.1.254. Both 10.50.0.25510.50.0.255 and 10.50.1.110.50.1.1 fall within this valid host range.

Step-by-Step Solution

1
Determine the subnet block range for 10.50.0.0/2310.50.0.0/23
A /23/23 prefix provides 232−23=5122^{32-23} = 512 total IP addresses (256×2256 \times 2). The block starts at 10.50.0.010.50.0.0 and ends at 10.50.1.25510.50.1.255.
Calculating the total size establishes the exact boundary of the assigned network segment.
2
Identify network and broadcast addresses
Network ID = 10.50.0.010.50.0.0, Broadcast Address = 10.50.1.25510.50.1.255.
The first address in the block is reserved as the network identifier and the last address is reserved for subnet-wide broadcasts.
3
Calculate the usable host IP address range
Usable host range spans from 10.50.0.110.50.0.1 to 10.50.1.25410.50.1.254.
Usable addresses exclude the reserved network ID and broadcast address.
4
Evaluate candidate IP addresses against the usable range
10.50.0.25510.50.0.255 and 10.50.1.110.50.1.1 fall inside [10.50.0.1,10.50.1.254][10.50.0.1, 10.50.1.254], while 10.50.1.25510.50.1.255 is the broadcast address and 10.50.2.110.50.2.1 is in the next subnet.
Matching each address identifies which ones are valid for host assignment.

Key Concept

IPv4 CIDR Subnetting and Usable Host Boundaries
Question 413Question →

A network administrator configures two separate VLANs, VLAN 10 and VLAN 20, on a single managed switch. Host A is connected to a port assigned to VLAN 10, and Host B is connected to a port assigned to VLAN 20. Which networking principle explains why Host A cannot send Layer 2 broadcast frames directly to Host B?

Show answer & explanation

Answer: Each VLAN defines a separate Layer 2 broadcast domain, preventing broadcast traffic from crossing VLAN boundaries without a Layer 3 routing device.

Answer

Each VLAN defines a separate Layer 2 broadcast domain, preventing broadcast traffic from crossing VLAN boundaries without a Layer 3 routing device.
Virtual Local Area Networks (VLANs) function by dividing a physical Layer 2 switch into separate logical broadcast domains. Because Layer 2 broadcast traffic is confined within its assigned VLAN, hosts on different VLANs cannot communicate directly at Layer 2 and require a router or Layer 3 switch for inter-VLAN routing.

Step-by-Step Solution

1
Identify the primary purpose of creating Virtual Local Area Networks (VLANs) on a switch.
VLANs segment a single physical Layer 2 switch into multiple logical networks.
Logically partitioning a switch groups ports into distinct broadcast domains.
2
Analyze how Layer 2 broadcast frames propagate within a VLAN.
Broadcast frames are forwarded only to ports assigned to the same VLAN as the ingress port.
Switches restrict Layer 2 broadcast traffic to its originating broadcast domain to maintain network efficiency and security.
3
Determine the requirement for inter-VLAN communication.
Inter-VLAN communication requires a Layer 3 device (such as a router or Layer 3 switch).
Since VLANs operate as separate subnets at Layer 3, packets moving between them must be routed.

Key Concept

VLAN Broadcast Domain Isolation
Estimated Time:45s
Question 414Question →

A network technician is auditing an enterprise Layer 2 campus network consisting of three interconnected switches: Switch-Alpha, Switch-Beta, and Switch-Gamma. The Spanning Tree Protocol (STP) bridge priorities are configured as follows: Switch-Alpha is set to 4096, Switch-Beta is set to 32768, and Switch-Gamma is set to 16384. Additionally, an IEEE 802.1Q trunk link between Switch-Alpha and Switch-Beta has a native VLAN mismatch (configured for VLAN 10 on Switch-Alpha and VLAN 1 on Switch-Beta). Hosts belonging to VLAN 20 are connected across Switch-Beta and Switch-Gamma, with no Layer 3 routing device present on the network. Which TWO of the following statements accurately describe the behavior and operational state of this switching environment?

Select all that apply

Show answer & explanation

Answer: Switch-Alpha will be elected as the Spanning Tree Protocol (STP) root bridge because it possesses the lowest numerical priority value.; Untagged frames sent from VLAN 10 on Switch-Alpha will be processed into VLAN 1 upon arrival at Switch-Beta, leading to cross-VLAN traffic leakage.

Answer

The two correct statements are: Switch-Alpha will be elected as the Spanning Tree Protocol (STP) root bridge because it possesses the lowest numerical priority value, and untagged frames sent from VLAN 10 on Switch-Alpha will be processed into VLAN 1 upon arrival at Switch-Beta, leading to cross-VLAN traffic leakage.
Switch-Alpha is elected as the STP root bridge because STP root election prioritizes the lowest numerical bridge priority value (4096 < 16384 < 32768). Furthermore, 802.1Q trunking sends frames belonging to the native VLAN untagged across the link; when Switch-Alpha sends untagged VLAN 10 traffic, Switch-Beta receives it untagged and implicitly assigns it to its own native VLAN (VLAN 1), causing cross-VLAN traffic leakage.

Step-by-Step Solution

1
Analyze Spanning Tree Protocol (STP) Root Bridge Election
Switch-Alpha (priority 4096) has a lower priority than Switch-Gamma (16384) and Switch-Beta (32768).
STP elects the switch with the lowest numerical Bridge ID (Priority + MAC address) as the root bridge.
2
Analyze 802.1Q Native VLAN Tagging and Mismatch Symptoms
Frames from VLAN 10 sent untagged by Switch-Alpha are received by Switch-Beta and assigned to its native VLAN (VLAN 1).
802.1Q trunks do not tag frames originating from the configured native VLAN; mismatched native VLAN settings cause untagged frames to leak into the recipient switch's native VLAN.
3
Evaluate Inter-VLAN Communication and Layer 2 Forwarding Boundaries
Inter-VLAN communication between VLAN 10 and VLAN 20 requires Layer 3 routing, and 802.1Q tags remain at Layer 2.
VLANs divide physical switches into logical Layer 2 broadcast domains that strictly require Layer 3 routing for inter-VLAN forwarding.

Key Concept

Spanning Tree Root Election and 802.1Q Native VLAN Trunk Operations
Estimated Time:2m 0s
Question 415Question →

A network administrator needs to ensure that a newly deployed network printer always receives the exact same IP address from the network server whenever it boots up. To save administrative overhead, the administrator does not want to manually configure static IP parameters on the printer's console. Which of the following DHCP features should be configured to meet this requirement?

Show answer & explanation

Answer: DHCP reservation

Answer

DHCP reservation
A DHCP reservation allows network administrators to allocate a fixed IP address to a specific client based on its unique MAC address while maintaining centralized automated IP address management via DHCP.

Step-by-Step Solution

1
Identify the goal
The requirement is to automatically assign a fixed, unchanging IP address to a client without manually setting a static IP on the client device.
Centralizing static IP management reduces manual configuration errors and administrative overhead.
2
Evaluate available DHCP service features
A DHCP reservation links the client device's unique Layer 2 MAC address to a fixed Layer 3 IP address within the DHCP lease database.
When the printer broadcasts a DHCP Request, the server recognizes its MAC address and issues the pre-assigned reserved IP address.

Key Concept

DHCP Reservation vs. Static IP Assignment
Question 416Question →

A user pairs a wireless Bluetooth headset directly with a smartphone to listen to audio while working at a desk. Which of the following network types best categorizes this personal, short-range wireless connection?

Show answer & explanation

Answer: Personal Area Network (PAN)

Answer

Personal Area Network (PAN)
A Personal Area Network (PAN) is specifically designed for short-range communication among personal devices centered around an individual's immediate workspace.

Step-by-Step Solution

1
Identify the physical scale and communication technology described in the scenario.
The connection uses Bluetooth between personal devices within immediate physical proximity (under 10 meters).
Network types are categorized based on their geographic scope and intended operational coverage.
2
Select the network classification matching personal short-range connections.
Connections focused around a single individual's personal space constitute a Personal Area Network (PAN).
PAN technology (such as Bluetooth or Zigbee) is designed specifically for interconnecting personal devices in close range.

Key Concept

Personal Area Network (PAN) scope and characteristics
Question 417Question →

Match each Layer 2 switching concept on the left with its primary operational characteristic or function on the right.

Click a left item, then click its matching right item

Items

IEEE 802.1Q
Native VLAN
Port Security
Spanning Tree Protocol (STP)

Matches

Show answer & explanation

Answer

IEEE 802.1Q matches with inserting a 4-byte frame header tag for VLAN identification; Native VLAN matches with carrying untagged frame traffic across trunk links; Port Security matches with restricting switchport traffic based on authorized MAC addresses; Spanning Tree Protocol matches with preventing Layer 2 loops by blocking redundant paths.
IEEE 802.1Q defines the 4-byte VLAN tagging standard for trunking; Native VLAN processes untagged frames across trunk links; Port Security restricts switchport access using MAC addresses; and Spanning Tree Protocol blocks redundant ports to avoid network loops.

Step-by-Step Solution

1
Identify the standard protocol used for VLAN trunk tagging.
IEEE 802.1Q defines inserting a 4-byte header tag into Ethernet frames for VLAN identification across switches.
Trunk links must distinguish frame traffic belonging to different VLAN IDs.
2
Determine how untagged frames are handled across trunk links.
The Native VLAN is designated on 802.1Q trunks to pass untagged frames.
This allows untagged control traffic and legacy frame types to transit trunk connections.
3
Identify the feature that controls port access via physical MAC addresses.
Port Security inspects incoming frames and permits only allowed MAC addresses based on defined limits.
Prevents unauthorized host connection and rogue switch addition at the access layer.
4
Determine the Layer 2 protocol responsible for loop prevention.
Spanning Tree Protocol (STP) monitors switch links and dynamically blocks redundant paths.
Prevents Layer 2 broadcast storms and MAC table instability.

Key Concept

Switching Concepts and VLAN Trunking Mechanisms
Question 418Question →

An industrial facility operates a legacy control network where every client automation terminal connects via an individual twisted-pair cable directly to a central multiport Ethernet switch. However, the media access control protocol running across the network relies on a token-passing scheme, where frame transmission rights are passed sequentially from workstation to workstation in a continuous deterministic cycle. Which option correctly identifies both the physical and logical topologies of this network architecture?

Show answer & explanation

Answer: Physical Star topology and Logical Ring topology

Answer

Physical Star topology and Logical Ring topology
Physical topology refers to the actual arrangement of physical cabling and central connecting devices (in this case, individual cables running from endpoints to a central switch forming a star). Logical topology refers to the path data takes across the network (in this case, token passing sequentially from node to node in a logical circle). Therefore, the deployment is a Physical Star, Logical Ring topology.

Step-by-Step Solution

1
Analyze the physical layout description.
Individual point-to-point cables running from each endpoint directly into a central multiport switch define a Physical Star layout.
Physical topology represents the actual physical cabling and hardware layout of the network infrastructure.
2
Analyze the data transmission mechanism.
Sequentially passing token rights from node to node in a deterministic cycle defines a Logical Ring topology.
Logical topology describes how data actually travels across the network protocol stack regardless of the underlying physical cable routes.
3
Synthesize the physical and logical characteristics.
The infrastructure combines a Physical Star with a Logical Ring.
Matching both physical cabling patterns and logical frame flow paths confirms the hybrid topology classification.

Key Concept

Distinction between Physical Topology (hardware deployment) and Logical Topology (data path/access control)
Estimated Time:2m 0s
Question 419Question →

A network router receives route advertisements for the same destination network from two different dynamic routing protocols: OSPF and RIP. The router automatically prefers the route learned through OSPF over the route learned through RIP. Which criterion does the router use first to compare the trustworthiness of these different routing protocols?

Show answer & explanation

Answer: Administrative distance

Answer

Administrative distance is used by routers to evaluate the trustworthiness of routes learned via different routing sources.
Administrative distance (AD) is a rating of trustworthiness of a routing information source on a scale from 0 to 255. A lower value signifies a more reliable protocol. Since OSPF has a lower administrative distance (110) than RIP (120), the router prefers the OSPF path.

Step-by-Step Solution

1
Identify the scenario requirement
The router has two routes to the same network from different sources (OSPF and RIP).
When multiple routes to a single network are learned via different protocols, the router must decide which source is more trustworthy.
2
Evaluate protocol preference metric
Administrative distance (AD) is evaluated first.
AD defines trustworthiness; lower AD values mean higher believability (OSPF AD = 110, RIP AD = 120).

Key Concept

Administrative Distance in Dynamic Routing
Question 420Question →

A network administrator is assigned the IPv4 address block 172.28.64.0/21172.28.64.0/21 to provision subnets for a multi-department enterprise deployment. Using Variable Length Subnet Masking (VLSM), subnets must be allocated sequentially starting from the lowest available IP address in order of capacity requirements (from largest to smallest):

- Subnet A: Requires capacity for up to 400400 usable host IP addresses.
- Subnet B: Requires capacity for up to 180180 usable host IP addresses.
- Subnet C: Requires capacity for up to 6060 usable host IP addresses.
- Subnet D: Requires capacity for exactly 22 usable host IP addresses.

Each subnet must be allocated using the smallest valid CIDR prefix length that accommodates its requirement. What is the total number of unallocated IP addresses remaining in the 172.28.64.0/21172.28.64.0/21 block after all four subnets are allocated?

Show answer & explanation

Answer: 1212

Answer

The total number of unallocated IP addresses remaining in the 172.28.64.0/21172.28.64.0/21 parent block after provisioning all four subnets is 1212.
The parent block 172.28.64.0/21172.28.64.0/21 contains 20482048 total IP addresses. Allocating the subnets sequentially requires: Subnet A (400400 hosts) →/23\rightarrow /23 (512512 IPs), Subnet B (180180 hosts) →/24\rightarrow /24 (256256 IPs), Subnet C (6060 hosts) →/26\rightarrow /26 (6464 IPs), and Subnet D (22 hosts) →/30\rightarrow /30 (44 IPs). Total allocated IP space is 512+256+64+4=836512 + 256 + 64 + 4 = 836 IP addresses. Subtracting 836836 from 20482048 leaves 12121212 total unallocated IP addresses.

Step-by-Step Solution

1
Calculate total IP address capacity of the parent block
172.28.64.0/21172.28.64.0/21 contains 232−21=211=20482^{32-21} = 2^{11} = 2048 total IP addresses
A /21/21 prefix uses 21 network bits, leaving 11 host bits for addressing.
2
Determine prefix length and block size for Subnet A
Prefix: /23/23, Block size: 512512 total IPs (172.28.64.0/23172.28.64.0/23)
To support 400400 hosts, 9 host bits are needed (29−2=510≥4002^9 - 2 = 510 \ge 400), giving a /23/23 prefix (512512 IPs).
3
Determine prefix length and block size for Subnet B
Prefix: /24/24, Block size: 256256 total IPs (172.28.66.0/24172.28.66.0/24)
To support 180180 hosts, 8 host bits are needed (28−2=254≥1802^8 - 2 = 254 \ge 180), giving a /24/24 prefix (256256 IPs).
4
Determine prefix length and block size for Subnet C
Prefix: /26/26, Block size: 6464 total IPs (172.28.67.0/26172.28.67.0/26)
To support 6060 hosts, 6 host bits are needed (26−2=62≥602^6 - 2 = 62 \ge 60), giving a /26/26 prefix (6464 IPs).
5
Determine prefix length and block size for Subnet D
Prefix: /30/30, Block size: 44 total IPs (172.28.67.64/30172.28.67.64/30)
To support 22 hosts (point-to-point link), 2 host bits are needed (22−2=2≥22^2 - 2 = 2 \ge 2), giving a /30/30 prefix (44 IPs).
6
Calculate remaining unallocated IP space
2048−(512+256+64+4)=2048−836=12122048 - (512 + 256 + 64 + 4) = 2048 - 836 = 1212 IP addresses
Subtracting the sum of all allocated CIDR block sizes from the total parent address space yields the remaining unallocated IP addresses.

Key Concept

Variable Length Subnet Masking (VLSM) allocation and host requirement calculations
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