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Networking Concepts

538 questions

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Question 241Question →

A network administrator is reviewing default firewall port permissions for common infrastructure protocols. Match each network service description on the left with its corresponding default port assignment and transport layer protocol on the right.

Click a left item, then click its matching right item

Items

Host name to IP address resolution service
Secure encrypted command-line remote administration
Outbound server-to-server email transmission
Encrypted web session traffic over TLS

Matches

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Answer

The correct pairings are: Host name to IP address resolution service matches UDP/TCP Port 53; Secure encrypted command-line remote administration matches TCP Port 22; Outbound server-to-server email transmission matches TCP Port 25; Encrypted web session traffic over TLS matches TCP Port 443.
Each service uses standard well-known port definitions: DNS name resolution uses UDP/TCP port 53; SSH encrypted remote administration uses TCP port 22; SMTP email relaying uses TCP port 25; HTTPS encrypted web communication uses TCP port 443.

Step-by-Step Solution

1
Identify the standard protocol associated with each service description.
Host name resolution corresponds to DNS; secure CLI administration corresponds to SSH; outbound mail relay corresponds to SMTP; secure web traffic corresponds to HTTPS.
Service descriptions directly reflect the fundamental functions of standard Network+ core protocols.
2
Map each protocol to its standard IANA default port assignment and transport protocol.
DNS maps to UDP/TCP port 53, SSH maps to TCP port 22, SMTP maps to TCP port 25, and HTTPS maps to TCP port 443.
These default well-known port allocations are standardized for enterprise firewall and ACL rules.

Key Concept

Standard IANA well-known port assignments and transport layer mechanisms for core network services.
Question 242Question →

An enterprise network deployment uses Per-VLAN Spanning Tree Plus (PVST+) across three interconnected Layer 2 switches (SW1, SW2, and SW3). The switches are configured with the following parameters:

- SW1: Base bridge priority set to 3276832768, MAC address `00:11:22:33:44:55`
- SW2: Base bridge priority set to 2867228672, MAC address `00:AA:BB:CC:DD:EE`
- SW3: Base bridge priority set to 2867228672, MAC address `00:AA:BB:CC:DD:11`

No explicit per-VLAN priority override commands have been configured for VLAN 100. Which switch is elected as the Spanning Tree Protocol (STP) Root Bridge for VLAN 100, and what is its calculated priority field within its Bridge Identifier (BID)?

Show answer & explanation

Answer: SW3 is elected as the Root Bridge with a total BID priority of 28772.

Answer

SW3 is elected as the Root Bridge for VLAN 100 with a total BID priority of 28772.
The Spanning Tree Protocol (STP) Root Bridge election evaluates the 8-byte Bridge Identifier (BID), which consists of a 2-byte Priority field and a 6-byte MAC address. In PVST+, the 2-byte Priority field is divided into a 4-bit Priority multiplier (in increments of 4096) and a 12-bit System ID Extension representing the VLAN ID. For VLAN 100, the effective priorities are: SW1 = 32768+100=3286832768 + 100 = 32868, SW2 = 28672+100=2877228672 + 100 = 28772, and SW3 = 28672+100=2877228672 + 100 = 28772. SW2 and SW3 tie with the lowest priority value of 28772. To break the tie, the MAC addresses are compared. Comparing `00:AA:BB:CC:DD:EE` (SW2) and `00:AA:BB:CC:DD:11` (SW3), SW3 has the lower MAC address because hexadecimal `11` is smaller than `EE`. Thus, SW3 becomes the Root Bridge for VLAN 100 with a total BID priority of 28772.

Step-by-Step Solution

1
Calculate the effective Bridge Priority for each switch for VLAN 100.
SW1 priority = 32768+100=3286832768 + 100 = 32868; SW2 priority = 28672+100=2877228672 + 100 = 28772; SW3 priority = 28672+100=2877228672 + 100 = 28772.
PVST+ incorporates the 12-bit System ID Extension (VLAN ID) into the 16-bit Priority field of the Bridge ID (BID).
2
Compare the calculated Bridge Priorities to find the lowest numerical value.
SW2 and SW3 tie with the lowest priority of 28772 (SW1 is eliminated with 32868).
STP selects the switch with the lowest Bridge Priority as the Root Bridge.
3
Break the tie between SW2 and SW3 by comparing their 48-bit MAC addresses.
SW3 (`00:AA:BB:CC:DD:11`) is lower than SW2 (`00:AA:BB:CC:DD:EE`).
When priority values are identical, STP uses the numerically lowest MAC address as the tiebreaker. Comparing byte-by-byte in hexadecimal: `11` (171017_{10}) is less than `EE` (23810238_{10}).

Key Concept

PVST+ System ID Extension and Root Bridge Election Priority/MAC Tiebreaker
Question 243Question →

Match each network core IP addressing service on the left with its primary function on the right.

Click a left item, then click its matching right item

Items

DHCP
DNS
NTP
IPAM

Matches

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Answer

DHCP pairs with dynamically assigning IP settings to host endpoints; DNS pairs with translating fully qualified domain names to IP addresses; NTP pairs with synchronizing clock times across infrastructure hardware; IPAM pairs with providing centralized tracking and inventory of IP address space.
DHCP provisions dynamic IP addresses to endpoints, DNS translates domain names into network IP addresses, NTP keeps network infrastructure device clocks synchronized, and IPAM tracks and manages enterprise IP inventory.

Step-by-Step Solution

1
Analyze each service listed on the left and identify its administrative role in IP network operations.
DHCP manages dynamic host configuration, DNS performs name resolution, NTP handles clock synchronization, and IPAM manages address space inventory.
Each core IP addressing service addresses a specific operational requirement in network management.

Key Concept

Core IP Addressing Services and Functions
Question 244Question →

A network technician is configuring a new subnet to accommodate a cluster of 4040 usable host interfaces. What is the minimum IPv4 prefix length (in CIDR notation, e.g., 2424 for /24/24) required to support this number of hosts?

Show answer & explanation

Answer: 26

Answer

The minimum IPv4 prefix length required is 26.
To support 4040 usable hosts, a subnet requires 66 host bits because 26−2=622^6 - 2 = 62 usable IP addresses (25−2=302^5 - 2 = 30 is too small). Subtracting 66 host bits from 3232 total IPv4 bits results in a prefix length of 2626 (or /26/26).

Step-by-Step Solution

1
Determine host capacity requirement
Need a subnet that provides at least 4040 usable host IP addresses.
Each subnet reserves two IP addresses: one for the network ID and one for the broadcast address.
2
Find host bits needed
66 host bits are required (26−2=622^6 - 2 = 62 usable hosts).
55 host bits only yield 3030 usable addresses, which fails the requirement of 4040 hosts.
3
Compute CIDR prefix length
32−6=2632 - 6 = 26.
An IPv4 address consists of 3232 total bits; subtracting the host bits gives the network prefix bits.

Key Concept

Host capacity formula and CIDR prefix length calculation
Question 245Question →

A network technician is configuring dynamic routing on an enterprise network to ensure efficient traffic forwarding across multiple subnets. The technician needs a interior gateway protocol that maintains a complete topological database of the network using link-state information. Which of the following dynamic routing protocols meets this requirement?

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Answer: Open Shortest Path First (OSPF)

Answer

Open Shortest Path First (OSPF) is an Interior Gateway Protocol that uses link-state advertisements to build a full map of the network topology and dynamically route Layer 3 traffic.
Open Shortest Path First (OSPF) is a link-state Interior Gateway Protocol (IGP) designed to exchange routing information within a single autonomous system. Each OSPF router collects link-state advertisements (LSAs) from neighbor routers to build a complete link-state database (LSDB), representing a map of the network topology.

Step-by-Step Solution

1
Identify the routing requirement stated in the scenario
The requirement specifies an Interior Gateway Protocol (IGP) that uses link-state information to build a complete topological map of the network.
Different routing protocols use distinct operational metrics and algorithms (e.g., link-state vs. distance-vector).
2
Evaluate the functional role of Open Shortest Path First (OSPF)
OSPF is an open standard Layer 3 link-state protocol that uses Link-State Advertisements (LSAs) to form neighbor relationships and calculate optimal paths.
OSPF satisfies both the link-state criteria and the interior network routing requirement.
3
Differentiate OSPF from non-routing protocols presented in the choices
Protocols such as STP, DHCP, and ARP perform switching, address assignment, and Layer 2 resolution functions rather than dynamic inter-network routing.
Only Layer 3 dynamic routing protocols exchange network reachability metrics to populate IP routing tables.

Key Concept

Link-state dynamic routing protocols (such as OSPF) flood network topology information to build complete network maps and calculate shortest paths across Layer 3 subnets.
Estimated Time:45s
Question 246Question →

A network administrator is auditing network traffic generated by an automated asset tracking system. The system continuously sends brief location update bursts from handheld scanners to a central server using User Datagram Protocol (UDP). During peak operational hours, some location update packets are dropped due to network congestion, but the server does not request retransmissions. Which characteristic of UDP explains why packet retransmission is not automatically performed at the Transport layer?

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Answer: UDP is a connectionless protocol that does not establish session state or track packet delivery using sequence numbers and acknowledgments.

Answer

UDP is connectionless and does not maintain session state, sequence numbers, or transmission acknowledgments at the Transport layer.
The correct choice highlights that UDP is a connectionless protocol. It does not perform a connection handshake, track payload delivery, or assign sequence numbers to datagrams. Consequently, it operates with minimal header overhead and lower latency, leaving any error recovery or retransmission functions to the application layer if required.

Step-by-Step Solution

1
Analyze the operational requirements described in the scenario.
The application transmits continuous data bursts via UDP, prioritizing minimal overhead and latency, while accepting packet loss during congestion.
Understanding why packet loss occurs without automatic recovery requires analyzing UDP protocol header structures and behaviors.
2
Evaluate protocol features of UDP at Layer 4.
UDP features a minimal 8-byte header lacking sequence numbers, acknowledgment numbers, and window size fields.
Without these fields, UDP cannot track missing datagrams or negotiate retransmissions natively.
3
Differentiate connection-oriented and connectionless Transport layer behaviors.
TCP handles packet sequencing and retransmissions at Layer 4, whereas UDP delegates reliability concerns (if needed) entirely to upper-layer applications.
Connectionless protocols provide best-effort delivery without connection handshakes or state tracking.

Key Concept

Connectionless transport protocols prioritize low overhead by omitting connection setup, sequencing, and flow control mechanisms.
Question 247Question →

A network technician is troubleshooting a custom client-server application. Packet analysis confirms that IP routing and TCP port assignments are functioning properly. However, the receiving system fails to display text correctly because it cannot translate the incoming ASCII-encoded payload into the required UTF-8 format. At which layer of the OSI model does this data syntax translation occur?

Show answer & explanation

Answer: Presentation Layer

Answer

Presentation Layer
The correct answer identifies the Presentation Layer (Layer 6). This layer serves as a translator for the network, handling tasks such as character encoding conversion (e.g., ASCII to Unicode/UTF-8), data compression, and data encryption/decryption so that applications can correctly interpret the payload.

Step-by-Step Solution

1
Analyze the problem scenario to identify the specific failure.
The issue involves translating character encoding formats from ASCII to UTF-8.
Identifying the functional requirement (character translation/formatting) isolates the responsible network model layer.
2
Map data syntax translation to its corresponding OSI model layer.
Layer 6 (Presentation Layer) is responsible for syntax conversion, data formatting, encryption, and compression.
The Presentation Layer ensures that data sent from an application on one host can be parsed and understood by the application on another host.

Key Concept

OSI Model Presentation Layer Functions
Question 248Question →

A network workstation initiates an NTP time synchronization request to an external time server. Arrange the outbound protocol data unit (PDU) encapsulation steps in the correct order, starting from the application data generation down to physical transmission.

Drag items to arrange them in the correct order

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Answer

The correct sequence starts with creating a Layer 4 UDP segment, encapsulating it into a Layer 3 IP packet, wrapping it in a Layer 2 Ethernet frame with an FCS trailer, and finally converting the frame into Layer 1 physical signals.
During outbound network transmission, data moves down the protocol stack. The Application data is first encapsulated with a Transport layer (Layer 4) UDP header to form a Segment. Next, the Network layer (Layer 3) adds an IP header to create a Packet. Then, the Data Link layer (Layer 2) appends an Ethernet header and FCS trailer to create a Frame. Finally, the Physical layer (Layer 1) serializes the Frame into signals across the physical medium.

Step-by-Step Solution

1
Identify Layer 4 encapsulation
NTP data gets a UDP header, forming a Segment.
Outbound data travels down the OSI stack; the transport layer adds port information first.
2
Identify Layer 3 encapsulation
The UDP Segment receives an IP header, forming a Packet.
The network layer adds IP addressing for end-to-end routing.
3
Identify Layer 2 encapsulation
The IP Packet receives an Ethernet header and FCS trailer, forming a Frame.
The data link layer adds local MAC addresses and frame check sequence for hop-to-hop delivery.
4
Identify Layer 1 serialization
The Frame is converted into physical bits/signals.
The physical layer converts structured frames into signals for transmission on physical media.

Key Concept

Outbound OSI Encapsulation Sequence (Data -> Segment -> Packet -> Frame -> Bits)
Estimated Time:2m 0s
Question 249Question →

A network administrator is reviewing system protocol definitions to clarify layer responsibilities within the OSI model. Which of the following functions specifically operate at Layer 5 (the Session Layer)? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: Establishing, maintaining, and gracefully terminating communication sessions between applications on distinct hosts; Controlling session dialogue modes (such as simplex, half-duplex, or full-duplex) and inserting synchronization checkpoints

Answer

The Session Layer (Layer 5) is responsible for establishing, managing, and terminating communication sessions between application entities, as well as managing dialogue control and inserting checkpoints into data streams.
Layer 5 (Session Layer) is dedicated to managing communication sessions between applications. Its primary duties include session setup, maintenance, and teardown, as well as dialogue control (duplex management) and checkpointing for session recovery.

Step-by-Step Solution

1
Analyze the core responsibilities of OSI Layer 5 (Session Layer).
Layer 5 coordinates the exchange of information between application sessions, managing session setup, teardown, dialogue mode, and synchronization checkpoints.
Differentiating Layer 5 from adjacent layers requires matching session control functions specifically to Layer 5.
2
Evaluate each provided option against OSI layer definitions.
Session establishment/termination and dialogue control belong to Layer 5. Encryption and data formatting belong to Layer 6. Segmentation and port-based flow control belong to Layer 4.
This confirms which options describe Layer 5 functions while identifying the mismapped adjacent layer functions.

Key Concept

OSI Layer 5 (Session Layer) Functions
Question 250Question →

A host computer on Subnet A transmits data to a server located on Subnet B through a Layer 3 router. As the data traverses the router to the destination network, which part of the Protocol Data Unit (PDU) is stripped and replaced with new addressing information at each hop?

Show answer & explanation

Answer: The Layer 2 frame header and trailer containing source and destination MAC addresses

Answer

The Layer 2 frame header and trailer containing source and destination MAC addresses
When a router forwards traffic between subnets, it de-encapsulates the incoming Layer 2 frame to extract the Layer 3 IP packet. After consulting its routing table for the next hop, it re-encapsulates the intact Layer 3 packet inside a brand-new Layer 2 frame with new source and destination MAC addresses appropriate for the next physical link.

Step-by-Step Solution

1
Identify the device performing the operation and its operating layer.
The router operates primarily at Layer 3 (Network Layer).
Routers make forwarding decisions based on destination IP addresses.
2
Analyze how encapsulation and de-encapsulation function during hop-by-hop routing.
When a router receives a frame, it de-encapsulates the Layer 2 frame, examines the Layer 3 packet IP header to find the next-hop interface, and re-encapsulates the IP packet into a new Layer 2 frame.
Layer 2 MAC addresses are local to individual broadcast domains and must be recreated for each physical subnet link.

Key Concept

Hop-by-hop Layer 2 re-framing during IP routing
Estimated Time:45s
Question 251Question →

Match each IP addressing and management service mechanism on the left with its primary functional purpose in an enterprise network on the right.

Click a left item, then click its matching right item

Items

DHCP Option 82
DNS SRV Record
IPAM Suite
NTP Stratum 1 Server

Matches

Show answer & explanation

Answer

DHCP Option 82 pairs with appending switch port and circuit metadata; DNS SRV Record pairs with publishing symbolic service names and port numbers; IPAM Suite pairs with centralizing tracking and allocation of address space; NTP Stratum 1 Server pairs with synchronizing time directly from hardware reference clocks.
Each IP addressing service component is accurately matched to its core technical definition: DHCP Option 82 appends circuit and switch port metadata at the relay agent; DNS SRV records resolve service instances and ports to hostnames; IPAM consolidates address inventory with DNS/DHCP administration; Stratum 1 NTP servers sync directly to physical reference timing devices.

Step-by-Step Solution

1
Analyze the functional role of DHCP Option 82
Identify that Relay Agent Information Option attaches layer 2 circuit and remote identifiers to relayed requests.
Relay agents add Option 82 headers so DHCP servers can make location-aware IP address assignments.
2
Analyze the functional role of DNS SRV records
Identify that SRV records contain protocol, service name, target hostname, priority, weight, and port assignments.
Services like Active Directory or VoIP rely on SRV records to discover server endpoints and service ports dynamically.
3
Analyze the functional role of IPAM software
Identify that IPAM orchestrates IP address inventory management with integrated DNS and DHCP services.
IPAM prevents IP address space collisions and provides auditing across multi-subnet environments.
4
Analyze the functional role of NTP Stratum levels
Identify that Stratum 1 devices attach directly to Stratum 0 physical timing standards.
Stratum 1 servers receive accurate time signals directly via dedicated hardware connections to reference timing sources.

Key Concept

Core operational roles and protocol characteristics of IP addressing, name resolution, network management, and clock synchronization services.
Question 252Question →

An automated logistics facility operates five separate warehouse buildings located within a single contiguous 200-acre industrial park owned by the company. The IT team must interconnect the core switches of all five buildings to support real-time inventory tracking with minimal latency. The design requires redundant paths so that if any single inter-building optical cable segment is severed, traffic automatically reroutes without disconnecting any building. However, budget constraints prevent installing dedicated fiber links between every possible building pair. Which network type classification and topology combination best satisfies these requirements?

Show answer & explanation

Answer: Campus Area Network (CAN) using a partial mesh topology

Answer

Campus Area Network (CAN) using a partial mesh topology
The deployment spans multiple buildings on a single corporate campus, placing it under the Campus Area Network (CAN) classification. To achieve fault tolerance against single fiber cuts while keeping costs within budget, a partial mesh topology is optimal because it provides redundant inter-building paths without requiring the expensive n(n−1)/2n(n-1)/2 links of a full mesh.

Step-by-Step Solution

1
Analyze the geographic scope of the deployment.
Connecting five buildings located within a single 200-acre corporate industrial park constitutes a Campus Area Network (CAN), as it spans multiple buildings across a contiguous, privately owned property.
Geographic network classifications are defined by physical extent: LAN is typically a single building, CAN covers multiple adjacent buildings on shared property, MAN covers a municipality, and WAN covers broad geographic regions.
2
Evaluate the redundancy and financial constraints for the physical topology.
A partial mesh topology provides multiple redundant interconnect pathways between critical nodes without requiring links between every possible node pair.
A full mesh topology for 5 nodes would require 10 dedicated links (n(n−1)/2n(n-1)/2), which violates the budget constraint. A star or bus topology lacks redundant links and creates single points of failure. A partial mesh balances cost with fault tolerance.
3
Synthesize the optimal network classification and topology choice.
Combining a CAN classification with a partial mesh topology fulfills both the geographic scope and high-availability criteria within budget limits.
This combination ensures continuous multi-building connectivity during single cable cuts while keeping installation costs reasonable.

Key Concept

Classification of network geographic scopes (CAN vs LAN/MAN/WAN) and selection of resilient physical topologies (Partial Mesh vs Full Mesh/Star/Bus).
Estimated Time:2m 0s
Question 253Question →

A network administrator needs to ensure that core infrastructure switches can automatically synchronize their system clocks with a centralized time server on the network. Which default port and transport protocol combination must be permitted through internal firewalls to allow Network Time Protocol (NTP) synchronization?

Show answer & explanation

Answer: UDP port 123

Answer

UDP port 123 is the standard port and transport protocol utilized by Network Time Protocol (NTP).
Network Time Protocol (NTP) uses UDP port 123 to provide clock synchronization across network-connected hardware. UDP is utilized because connection establishment overhead and TCP retransmissions would introduce latency variations that degrade timekeeping accuracy.

Step-by-Step Solution

1
Identify the required network service
The network service requested is Network Time Protocol (NTP), which synchronizes device clocks across a IP network.
Accurate timestamps across devices are essential for logging and event correlation.
2
Determine the transport protocol and port number for NTP
NTP standardized operations run on User Datagram Protocol (UDP) on well-known port 123.
UDP is chosen for NTP to avoid connection-establishment delays and retransmission delays that could distort time synchronization math.

Key Concept

Standard Network Port Assignments (NTP)
Question 254Question →

A network administrator is configuring inbound ACL rules on an edge firewall. The security baseline requires enabling secure, encrypted command-line management for distant switches while allowing internal infrastructure devices to synchronize their clocks against an external stratum-1 time server. Which TWO of the following port and transport protocol combinations must be permitted in the firewall policy? (Select TWO)

Select all that apply

Show answer & explanation

Answer: TCP port 22; UDP port 123

Answer

The required services are Secure Shell (SSH), which operates over TCP port 22, and Network Time Protocol (NTP), which operates over UDP port 123.
Secure Shell (SSH) provides secure, encrypted command-line sessions using TCP port 22. Network Time Protocol (NTP) synchronizes system clocks across network devices using connectionless UDP port 123.

Step-by-Step Solution

1
Determine the protocol and port required for secure remote command-line access.
SSH (Secure Shell) meets the requirement for encrypted terminal access and runs on TCP port 22.
Telnet operates on TCP port 23 but sends credentials and payload in unencrypted cleartext.
2
Determine the protocol, port, and transport layer for system clock synchronization.
NTP (Network Time Protocol) synchronizes device clocks using UDP port 123.
NTP requires low-overhead, latency-sensitive transport and is defined to run over connectionless UDP rather than TCP.

Key Concept

Port number and transport protocol mapping for SSH (TCP 22) and NTP (UDP 123).
Question 255Question →

Match each core IP addressing service or protocol feature on the left with its primary function or use case on the right.

Click a left item, then click its matching right item

Items

DHCP Relay Agent
DNS PTR Record
IP Address Management (IPAM)
NTP Stratum 1 Server

Matches

Show answer & explanation

Answer

DHCP Relay Agent matches forwarding broadcast messages across routers; DNS PTR Record matches mapping IP addresses to domain names for reverse resolution; IP Address Management (IPAM) matches centralized tracking of subnet utilization and allocations; NTP Stratum 1 Server matches synchronizing directly from a primary reference clock like GPS.
Each IP service maps directly to its specialized administrative role: DHCP relay agents enable multi-VLAN IP leasing across routers, PTR records implement reverse DNS resolution, IPAM centralizes address space tracking, and Stratum 1 NTP servers sync directly with physical reference time sources.

Step-by-Step Solution

1
Analyze the primary operation of a DHCP Relay Agent.
Routers drop client DHCP broadcasts by default. A relay agent receives these broadcasts on a local VLAN and forwards them to a remote server.
Enables centralizing DHCP servers across multiple VLANs without putting a server in every subnet.
2
Determine the function of a DNS PTR record.
PTR records perform reverse name resolution by mapping IP addresses back to domain names.
Forward DNS zones map names to IP addresses (A/AAAA records), whereas reverse DNS zones store PTR records.
3
Identify the role of IPAM in enterprise networks.
IPAM centralizes address space planning, tracks static and dynamic assignments, and monitors pool depletion.
Consolidates management across DNS, DHCP, and IP address inventory to prevent conflicts and track utilization.
4
Evaluate NTP Stratum hierarchy levels.
Stratum 1 servers attach directly to physical Stratum 0 time sources.
Stratum 0 represents high-precision hardware clocks (GPS/atomic), making directly attached network time servers Stratum 1.

Key Concept

Core IP Addressing Services
Question 256Question →

A system administrator observes syslog messages being sent from a local host to a remote log server. As the outgoing syslog data passes down the host's protocol stack during encapsulation, which of the following actions occur? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: The Transport layer encapsulates the syslog payload into a segment or datagram by adding transport port headers.; The Network layer encapsulates the transport data unit into a packet by adding source and destination IP addresses.

Answer

During outbound encapsulation, the Transport layer appends transport port headers to create a segment/datagram, and the Network layer appends IP addresses to form a packet.
Outbound encapsulation works down the protocol stack. The Transport layer adds port headers to form segments or datagrams, and the Network layer adds IP headers to form packets.

Step-by-Step Solution

1
Analyze Layer 4 processing during outbound encapsulation.
Application payload (syslog data) receives a Transport layer header with source and destination ports.
Layer 4 handles process-to-process communication by prepending port addresses.
2
Analyze Layer 3 processing during outbound encapsulation.
The resulting Transport datagram receives a Network layer header containing source and destination IP addresses.
Layer 3 handles host-to-host logical routing across network boundaries.

Key Concept

Outbound Encapsulation Sequence and Protocol Data Units
Estimated Time:45s
Question 257Question →

A network technician needs to configure a DNS server to enable hostnames to be resolved from known IP addresses during network troubleshooting and log auditing. Which DNS record type should be created to support this reverse lookup functionality?

Show answer & explanation

Answer: PTR record

Answer

PTR record
PTR (Pointer) records provide reverse DNS resolution by mapping IP addresses back to fully qualified domain names.

Step-by-Step Solution

1
Analyze the service requirement
The target requirement is resolving a host's domain name starting from an IP address (reverse DNS lookup).
Understanding whether the operational need is forward or reverse resolution determines the record type.
2
Select the appropriate DNS resource record type
PTR records store mapping entries specifically for reverse DNS resolution.
Forward records (A/AAAA) convert names to IP addresses, while PTR records map IP addresses to canonical hostnames.

Key Concept

DNS Reverse Lookup and PTR Records
Question 258Question →

Host A (IP address 10.1.10.45/24) is connected to an access port assigned to VLAN 10 on a Layer 2 switch. Host B (IP address 10.1.20.80/24) is connected to another access port assigned to VLAN 20 on the same switch. Host A attempts to send a packet to Host B, causing Host A to broadcast an ARP request for Host B's IP address. No Layer 3 router or Layer 3 switch interface is configured on the network. How does the Layer 2 switch process the incoming ARP broadcast frame?

Show answer & explanation

Answer: The switch floods the ARP broadcast frame exclusively to operational ports assigned to VLAN 10, preventing Host B from receiving the request.

Answer

The switch floods the ARP broadcast frame exclusively to operational ports assigned to VLAN 10, preventing Host B from receiving the request.
VLANs create distinct Layer 2 broadcast domains. When Host A generates an ARP broadcast request on VLAN 10, the Layer 2 switch floods the frame only out of ports assigned to VLAN 10. Since Host B is on VLAN 20 and no Layer 3 routing device exists to route traffic between the subnets, Host B will never receive the ARP request.

Step-by-Step Solution

1
Identify the ingress port and VLAN membership of the sender.
Host A is connected to an access port assigned to VLAN 10, placing its generated frames inside VLAN 10's broadcast domain.
Layer 2 switches associate incoming frames on access ports with the port's assigned VLAN ID.
2
Analyze frame destination MAC address and switch forwarding rules.
An ARP request uses the Layer 2 broadcast destination address (FF:FF:FF:FF:FF:FF). The switch floods broadcast frames to all active ports within the ingress VLAN.
Broadcast frames must be sent to all members of the broadcast domain, but VLAN boundaries strictly limit the scope of the flood.
3
Evaluate inter-VLAN reachability without a Layer 3 gateway.
Because Host B is on VLAN 20 and no router/Layer 3 device is present, the broadcast frame never reaches Host B, and inter-VLAN communication fails.
VLANs act as logically independent switch instances at Layer 2 and require Layer 3 routing (Router-on-a-Stick or Layer 3 switch SVI) to forward traffic between different subnets/VLANs.

Key Concept

VLAN Broadcast Domain Isolation
Estimated Time:1m 30s
Question 259Question →

A network technician is configuring an interface access list on a router to allow secure, encrypted command-line management connections to a remote Linux server. Which transport protocol and port combination must be permitted to allow Secure Shell (SSH) traffic?

Show answer & explanation

Answer: TCP port 22

Answer

TCP port 22 is the standard port and protocol for Secure Shell (SSH) remote management traffic.
Secure Shell (SSH) uses TCP port 22 by default to provide secure, encrypted command-line management traffic over a connection-oriented session.

Step-by-Step Solution

1
Identify the application protocol requested in the scenario.
The scenario specifies Secure Shell (SSH) for encrypted command-line management.
Determining the target application protocol is necessary to select the correct transport protocol and port number.
2
Determine the transport layer protocol requirement.
SSH requires a reliable, connection-oriented session provided by TCP.
Interactive terminal communications rely on error checking and session establishment.
3
Map SSH to its well-known default port number.
SSH uses TCP port 22 by default.
IANA assigns port 22 specifically for secure terminal access via SSH.

Key Concept

Standard Default Ports and Protocols
Estimated Time:45s
Question 260Question →

A network administrator is creating a new subnet dedicated to a small group of administrative servers using the IPv4 prefix length /28/28. What is the maximum number of usable host IP addresses available for devices within this subnet?

Show answer & explanation

Answer: 14

Answer

14 usable host IP addresses
With a /28/28 prefix, 44 bits remain for the host portion (32−28=432 - 28 = 4). The total number of IP addresses is 24=162^4 = 16. Because the network address (all host bits 0) and the broadcast address (all host bits 1) cannot be assigned to hosts, subtracting 22 yields exactly 1414 usable host IP addresses.

Step-by-Step Solution

1
Determine the number of host bits
32−28=432 - 28 = 4 host bits
An IPv4 address consists of 3232 bits. Subtracting the prefix length leaves the host portion of the address.
2
Calculate total IP addresses in the subnet
24=162^4 = 16 total IP addresses
The total number of IP addresses in a CIDR block is calculated as 2host bits2^{\text{host bits}}.
3
Subtract reserved addresses to find usable host IP count
16−2=1416 - 2 = 14 usable host IPs
The first address (network ID) and the last address (broadcast address) cannot be assigned to individual host interfaces.

Key Concept

IPv4 Subnet Usable Host Capacity Calculation
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