Examkin

Networking Concepts

538 questions

All types538Multiple choice279Multiple select117Numeric19Matching95Ordering28
Question 61Question →

An engineer is mapping network service traffic for perimeter security monitoring. Match each network protocol on the left with its standard default port number and transport layer protocol on the right.

Click a left item, then click its matching right item

Items

Server Message Block (SMB)
Internet Message Access Protocol (IMAP)
SNMP Trap notifications
Network Time Protocol (NTP)

Matches

Show answer & explanation

Answer

Server Message Block (SMB) matches Port 445 / TCP; Internet Message Access Protocol (IMAP) matches Port 143 / TCP; SNMP Trap notifications match Port 162 / UDP; Network Time Protocol (NTP) matches Port 123 / UDP.
Each protocol is accurately matched to its assigned IANA default port number and transport layer protocol: SMB uses TCP port 445, IMAP uses TCP port 143, SNMP traps use UDP port 162, and NTP uses UDP port 123.

Step-by-Step Solution

1
Identify transport requirements for connection-oriented file sharing and email retrieval protocols.
SMB requires reliable TCP transmission over port 445, and standard IMAP requires TCP over port 143.
File transfers and email access require connection-oriented transport guarantees provided by TCP.
2
Identify transport requirements for time synchronization and asynchronous trap alert protocols.
NTP uses lightweight connectionless UDP port 123, and SNMP Traps use UDP port 162 to deliver event messages to management systems.
Time synchronization and automated agent traps rely on low-overhead UDP transport.

Key Concept

Standard default port numbers and transport layer protocols (TCP/UDP) for enterprise networking services.
Question 62Question →

A network administrator is documenting Layer 2 switch interface behaviors and VLAN features for an enterprise network deployment. Match each switching concept on the left with its correct operational description on the right.

Click a left item, then click its matching right item

Items

Voice VLAN
Access Port
Trunk Port
Broadcast Storm Control

Matches

Show answer & explanation

Answer

Voice VLAN matches with configuring an auxiliary VLAN path with QoS markings; Access Port matches with transmitting traffic for a single assigned VLAN without 802.1Q headers; Trunk Port matches with multiplexing traffic from multiple VLANs using 802.1Q tagging; Broadcast Storm Control matches with dropping excessive Layer 2 broadcast frames exceeding defined thresholds.
Each feature corresponds strictly to its operational role: Voice VLAN provides dedicated bandwidth and QoS prioritization for VoIP telephony; Access Port connects host endpoints inside a single untagged VLAN; Trunk Port multiplexes multi-VLAN frames across switches using 802.1Q header encapsulation; Broadcast Storm Control enforces bandwidth caps on broadcast traffic to prevent network disruption.

Step-by-Step Solution

1
Identify the primary purpose of Voice VLAN technology on enterprise access switches.
Voice VLAN delivers a secondary VLAN tag on a single port for IP phone traffic and applies Quality of Service (QoS) markings.
This separates voice traffic from workstation data traffic while maintaining call quality.
2
Differentiate between Access Port and Trunk Port forwarding behaviors.
Access ports process untagged frames for a single broadcast domain, whereas trunk ports aggregate multi-VLAN traffic across inter-switch links via 802.1Q headers.
End-user workstations require untagged access ports, while switch-to-switch links require trunks.
3
Determine the functional role of Broadcast Storm Control.
Storm control monitors ingress traffic levels and discards frames when broadcast, multicast, or unknown unicast traffic exceeds configured limits.
This mitigates Layer 2 loops or rogue device traffic from overwhelming switch CPU and link bandwidth.

Key Concept

Switch Port Modes, VLAN Allocation, and Traffic Control Features
Estimated Time:1m 30s
Question 63Question →

A logistics company operates four separate fulfillment centers located within a 30-mile radius across the same city. The network engineering team connects these facilities using leased dark fiber provided by a municipal telecommunications provider to enable high-speed inter-building communication across the city. Which of the following network types best classifies this deployment?

Show answer & explanation

Answer: Metropolitan Area Network (MAN)

Answer

Metropolitan Area Network (MAN)
A Metropolitan Area Network (MAN) is designed to connect facilities distributed across a city or metropolitan region (typically covering tens of miles). It relies on municipal or telecommunications provider links such as leased dark fiber.

Step-by-Step Solution

1
Analyze the geographic scope of the scenario
The network spans multiple facilities across a 30-mile radius within a single city.
Geographic footprint dictates the primary network type classification (PAN, LAN, CAN, MAN, WAN).
2
Evaluate the underlying connection infrastructure
Leased municipal dark fiber from a local telecommunications provider is utilized.
Networks spanning an entire city using city-wide service provider or municipal infrastructure are classified as Metropolitan Area Networks.
3
Match the scope and infrastructure to the correct network classification
A city-wide network larger than a CAN (campus) but smaller than a country-wide WAN is a MAN.
A Metropolitan Area Network specifically covers a city or town-sized geographical region.

Key Concept

Geographic Network Classifications (MAN vs CAN vs SAN vs PAN)
Estimated Time:1m 0s
Question 64Question →

A network host receives an incoming Ethernet frame containing an IP packet directed to a local service. Which TWO of the following actions correctly describe the de-encapsulation steps performed as data moves up the destination protocol stack?

Select all that apply

Show answer & explanation

Answer: The Network Interface Card verifies frame integrity using the Frame Check Sequence trailer and strips the Layer 2 header and trailer before passing the payload to the Network layer.; The Network layer examines the Protocol field in the IP header to identify the appropriate Layer 4 protocol to receive the decapsulated payload.

Answer

The Network Interface Card verifies frame integrity using the Frame Check Sequence trailer and strips the Layer 2 header and trailer before passing the payload to the Network layer, and the Network layer examines the Protocol field in the IP header to identify the appropriate Layer 4 protocol to receive the decapsulated payload.
During de-encapsulation, each layer inspects and removes its corresponding header or trailer before passing the remaining payload to the layer above. At Layer 2, frame integrity is verified using the Frame Check Sequence (FCS), after which the Layer 2 header and trailer are removed. At Layer 3, the Network layer inspects the Protocol field within the IP header to determine which Transport layer protocol (such as TCP or UDP) receives the payload.

Step-by-Step Solution

1
Analyze Layer 2 frame processing and de-encapsulation
The host physical interface receives incoming signals, builds the frame, verifies checksum integrity using the Frame Check Sequence (FCS) in the trailer, and strips the Ethernet header and trailer.
De-encapsulation progresses from lower layers to higher layers, with each layer removing its respective control headers and trailers.
2
Analyze Layer 3 header evaluation and protocol demultiplexing
The Network layer receives the IP packet, verifies target logical addressing, and inspects the IP header Protocol field to pass the payload to the appropriate Layer 4 handler (such as TCP or UDP).
The Protocol field in the IP header specifies the upper-layer protocol driver that must process the enclosed payload.

Key Concept

De-encapsulation protocol stack unwrapping and header field demultiplexing
Estimated Time:1m 30s
Question 65Question →

A client workstation on a corporate LAN transmits web traffic to an external server across multiple subnets. When an intermediate router receives the Ethernet frame on its ingress interface, which of the following best describes the encapsulation and de-encapsulation actions performed by the router prior to forwarding the packet out its egress interface?

Show answer & explanation

Answer: The router strips the incoming Layer 2 frame header and trailer to inspect the Layer 3 packet, then encapsulates the original Layer 3 packet into a new Layer 2 frame formatted for the egress link.

Answer

The router removes the incoming Layer 2 frame header and trailer, inspects the Layer 3 packet header to make a forwarding decision, decrements the Time to Live (TTL), and encapsulates the untouched IP packet inside a new Layer 2 frame for the outbound interface.
When traffic crosses a router (Layer 3 device), the router strips the incoming Layer 2 frame header and trailer because MAC addresses are only locally significant. The router inspects the Layer 3 IP header to determine the next hop, decrements the TTL, and then wraps the original IP packet inside a new Layer 2 header and trailer formatted specifically for the next hop's data link protocol.

Step-by-Step Solution

1
De-encapsulate Layer 2
The router receives the physical signals on its ingress interface, validates the Frame Check Sequence (FCS), and strips the Layer 2 Ethernet header and trailer.
Layer 2 encapsulation is only valid for local communication within the ingress network segment.
2
Inspect Layer 3 IP Header
The router reads the destination IP address in the Layer 3 packet header, consults its routing table, decrements the TTL field, and re-calculates the IP header checksum.
Layer 3 payload (the IP packet, including Layer 4 TCP and upper-layer data) remains intact to maintain end-to-end transport integrity.
3
Encapsulate for Egress Layer 2
The router adds a new Layer 2 header (with the egress interface MAC as source and next-hop MAC as destination) and a new trailer around the IP packet.
New Layer 2 encapsulation is required to transmit the packet across the next physical hop toward the destination.

Key Concept

Hop-by-hop Layer 2 de-encapsulation and re-encapsulation by Layer 3 devices
Estimated Time:1m 30s
Question 66Question →

A network technician is configuring switch ports and VLAN settings across an enterprise network. Match each switching concept or feature to its primary operational behavior.

Click a left item, then click its matching right item

Items

802.1Q Trunking
Native VLAN
Access Port
Sticky MAC Learning

Matches

Show answer & explanation

Answer

802.1Q Trunking matches carrying traffic for multiple VLANs via 4-byte header tagging; Native VLAN matches processing untagged frames on a trunk link; Access Port matches carrying single VLAN untagged traffic to endpoints; Sticky MAC Learning matches converting dynamically learned MAC addresses into permanent switch port configuration entries.
Each switching concept accurately maps to its operational definition: 802.1Q trunking enables multi-VLAN transport via 4-byte frame tagging, native VLAN handles untagged trunk traffic, access ports connect end devices within a single VLAN, and sticky MAC learning dynamically saves secure MAC addresses to configuration memory.

Step-by-Step Solution

1
Identify the mechanism for multiplexing multiple VLANs across a single inter-switch link.
802.1Q trunking inserts a 4-byte VLAN tag into the frame header to identify traffic belonging to different VLANs.
Trunking is specifically designed to carry traffic for multiple VLANs across point-to-point switch connections.
2
Identify how untagged frames arriving on a trunk port are processed.
Untagged frames on an 802.1Q trunk link are automatically mapped to the designated Native VLAN.
The IEEE 802.1Q standard utilizes a native VLAN for backward compatibility with untagged legacy traffic.
3
Determine the switch port configuration used for connecting individual endpoint hosts.
An Access Port belongs to a single VLAN and delivers standard untagged frames to connected endpoint devices.
Endpoints generally do not inspect 802.1Q tags, so access ports strip internal tags before frame egress.
4
Identify the port security mechanism that saves dynamically learned MAC addresses.
Sticky MAC Learning converts dynamically learned MAC addresses into static entries recorded in the switch configuration.
This avoids manual MAC address entry while enforcing port security parameters across device reboots.

Key Concept

Switching Concepts and VLAN Operations
Estimated Time:1m 30s
Question 67Question →

A host operating system receives an incoming transmission containing web application traffic. As the destination network stack processes the protocol data unit (PDU) through de-encapsulation, which of the following operations occur specifically at Layer 3 (Network Layer) before the payload is passed up to Layer 4? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: The destination IP address in the packet header is verified against the receiving host's configured address.; The Protocol field within the IP header is inspected to identify which Transport layer protocol should receive the encapsulated payload.

Answer

The correct Layer 3 operations are verifying the destination IP address against the local host address and inspecting the IP header Protocol field to determine the target Layer 4 protocol handler.
During de-encapsulation at Layer 3 (Network Layer), the network stack inspects the IP header to verify that the packet's destination IP matches the host interface address. It also checks the IP header's Protocol field (such as 6 for TCP or 17 for UDP) to determine which Layer 4 protocol module should receive the stripped payload.

Step-by-Step Solution

1
Analyze Layer 3 IP header processing during de-encapsulation.
Identify that Layer 3 processes the IPv4 or IPv6 header containing source/destination IP addresses and protocol demultiplexing information.
De-encapsulation strips headers bottom-up; Layer 3 processes IP header fields before stripping them to yield the Layer 4 segment.
2
Evaluate destination addressing at Layer 3.
Verifying destination IP address ensures the host is the intended recipient.
If the destination IP does not match the receiving host or allowed multicast/broadcast, the packet is dropped.
3
Determine how Layer 3 hands off data to Layer 4.
Reading the Protocol field in the IP header tells Layer 3 which protocol (e.g., TCP or UDP) receives the payload.
Demultiplexing at Layer 3 relies on the IP Protocol field to pass the payload up to Layer 4.

Key Concept

Layer 3 De-encapsulation and Header Demultiplexing
Question 68Question →

A network engineering team is deploying core IP addressing and network management services for domain-joined hosts across multiple routed subnets. To ensure accurate logging correlation across distributed systems and proper name resolution for dual-stack hosts, which of the following protocols or resource records should be implemented? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: NTP operating on UDP port 123 to maintain consistent time synchronization across network hosts and security devices.; AAAA resource records configured in the DNS server to map internal hostnames to assigned IPv6 addresses.

Answer

The correct implementations are NTP operating on UDP port 123 for network-wide time synchronization and AAAA records in DNS for mapping hostnames to IPv6 addresses.
NTP operating on UDP port 123 provides centralized time synchronization required for log correlation across network systems. In addition, AAAA resource records in DNS provide standard hostname-to-IPv6 address resolution for dual-stack environments.

Step-by-Step Solution

1
Analyze time synchronization requirement
NTP operates on UDP port 123 to provide clock synchronization across log sources and network endpoints.
Accurate event correlation and log auditing across distributed subnets require synchronized clocks using NTP.
2
Analyze host name resolution requirement for dual-stack hosts
DNS AAAA records map hostnames directly to 128-bit IPv6 addresses.
Standard forward DNS resolution for IPv6 protocol addressing uses AAAA resource records.
3
Evaluate distractor options against protocol specifications and placement rules
DHCP uses UDP ports 67/68 (not TCP), PTR records perform reverse lookups (not CNAME), and relay agents belong on client default gateways (not the server interface).
Identifying accurate protocol ports, record types, and deployment locations eliminates incorrect service configurations.

Key Concept

IP Addressing Services (NTP operations, DNS record types, and DHCP relay architectural requirements)
Question 69Question →

Four routers on a shared Layer 2 Ethernet segment are configured to run OSPF using the broadcast network type. The administrator configures the OSPF interface priorities and router IDs as follows:

RouterOSPF Interface PriorityRouter ID
Router A1192.168.1.10
Router B210.10.10.1
Router C010.20.20.20
Router D210.10.10.2

Assuming all routers boot simultaneously and form adjacencies, which router will be elected as the Designated Router (DR) for this multi-access segment?

Show answer & explanation

Answer: Router D

Answer

Router D will be elected as the Designated Router because it shares the highest priority (2) and has a higher Router ID than Router B.
In OSPF broadcast networks, the Designated Router election evaluates highest interface priority first. Routers with priority 0 are completely excluded. Routers B and D tie with a priority of 2. To break the tie, OSPF selects the router with the highest Router ID. Comparing 10.10.10.2 (Router D) against 10.10.10.1 (Router B), Router D has the higher Router ID and is elected DR.

Step-by-Step Solution

1
Filter out ineligible routers based on OSPF interface priority.
Router C (priority 0) is excluded from DR and BDR election.
Setting an OSPF priority to 0 makes a router ineligible to become a DR or BDR.
2
Compare the remaining routers by highest OSPF interface priority.
Router B and Router D tie with the highest priority of 2, while Router A has a lower priority of 1.
OSPF DR election first evaluates the highest interface priority value.
3
Break the tie between Router B and Router D using Router IDs.
Router D wins with Router ID 10.10.10.2 compared to Router B's 10.10.10.1.
When interface priorities are equal, the router with the highest numeric Router ID is elected.

Key Concept

OSPF Designated Router (DR) election criteria (Highest Priority > Highest Router ID; Priority 0 excludes).
Estimated Time:1m 0s
Question 70Question →

Match each IPv6 base header field to its correct functional description.

Click a left item, then click its matching right item

Items

Payload Length
Flow Label
Next Header
Hop Limit

Matches

Show answer & explanation

Answer

Payload Length pairs with describing payload size excluding the 40-byte base header; Flow Label pairs with identifying specific packet flows for path handling; Next Header pairs with specifying the encapsulated protocol or extension header; Hop Limit pairs with decrementing at each hop to prevent loops.
Each item matches its corresponding function: Payload Length measures bytes beyond the fixed 40-byte base header, Flow Label tags specific packet streams for uniform routing treatment, Next Header identifies protocol payloads or extension headers, and Hop Limit prevents loops by decrementing at each hop.

Step-by-Step Solution

1
Identify the purpose of Payload Length in the IPv6 header.
Payload Length measures the bytes after the 40-byte base header.
Unlike IPv4's Total Length field, IPv6 Payload Length excludes the base header itself.
2
Determine the function of the Flow Label field.
Flow Label maintains flow consistency for real-time packets.
It allows routers to service non-fragmented sequence packets along identical paths.
3
Determine the function of the Next Header field.
Next Header specifies payload or extension headers.
It acts as a pointer to the upper-layer protocol (e.g., TCP=6, UDP=17) or an IPv6 extension header.
4
Identify the function of Hop Limit.
Hop Limit decrements to stop loop propagation.
It performs the exact functionality of IPv4 TTL.

Key Concept

IPv6 Base Header Architecture
Question 71Question →

An organization is deploying a high-speed fiber-optic network to interconnect several municipal government buildings located across a single city boundary to share central voice and video communications. Which of the following network scope classifications best describes this deployment?

Show answer & explanation

Answer: Metropolitan Area Network (MAN)

Answer

Metropolitan Area Network (MAN)
A Metropolitan Area Network (MAN) is designed to connect facilities and resources distributed across a single city or metropolitan area, typically utilizing leased telecom lines or dedicated dark fiber.

Step-by-Step Solution

1
Analyze the geographic parameters given in the scenario.
The network interconnects multiple facilities distributed across a single city.
Network types (PAN, LAN, CAN, MAN, WAN) are categorized primarily by geographic coverage.
2
Match the geographic coverage to the correct network classification.
A network that spans across a town or city is classified as a Metropolitan Area Network (MAN).
LANs serve localized sites, CANs serve connected campus buildings, MANs serve entire cities, and WANs serve broad geographical distances.

Key Concept

Geographic scope classifications of network types (MAN vs LAN, WAN, SAN)
Question 72Question →

A network engineering team is evaluating WAN deployment models to interconnect several mission-critical data center sites. They are considering implementing a full-mesh physical topology to maximize overall system availability.

Which of the following statements accurately describe the operational characteristics and design trade-offs of a full-mesh network topology? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: The total number of direct physical point-to-point links required to connect nn sites is calculated using the formula n(n−1)2\frac{n(n - 1)}{2}.; It provides complete redundancy, ensuring network traffic can be rerouted if any individual link experiences a physical outage.

Answer

The correct statements are that the formula n(n−1)2\frac{n(n - 1)}{2} determines the required link count for nn nodes, and that full-mesh topologies offer complete redundancy against single link failures.
A full-mesh network provides maximum redundancy by connecting every node directly to every other node using n(n−1)2\frac{n(n - 1)}{2} links. This structural redundancy ensures that traffic can be dynamically rerouted around any single point-to-point link failure.

Step-by-Step Solution

1
Analyze full-mesh connectivity requirements
Each node must have a dedicated connection to every other (n−1)(n - 1) node, yielding n(n−1)2\frac{n(n - 1)}{2} total physical links.
This formula accurately accounts for bi-directional unique links between all nodes without double-counting.
2
Evaluate fault tolerance and redundancy
High availability is maintained because alternate physical paths exist between all sites.
Redundant links prevent any single line outage from partitioning the network.
3
Identify misconceptions regarding routing and scaling
Full-mesh physical topologies do not flatten Layer 3 boundaries, and cabling grows quadratically (O(n2)O(n^2)), not linearly.
Confusing physical link layout with Layer 2 broadcast domains or assuming linear growth leads to improper network design assumptions.

Key Concept

Full-Mesh Topology Link Calculation and Redundancy Characteristics
Estimated Time:1m 30s
Question 73Question →

Match each network service requirement on the left with its corresponding default port number and transport layer protocol on the right.

Click a left item, then click its matching right item

Items

Directory service queries requiring implicit encryption via TLS
Network clock synchronization across distributed network infrastructure
Encrypted remote command-line administration and secure file transfer
Initial key exchange and Security Association negotiation for IPsec VPNs

Matches

Show answer & explanation

Answer

Directory queries via TLS matches TCP port 636; Network clock synchronization matches UDP port 123; Encrypted remote administration matches TCP port 22; Initial key exchange for IPsec VPNs matches UDP port 500.
Each service requirement maps strictly to its standard port and transport protocol: encrypted LDAP queries require TCP port 636, time synchronization requires UDP port 123, secure remote management requires TCP port 22, and IPsec key exchange requires UDP port 500.

Step-by-Step Solution

1
Analyze secure directory service protocol requirements.
Identified LDAPS operating over TCP port 636.
Standard unencrypted LDAP uses TCP 389, while implicitly encrypted LDAPS mandates TCP port 636.
2
Analyze time synchronization protocol requirements.
Identified NTP operating over UDP port 123.
Clock synchronization requires low overhead without connection setup delays, necessitating UDP port 123.
3
Analyze secure CLI and file administration protocol requirements.
Identified SSH/SFTP operating over TCP port 22.
Secure Shell provides reliable encrypted interactive shell access and SFTP transport via TCP port 22.
4
Analyze IPsec VPN key negotiation protocol requirements.
Identified IKE operating over UDP port 500.
Internet Key Exchange Phase 1 Security Association setup uses ISAKMP over UDP port 500.

Key Concept

Standard Network Port Assignments and Transport Layer Selection
Question 74Question →

A network administrator is configuring a Domain Name System (DNS) server to support an automated security monitoring system. The monitoring system captures IP addresses from network traffic logs and performs reverse DNS lookups to convert those IP addresses back into hostnames. Which DNS record type must be created on the server to enable this reverse resolution functionality?

Show answer & explanation

Answer: PTR record

Answer

PTR record
A Pointer (PTR) record is specifically designed for reverse DNS lookups, mapping an IP address (in the in-addr.arpa or ip6.arpa domain) back to a Fully Qualified Domain Name (FQDN). This fulfills the requirement for the security monitoring tool to resolve logged IP addresses into readable hostnames.

Step-by-Step Solution

1
Identify the primary functional requirement from the scenario
The system needs to perform reverse DNS lookups (mapping an IP address to a hostname).
Security logging and network management tools use reverse DNS to identify hostnames associated with captured IP addresses.
2
Evaluate DNS record types against the reverse resolution requirement
Pointer (PTR) records reside in reverse lookup zones (in-addr.arpa for IPv4, ip6.arpa for IPv6) and explicitly map IP addresses to hostnames.
Forward lookup records (A, AAAA, CNAME) map hostnames to IP addresses or other hostnames, whereas PTR records perform the inverse mapping.

Key Concept

DNS Record Types and Reverse DNS Resolution
Question 75Question →

A network administrator is allocating subnets from the assigned address block 192.168.100.0/24192.168.100.0/24 using Variable Length Subnet Masking (VLSM). Subnets must be provisioned sequentially starting from the beginning of the block, ordered from largest to smallest host capacity requirement:

1. Sales Division: requires 5858 usable host IP addresses
2. Marketing Division: requires 2828 usable host IP addresses
3. Executive Staff: requires 1212 usable host IP addresses

If the Sales Division is assigned the first subnet starting at 192.168.100.0192.168.100.0, what is the range of usable host IP addresses for the Marketing Division?

Show answer & explanation

Answer: 192.168.100.65192.168.100.65 through 192.168.100.94192.168.100.94

Answer

The usable host IP address range for the Marketing Division is 192.168.100.65192.168.100.65 through 192.168.100.94192.168.100.94.
The Sales Division takes 192.168.100.0/26192.168.100.0/26 (6464 addresses), placing the next available network block at 192.168.100.64192.168.100.64. The Marketing Division requires 2828 usable hosts, requiring a block of at least 3030 IP addresses. A /27/27 subnet provides 3232 total addresses (252^5), creating a subnet range of 192.168.100.64192.168.100.64 to 192.168.100.95192.168.100.95. Omitting the reserved network ID (192.168.100.64192.168.100.64) and broadcast address (192.168.100.95192.168.100.95) leaves valid usable host addresses from 192.168.100.65192.168.100.65 through 192.168.100.94192.168.100.94.

Step-by-Step Solution

1
Calculate subnet mask and address range for the Sales Division (58 hosts).
Sales needs 58+2=6058 + 2 = 60 total addresses. The smallest power of 22 greater than or equal to 6060 is 26=642^6 = 64 addresses (/26/26 prefix). Network ID: 192.168.100.0/26192.168.100.0/26, Broadcast: 192.168.100.63192.168.100.63.
Two addresses are reserved for network ID and broadcast address.
2
Determine the starting network address for the Marketing Division.
The next available IP address after the Sales broadcast address is 192.168.100.64192.168.100.64.
VLSM allocations must be contiguous without overlapping existing subnets.
3
Calculate prefix length and address block size for the Marketing Division (28 hosts).
Marketing needs 28+2=3028 + 2 = 30 total addresses. The smallest power of 22 greater than or equal to 3030 is 25=322^5 = 32 addresses (/27/27 prefix). Subnet block: 192.168.100.64/27192.168.100.64/27.
A /27/27 prefix provides 3232 total addresses (3030 usable).
4
Identify the usable host IP range for the Marketing Division.
Network ID: 192.168.100.64192.168.100.64. Broadcast address: 192.168.100.64+31=192.168.100.95192.168.100.64 + 31 = 192.168.100.95. Usable host range: 192.168.100.65192.168.100.65 to 192.168.100.94192.168.100.94.
Usable host range spans from Network ID + 1 up to Broadcast address - 1.

Key Concept

Variable Length Subnet Masking (VLSM) Host Capacity Allocation
Question 76Question →

A network administrator is deploying a live voice over IP (VoIP) application that requires low latency and can tolerate occasional data loss. Which Transport layer protocol should the administrator select to meet these requirements?

Show answer & explanation

Answer: User Datagram Protocol (UDP)

Answer

User Datagram Protocol (UDP) should be used because it is a connectionless Transport layer protocol optimized for low latency without packet retransmission overhead.
User Datagram Protocol (UDP) is connectionless, meaning it transmits datagrams without establishing a preliminary connection or requesting acknowledgments. This minimal overhead makes UDP the preferred choice for real-time streaming and voice applications where speed is critical.

Step-by-Step Solution

1
Analyze the application traffic requirements
Real-time VoIP traffic prioritizes speed and minimal delay over guaranteed delivery.
Retransmitting delayed voice packets creates jitter and poor call quality.
2
Identify the appropriate Transport layer protocol
Connectionless protocols like UDP send data without establishing a connection or waiting for acknowledgments.
This eliminates handshaking overhead and latency, perfectly matching VoIP requirements.

Key Concept

Connection-Oriented vs Connectionless Transport Protocols
Estimated Time:45s
Question 77Question →

A network administrator is deploying a database synchronization service across a wide area network (WAN). The application requires guaranteed packet delivery, strict in-order sequence processing, and flow control to prevent buffer exhaustion at the receiving endpoint, while being tolerant of minor latency delays. Which transport layer protocol and operational characteristic best satisfy these application requirements?

Show answer & explanation

Answer: TCP, because it establishes a virtual connection using a three-way handshake and uses sequence numbers and acknowledgments to ensure reliable, ordered delivery.

Answer

TCP is the correct protocol selection because it is connection-oriented, utilizing a three-way handshake, sequence numbers, acknowledgments, and windowing to guarantee data delivery and order for critical database synchronization.
Database synchronization requires high data integrity, complete delivery, and strict packet ordering. TCP (Transmission Control Protocol) is a connection-oriented Layer 4 protocol that establishes a reliable session via a three-way handshake (SYN, SYN-ACK, ACK). It tracks data stream progress using sequence and acknowledgment numbers and employs sliding window mechanisms for flow control, making it the ideal protocol despite its higher header overhead.

Step-by-Step Solution

1
Analyze the application requirements from the scenario.
The scenario requires guaranteed delivery, strict packet ordering, and flow control, while tolerating minor latency.
Identifying necessary features dictates whether a connection-oriented or connectionless protocol is required.
2
Compare TCP and UDP protocol capabilities against identified requirements.
TCP provides connection-oriented reliability, error recovery, flow control, and sequence ordering at Layer 4. UDP provides connectionless, unacknowledged, low-overhead transmission.
TCP matches all requirements (guaranteed delivery and ordering), whereas UDP prioritizes speed over reliability.
3
Select the option specifying TCP with accurate architectural descriptions.
TCP utilizes a three-way handshake (SYN, SYN-ACK, ACK) to set up sessions and uses sequence numbers for ordering.
This accurate description directly fulfills the requirements of database synchronization.

Key Concept

Connection-Oriented vs Connectionless Transport Protocols
Question 78Question →

A network engineer is troubleshooting an enterprise application connectivity issue using a packet analyzer. The capture indicates that the host successfully establishes a lower-level transport connection via a TCP three-way handshake. Following segment reassembly at Layer 4, the data stream is handed up the stack. However, the connection drops because the host fails to negotiate character encoding syntax and payload encryption protocols required to interpret the message structure. At which layer of the Open Systems Interconnection (OSI) model does this specific negotiation failure occur, and what general term describes the Protocol Data Unit (PDU) at this stage?

Show answer & explanation

Answer: Presentation Layer (Layer 6); the PDU is referred to as Data

Answer

The negotiation failure occurs at the Presentation Layer (Layer 6), where the PDU is generically referred to as Data.
The correct answer identifies the Presentation Layer (Layer 6) as responsible for data representation, encryption/decryption, compression, and syntax conversion. Upper OSI layers (Layers 5 through 7) process the payload referred to as Data.

Step-by-Step Solution

1
Analyze the scenario description to identify the failing network operation.
The failure occurs during character encoding syntax negotiation and payload encryption formatting after TCP segment reassembly.
Character set translation (e.g., ASCII/EBCDIC), data compression, and cryptographic formatting (e.g., TLS syntax/presentation handling) are core functions of the OSI Presentation Layer.
2
Determine the corresponding OSI model layer for syntax, formatting, and encryption.
Layer 6 (Presentation Layer) is identified as the layer responsible.
Layer 6 acts as the data translator for the network, ensuring payload structures are formatted correctly for consumption by Layer 7.
3
Identify the proper Protocol Data Unit (PDU) nomenclature at this stage of processing.
At Layers 5, 6, and 7, the encapsulated payload is classified generically as Data.
Specific PDU naming conventions apply to lower layers (Layer 2 = Frame, Layer 3 = Packet, Layer 4 = Segment), while upper layers process unified Data.

Key Concept

OSI Layer 6 (Presentation Layer) functions and PDU classification
Estimated Time:2m 0s
Question 79Question →

A network technician is using a packet analyzer to inspect network traffic on a local Ethernet segment. The analyzer displays frame headers containing source and destination Media Access Control (MAC) addresses alongside a Logical Link Control (LLC) header. At which layer of the OSI model does this header encapsulation and addressing occur?

Show answer & explanation

Answer: Data Link layer

Answer

The Data Link layer
The Data Link layer (Layer 2) of the OSI model operates with frames and uses Media Access Control (MAC) addresses for local node-to-node delivery. It is split into two sublayers: LLC (which interfaces with Layer 3) and MAC (which manages physical media access).

Step-by-Step Solution

1
Identify the data structures mentioned in the packet capture
The packet capture shows Media Access Control (MAC) addresses and a Logical Link Control (LLC) header.
Analyzing specific header elements allows accurate mapping to the corresponding OSI model layer.
2
Map the MAC and LLC components to the OSI model hierarchy
LLC and MAC are the two standardized sublayers of Layer 2 (Data Link layer).
Layer 2 provides physical addressing and structures raw bits into readable frames for local network transport.

Key Concept

OSI Data Link Layer Functions and Sublayers
Question 80Question →

A network engineer is inspecting an interface configuration on a core router. The uncompressed 128-bit IPv6 address assigned to the interface is 2001:0db8:0000:0000:0008:0000:0000:00012001:0db8:0000:0000:0008:0000:0000:0001. According to standard IPv6 address compression rules (RFC 5952), which of the following is the correct shorthand representation of this address?

Show answer & explanation

Answer: 2001:db8::8:0:0:1

Answer

The correct compressed IPv6 address is 2001:db8::8:0:0:1.
The address 2001:db8::8:0:0:12001:db8::8:0:0:1 correctly applies RFC 5952 guidelines: leading zeros within each hextet are truncated (0db8→db80db8 \rightarrow db8, 0008→80008 \rightarrow 8, 0000→00000 \rightarrow 0), and the double-colon (::) is used exactly once to replace the first contiguous block of zero hextets (0:00:0).

Step-by-Step Solution

1
Omit leading zeros in each 16-bit hextet.
The original address 2001:0db8:0000:0000:0008:0000:0000:00012001:0db8:0000:0000:0008:0000:0000:0001 simplifies to 2001:db8:0:0:8:0:0:12001:db8:0:0:8:0:0:1.
Leading zeros within a four-digit hexadecimal group do not add value and can be safely suppressed.
2
Identify consecutive zero hextet runs to apply the double-colon (::) operator.
There are two zero runs of equal length (two groups of zeros each): 0:00:0 after db8db8, and 0:00:0 after 88.
RFC 5952 specifies that when zero runs are of equal length, the first contiguous run of all-zero hextets must be replaced with '::'.
3
Apply the double colon to the first run of zero hextets only.
The final compressed address is written as 2001:db8::8:0:0:12001:db8::8:0:0:1.
The double colon (::) can only be used once in an IPv6 address to prevent ambiguity during address expansion.

Key Concept

IPv6 Address Compression Rules (RFC 5952)
PreviousPage 4 / 27Next
Networking Concepts Practice Questions — CompTIA Network+ — Page 4 | Examkin