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Question 141Question →

A network engineer is analyzing the structural design differences between IPv4 and IPv6 packet headers. In IPv4, optional parameters increase the header size dynamically up to 60 bytes, requiring routers to process variable-length headers. How does the IPv6 header architecture efficiently support optional control data while keeping the base IPv6 header fixed at a constant 40 bytes?

Show answer & explanation

Answer: By inserting optional Extension Headers sequentially between the base IPv6 header and the upper-layer payload using the Next Header field

Answer

IPv6 handles optional control parameters by inserting optional Extension Headers sequentially between the fixed 40-byte base header and the upper-layer payload, linked via the Next Header field.
The IPv6 protocol uses a fixed 40-byte base header to streamline routing lookup and performance. Optional data—such as hop-by-hop options, routing headers, fragmentation information, and IPsec security payloads—are offloaded to Extension Headers. These are chained sequentially after the base header using the Next Header field.

Step-by-Step Solution

1
Examine the IPv6 base header architecture
The IPv6 base header is strictly fixed at 40 bytes to allow efficient hardware-based processing by intermediate routers.
Eliminating variable-length headers in the base structure removes processing overhead.
2
Analyze how optional information (such as fragmentation or security) is appended
Options are moved into dedicated Extension Headers placed after the base header.
Intermediate routers generally only need to inspect the base header unless specific extension headers require processing.
3
Identify the linking mechanism between headers
The Next Header field in the base header specifies the type of the first extension header (or upper-layer protocol), forming a daisy chain.
This allows flexible, modular header expansion without changing the base header size.

Key Concept

IPv6 Extension Headers and Next Header Chaining
Question 142Question →

A network technician needs to configure a switch interface that connects to an upstream distribution switch. The link must carry traffic for multiple VLANs concurrently over the single physical link using IEEE 802.1Q tagging. Which interface configuration command must be applied to achieve this functionality?

Show answer & explanation

Answer: switchport mode trunk

Answer

The interface must be configured using `switchport mode trunk` to carry traffic for multiple VLANs over a single physical link.
The command `switchport mode trunk` configures an Ethernet interface to operate as an 802.1Q trunk link, allowing frames from multiple VLANs to be tagged and multiplexed across a single physical link between network devices.

Step-by-Step Solution

1
Identify the requirement for cross-switch link multi-VLAN traffic transport
The link must multiplex frames from multiple broadcast domains using IEEE 802.1Q header tags.
Standard access ports strips or drops tags and belong to only one VLAN.
2
Select the appropriate switchport interface command
Applying `switchport mode trunk` explicitly sets the operational mode to trunking.
Trunk ports inspect 802.1Q tags and pass traffic for all allowed VLANs across the switch interconnect.

Key Concept

VLAN Trunking and 802.1Q Configuration
Question 143Question →

A network administrator is updating the configuration of an IEEE 802.1Q trunk link on interface `GigabitEthernet1/0/48` connecting two switch stacks. The interface currently passes traffic for VLANs 10, 20, and 30. The administrator must permit VLAN 50 across the link without disrupting the traffic of the existing VLANs. Which command should be applied under interface configuration mode to accomplish this requirement?

Show answer & explanation

Answer: switchport trunk allowed vlan add 50

Answer

The command 'switchport trunk allowed vlan add 50' appends VLAN 50 to the existing list of allowed VLANs on an 802.1Q trunk interface without replacing current entries.
The command containing the 'add' parameter appends VLAN 50 to the existing allowed list (VLANs 10, 20, and 30). This allows VLAN 50 traffic to cross the trunk while keeping active production traffic on the existing VLANs uninterrupted.

Step-by-Step Solution

1
Identify the goal for trunk VLAN modification
VLAN 50 must be appended to the allowed VLAN trunk list alongside existing VLANs 10, 20, and 30.
Omitting the 'add' parameter causes the command to overwrite the existing allowed VLAN list entirely.
2
Select the proper Cisco IOS trunk configuration syntax
Use 'switchport trunk allowed vlan add 50' on interface GigabitEthernet1/0/48.
The 'add' keyword preserves currently permitted VLANs while expanding trunk access to include the new VLAN ID.

Key Concept

802.1Q Trunk Allowed VLAN List Modification
Estimated Time:1m 15s
Question 144Question →

A network administrator is troubleshooting an IEEE 802.1Q trunk link between two switches, SW-Access-A and SW-Access-B. On SW-Access-A, interface GigabitEthernet1/0/1 is configured as an 802.1Q trunk with native VLAN 20 and allowed VLANs 10, 20, and 30. On SW-Access-B, interface GigabitEthernet1/0/1 is configured as an 802.1Q trunk with native VLAN 1 and allowed VLANs 10, 20, and 30. Which of the following operational outcomes will occur on this trunk link? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: Untagged frames transmitted from VLAN 20 on SW-Access-A will be assigned to VLAN 1 upon reception on SW-Access-B.; Link-layer protocols such as CDP or LLDP will log native VLAN mismatch warnings on both switches.

Answer

The correct outcomes are: untagged frames transmitted from VLAN 20 on SW-Access-A will be assigned to VLAN 1 upon reception on SW-Access-B, and link-layer protocols such as CDP or LLDP will log native VLAN mismatch warnings on both switches.
When native VLAN IDs differ across an 802.1Q trunk link, untagged traffic sent by one switch's native VLAN is implicitly placed into the receiving switch's native VLAN, causing cross-VLAN traffic leakage. Additionally, neighbor discovery protocols like CDP or LLDP continuously exchange native VLAN parameters and log errors when a mismatch is detected.

Step-by-Step Solution

1
Analyze how 802.1Q trunking handles native VLAN traffic on transmission.
SW-Access-A strips 802.1Q tags from frames belonging to its native VLAN (VLAN 20) before forwarding them out GigabitEthernet1/0/1.
By definition in the IEEE 802.1Q standard, frames matching the local interface's native VLAN are sent untagged across trunk links.
2
Determine how the receiving switch processes untagged ingress frames.
SW-Access-B receives the untagged frames and places them into its locally configured native VLAN (VLAN 1).
An 802.1Q switch implicitly assigns all incoming untagged trunk frames to whatever VLAN ID is configured as native on that specific ingress interface.
3
Evaluate link-layer management protocol behaviors for misconfigured trunk parameters.
CDP or LLDP packets report mismatched native VLAN IDs (20 vs 1), generating console notification logs.
Discovery protocols periodically advertise interface metadata including native VLAN IDs, allowing network equipment to detect asymmetric trunk configurations.
4
Evaluate the impact on tagged VLAN traffic (VLAN 10 and VLAN 30).
Tagged traffic for allowed VLANs (10 and 30) continues to traverse the trunk properly and remains isolated.
Explicit 802.1Q tags are unaffected by native VLAN mismatches, and inter-VLAN routing remains required to move traffic between different VLAN numbers.

Key Concept

802.1Q Native VLAN Operation and Mismatch Behaviors
Estimated Time:2m 0s
Question 145Question →

A network administrator is configuring inbound firewall rules on a security zone interface separating managed infrastructure switches from a central Network Management System (NMS). The switches are configured to send asynchronous, unprompted alert notifications to the NMS whenever a physical interface state changes. Which port and transport layer protocol combination must be permitted on the firewall to allow these unsolicited switch notifications to reach the central management server?

Show answer & explanation

Answer: UDP port 162

Answer

UDP port 162 is the standard port and protocol required to receive unsolicited Simple Network Management Protocol (SNMP) traps and notifications on an NMS.
Simple Network Management Protocol (SNMP) uses two distinct UDP ports depending on the direction and nature of the communication. When managed network devices send unprompted, asynchronous event alerts (SNMP Traps or InformRequests) to a Network Management System (NMS), the traffic is destined for UDP port 162 on the NMS. Therefore, the perimeter firewall must permit inbound UDP port 162 traffic from the switches to the NMS.

Step-by-Step Solution

1
Identify the protocol and communication direction required by the scenario
The network switches are sending asynchronous, unsolicited event alerts to a central Network Management System (NMS), which corresponds to Simple Network Management Protocol (SNMP) Traps.
Unprompted event notifications initiated by managed agents use SNMP Trap messages.
2
Differentiate between SNMP polling ports and SNMP trap ports
SNMP agent polling (queries initiated by the NMS) uses UDP port 161, whereas SNMP trap messages (alerts initiated by managed devices) are sent to destination UDP port 162 on the NMS collector.
Separating polling requests from event traps prevents trap notifications from interfering with routine request-response polling traffic.
3
Determine the transport protocol requirement
SNMP standard operations utilize UDP for low-overhead, connectionless transport.
UDP provides the required lightweight transport mechanism for SNMP messages without session establishment overhead.

Key Concept

SNMP Port Differentiation (UDP 161 vs UDP 162)
Estimated Time:1m 30s
Question 146Question →

A network engineer is troubleshooting a cross-platform file transfer utility operating between a legacy mainframe and a modern Linux server. While the network connection establishes without error, transferred text files display corrupt characters because the source host formats data using EBCDIC character encoding while the destination expects ASCII. Which OSI model layer is directly responsible for handling character set translation and data syntax formatting between disparate systems?

Show answer & explanation

Answer: Presentation Layer

Answer

Presentation Layer
The Presentation Layer (Layer 6) serves as the network's data translator. It handles tasks such as character code conversion (e.g., EBCDIC to ASCII), data compression, encryption, and syntax formatting, enabling distinct computer systems with different internal data representations to communicate effectively.

Step-by-Step Solution

1
Analyze the reported technical issue described in the scenario.
The issue involves mismatched character encodings (EBCDIC versus ASCII) and syntax formatting between two different computer architectures.
Identifying the functional requirement (character translation and data representation) narrows down which OSI layer handles this specific task.
2
Map the functional requirement to OSI Model layer responsibilities.
Layer 6 (Presentation Layer) is specifically defined to manage data representation, character code translation, formatting, syntax standardization, encryption, and compression.
The Presentation Layer acts as the translator for the network, ensuring heterogeneous systems can interpret application data correctly.

Key Concept

OSI Model Presentation Layer Functions
Estimated Time:1m 0s
Question 147Question →

Match each network protocol or service requirement on the left with its standard default port number and primary transport protocol on the right.

Click a left item, then click its matching right item

Items

Simple Network Management Protocol (SNMP) polling queries
Secure File Transfer Protocol (SFTP) administrative file transfers
Domain Name System (DNS) zone transfers between primary and secondary servers
Network Time Protocol (NTP) clock synchronization

Matches

Show answer & explanation

Answer

Simple Network Management Protocol (SNMP) polling queries matches UDP port 161; Secure File Transfer Protocol (SFTP) administrative file transfers matches TCP port 22; Domain Name System (DNS) zone transfers between primary and secondary servers matches TCP port 53; Network Time Protocol (NTP) clock synchronization matches UDP port 123.
Each service is correctly paired with its transport protocol and port: SNMP polling query traffic uses UDP port 161; SFTP leverages SSH encapsulation on TCP port 22; DNS zone transfers switch to TCP port 53 for reliability; and NTP time synchronization operates over UDP port 123.

Step-by-Step Solution

1
Identify the transport protocol and port requirements for management and polling services.
SNMP queries use UDP port 161 (SNMP traps use UDP 162).
Management polling relies on lightweight UDP communications.
2
Identify encrypted administrative file transfer protocols.
SFTP operates inside an SSH session on TCP port 22.
Unlike legacy FTP (TCP 20/21) or FTPS (TCP 989/990), SFTP shares SSH's underlying port.
3
Differentiate standard DNS queries from DNS zone transfers.
DNS zone transfers require reliable bulk data transfer over TCP port 53.
Standard client queries use UDP port 53, but large database transfers between DNS servers require TCP port 53.
4
Identify network synchronization protocol ports.
NTP utilizes UDP port 123.
Time synchronization relies on low-latency UDP packet delivery.

Key Concept

Common Network Ports and Transport Layer Dependencies
Question 148Question →

A network technician is reviewing Layer 2 switch forwarding modes and port auto-negotiation behaviors. Match each switching operation or port feature on the left with its corresponding functional description on the right.

Click a left item, then click its matching right item

Items

Store-and-Forward Switching
Cut-Through Switching
Fragment-Free Switching
Auto-MDIX

Matches

Show answer & explanation

Answer

Store-and-Forward Switching matches buffering the full frame and verifying the FCS CRC. Cut-Through Switching matches reading only the destination MAC address for minimal latency. Fragment-Free Switching matches inspecting the first 64 bytes to eliminate collision fragments. Auto-MDIX matches automatically detecting cable pinouts and swapping transmit/receive pairs.
Store-and-Forward buffers the full frame for FCS validation; Cut-Through forwards after inspecting only the 6-byte destination MAC; Fragment-Free inspects 64 bytes to drop collision fragments; Auto-MDIX automatically configures internal transmit/receive pinouts based on connected cabling.

Step-by-Step Solution

1
Identify full-frame validation switching behavior
Store-and-Forward buffers the entire frame to calculate the Frame Check Sequence CRC, ensuring invalid frames are dropped.
Full frame buffering provides maximum reliability at the expense of slight latency.
2
Identify ultra-low latency forwarding behavior
Cut-Through reads only the destination MAC address (6 bytes) before immediately forwarding.
Minimal preamble/address inspection yields the lowest possible forwarding delay.
3
Identify collision fragment mitigation behavior
Fragment-Free inspects the first 64 bytes, which is where collision fragments occur in half-duplex Ethernet.
This serves as a compromise between cut-through speed and basic collision filtering.
4
Identify physical interface pinout automation
Auto-MDIX electronically adjusts pin functions to allow straight-through or crossover cables interchangeably.
Automated physical layer detection eliminates wiring mismatched connections.

Key Concept

Switch Frame Forwarding Methods and Auto-MDIX
Estimated Time:1m 30s
Question 149Question →

A network router has learned multiple routes to the same destination network from different routing sources. Arrange the following route sources in order of administrative distance priority, from the most trustworthy source (lowest administrative distance) to the least trustworthy source (highest administrative distance).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from most trustworthy (lowest AD) to least trustworthy (highest AD) is: Directly connected interface (AD 0), Static route (AD 1), OSPF dynamic route (AD 110), and RIP dynamic route (AD 120).
Administrative distance (AD) rates the trustworthiness of routing information received from different routing sources on a scale where lower values are preferred. Directly connected routes are most trusted (AD 0), followed by static routes (AD 1), OSPF dynamic routes (AD 110), and RIP dynamic routes (AD 120).

Step-by-Step Solution

1
Identify the Administrative Distance (AD) value for directly connected networks.
Directly connected interfaces have an AD of 0.
Direct connections represent local active hardware links and are the most reliable.
2
Identify the AD value for manually configured static routes.
Static routes have a default AD of 1.
Explicit administrative configuration takes precedence over all dynamic protocols.
3
Compare default AD values of dynamic interior gateway protocols.
OSPF has an AD of 110, whereas RIP has an AD of 120.
Lower numerical AD values indicate higher trustworthiness when selecting routes from different protocols.

Key Concept

Administrative Distance Priority in Route Selection
Estimated Time:45s
Question 150Question →

A network administrator is designing an IP addressing scheme for a new corporate branch facility using the assigned block 10.45.128.0/1910.45.128.0/19. The engineering division requires a dedicated subnet capable of supporting at least 1,1001,100 usable host IP addresses. Which CIDR prefix represents the smallest possible subnet allocation that satisfies this host requirement while conserving IP space, and what is the last usable host IP address in that subnet?

Show answer & explanation

Answer: /21/21 prefix with a last usable host address of 10.45.135.25410.45.135.254

Answer

The /21/21 prefix with a last usable host address of 10.45.135.25410.45.135.254
The correct selection uses a /21/21 prefix, which grants 11 host bits (211−2=2,0462^{11} - 2 = 2,046 usable host addresses), satisfying the requirement for at least 1,100 hosts while conserving address space. With a 3rd octet block size of 8, the subnet range is 10.45.128.010.45.128.0 to 10.45.135.25510.45.135.255, making 10.45.135.25410.45.135.254 the final usable host IP address.

Step-by-Step Solution

1
Determine the minimum number of host bits required for 1,100 hosts.
Using the formula 2h−2≥11002^h - 2 \ge 1100, h=10h=10 yields 1,0221,022 usable hosts (insufficient), while h=11h=11 yields 2,0462,046 usable hosts. Therefore, 1111 host bits are needed.
Two addresses (network ID and broadcast address) must be subtracted from the total address count.
2
Calculate the required CIDR prefix length.
32−11=2132 - 11 = 21, so a /21/21 prefix is required.
An IPv4 address consists of 32 total bits.
3
Determine the block size and subnet range starting from 10.45.128.010.45.128.0.
With 1111 host bits (33 bits in the 3rd octet and 88 bits in the 4th octet), the 3rd octet block increment is 23=82^3 = 8. The subnet spans 10.45.128.010.45.128.0 to 10.45.135.25510.45.135.255.
The 3rd octet starts at 128 and increments by 8, giving a range of 128 through 135.
4
Identify the broadcast address and last usable host address.
Broadcast address is 10.45.135.25510.45.135.255, and the last usable host IP is 10.45.135.25410.45.135.254.
The last usable host IP is immediately prior to the subnet broadcast address.

Key Concept

Calculating required subnet mask size based on host capacity constraints and determining subnet host address bounds.
Question 151Question →

A network administrator is configuring a subnet for a small office branch using the subnet mask 255.255.255.240255.255.255.240. What is the maximum number of usable host IP addresses available on this subnet?

Show answer & explanation

Answer: 14

Answer

14 usable host IP addresses
A subnet mask of 255.255.255.240255.255.255.240 provides 4 host bits, resulting in 24=162^4 = 16 total IP addresses. Subtracting 2 reserved addresses (the network address and broadcast address) leaves exactly 14 usable host IP addresses.

Step-by-Step Solution

1
Determine the number of host bits from the subnet mask.
The dotted-decimal mask 255.255.255.240255.255.255.240 converted to binary has 28 network bits and 4 host bits (32−28=432 - 28 = 4).
The last octet value 240240 is 11110000211110000_2, leaving 4 trailing zeros for host addressing.
2
Calculate total IP addresses in the subnet.
24=162^4 = 16 total IP addresses.
The total address capacity of a subnet is calculated using 2h2^h, where hh is the number of host bits.
3
Calculate usable host IP addresses.
16−2=1416 - 2 = 14 usable host IP addresses.
Two IP addresses in every standard subnet are reserved: the network ID (all host bits 0) and the broadcast address (all host bits 1).

Key Concept

Calculating usable IPv4 host addresses from a subnet mask by subtracting network and broadcast addresses
Question 152Question →

A network administrator is auditing an enterprise multi-protocol routing environment. Match each dynamic routing protocol on the left with its corresponding default administrative distance, primary metric calculation method, and underlying algorithm on the right.

Click a left item, then click its matching right item

Items

Internal EIGRP
eBGP
OSPF
RIPv2

Matches

Show answer & explanation

Answer

Internal EIGRP matches Administrative Distance 90 with composite bandwidth/delay metrics; eBGP matches Administrative Distance 20 with path-vector AS-Path metrics; OSPF matches Administrative Distance 110 with cost metrics via Dijkstra; RIPv2 matches Administrative Distance 120 with hop count metrics via Bellman-Ford.
Each dynamic routing protocol correctly maps to its standardized default Administrative Distance (eBGP=20, Internal EIGRP=90, OSPF=110, RIPv2=120) and matching metric/algorithm characteristics.

Step-by-Step Solution

1
Determine the default Administrative Distance (AD) for each dynamic routing protocol.
eBGP has AD 20, Internal EIGRP has AD 90, OSPF has AD 110, and RIPv2 has AD 120.
Administrative Distance defines source trustworthiness when different protocols learn identical destination prefixes.
2
Associate each protocol with its underlying path computation algorithm and metric scheme.
Internal EIGRP uses DUAL with bandwidth/delay; eBGP uses Path Vector with AS-Path; OSPF uses Dijkstra SPF with interface cost; RIPv2 uses Bellman-Ford with hop count.
Each protocol type utilizes distinct algorithmic mechanisms to compute loop-free paths.

Key Concept

Routing protocol Administrative Distance hierarchies, underlying path calculation algorithms, and metric evaluation mechanisms.
Question 153Question →

Match each VLAN configuration term on the left with its corresponding operational behavior on the right.

Click a left item, then click its matching right item

Items

Native VLAN
Access Port
802.1Q Tagging
Trunk Port

Matches

Show answer & explanation

Answer

Native VLAN matches with carrying untagged frame traffic across an 802.1Q trunk link without modifying the Ethernet frame header. Access Port matches with connecting to an end-host device and carrying traffic for only one assigned VLAN using standard untagged frames. 802.1Q Tagging matches with inserting a 4-byte header into the Ethernet frame to identify the source VLAN across switch-to-switch links. Trunk Port matches with carrying traffic for multiple VLANs simultaneously over a single physical link between network switches.
Each concept aligns directly with its fundamental operation: Native VLAN handles untagged trunk traffic, Access Ports connect end-user devices on a single VLAN, 802.1Q Tagging inserts a 4-byte VLAN identifier into frames, and Trunk Ports carry traffic for multiple VLANs across a single physical link.

Step-by-Step Solution

1
Identify the purpose of Native VLAN on 802.1Q trunks.
Determine that Native VLAN handles untagged frame traffic across trunk connections.
By default in 802.1Q trunking, frames belonging to the designated native VLAN are transmitted without an 802.1Q tag.
2
Distinguish between Access Ports and Trunk Ports.
Match Access Port to single-VLAN host connectivity and Trunk Port to multi-VLAN inter-switch links.
Access ports service host devices on a single broadcast domain, while trunk ports multiplex multiple VLANs across network switches.
3
Identify the mechanism of 802.1Q frame encapsulation.
Match 802.1Q Tagging with inserting a 4-byte header into Ethernet frames.
The 802.1Q standard identifies frame VLAN membership by inserting a 4-byte field containing the VLAN ID into frame headers.

Key Concept

VLAN Configuration and 802.1Q Trunking Fundamentals
Question 154Question →

A network architect is assigned the IPv4 address block 172.24.160.0/19172.24.160.0/19 to provision subnets for a new regional data center. To optimize routing and broadcast domain sizes, the architect must partition the entire block into equal-sized subnets such that each subnet can accommodate a minimum of 120120 usable host IP addresses. Using the longest possible CIDR prefix mask that satisfies this host requirement, what is the total maximum number of usable host IP addresses available across all subnets created?

Show answer & explanation

Answer: 8064

Answer

8064 usable host IP addresses
To support at least 120 usable host IP addresses per subnet while utilizing the longest possible CIDR prefix, a /25 subnet mask (7 host bits) is required, providing 27−2=1262^7 - 2 = 126 usable host addresses per subnet. Partitioning the parent 172.24.160.0/19172.24.160.0/19 block into /25 subnets yields 225−19=26=642^{25-19} = 2^6 = 64 subnets. Multiplying 64 subnets by 126 usable host IP addresses per subnet results in 8064 total usable host IP addresses.

Step-by-Step Solution

1
Determine the host bit requirement (hh) for at least 120 usable hosts per subnet.
7 host bits are required because 27−2=126≥1202^7 - 2 = 126 \ge 120 usable host addresses.
6 host bits (26−2=622^6 - 2 = 62) are insufficient to meet the 120 host requirement.
2
Calculate the subnet prefix length (CIDR notation).
Prefix length is /25/25 (32−7=2532 - 7 = 25).
Subtracting 7 host bits from the 32 total IPv4 bits leaves a 25-bit network prefix.
3
Calculate total /25/25 subnets created from the parent /19/19 block.
64 subnets (225−19=26=642^{25 - 19} = 2^6 = 64).
The difference between the new prefix (/25) and parent prefix (/19) is 6 subnet bits.
4
Calculate total usable host IP addresses across all 64 subnets.
8064 usable host IP addresses (64×126=806464 \times 126 = 8064).
Each of the 64 subnets reserves 2 IP addresses (network ID and broadcast address) out of 128 total addresses.

Key Concept

Partitioning IPv4 CIDR blocks using VLSM/subnetting and calculating total aggregate usable host space.
Question 155Question →

A network administrator is setting up an 802.1Q trunk link connecting two switches, SW-Dist01 and SW-Access01, on interface GigabitEthernet1/0/24.

On SW-Dist01, the interface is configured as follows:

switchport mode trunk
switchport trunk native vlan 50
switchport trunk allowed vlan 10,20,50

On SW-Access01, the interface is configured as follows:

switchport mode trunk
switchport trunk native vlan 1
switchport trunk allowed vlan 10,20,50

Based on these configurations, which of the following operational outcomes will occur on this link? (Select TWO)

Select all that apply

Show answer & explanation

Answer: Untagged frames transmitted from VLAN 50 on SW-Dist01 will be received into VLAN 1 on SW-Access01, creating cross-VLAN traffic leakage.; Cisco Discovery Protocol (CDP) or log monitoring services will generate native VLAN mismatch warning messages on both switches.

Answer

Untagged frames transmitted from VLAN 50 on SW-Dist01 will be received into VLAN 1 on SW-Access01 (causing cross-VLAN traffic leakage), and system management protocols (like CDP) will generate native VLAN mismatch warnings on both switches.
On an 802.1Q trunk, traffic matching the local native VLAN is forwarded untagged. When SW-Dist01 sends untagged VLAN 50 frames, SW-Access01 receives them without a tag and classifies them into its own native VLAN (VLAN 1), causing cross-VLAN traffic leakage. Furthermore, protocols such as CDP exchange VLAN configuration metadata and trigger explicit native VLAN mismatch logs when opposing ends do not match.

Step-by-Step Solution

1
Analyze frame egress behavior on SW-Dist01 for native VLAN 50
SW-Dist01 sends traffic originating from VLAN 50 across the 802.1Q trunk link untagged because native VLAN traffic is sent without an 802.1Q header tag by default.
802.1Q trunking rules dictate that traffic matching the port's configured native VLAN is sent unencapsulated.
2
Analyze frame ingress behavior on SW-Access01 for untagged frames
SW-Access01 receives the untagged frames and places them directly into its own native VLAN, which is VLAN 1.
An 802.1Q switch port assigns all incoming untagged frames to whichever VLAN ID is configured as its native VLAN.
3
Evaluate protocol monitoring and log detection mechanisms
Neighbor discovery protocols like CDP detect the discrepancy (VLAN 50 vs. VLAN 1) and periodically generate native VLAN mismatch alerts in the console/syslog logs.
CDP advertises local native VLAN configurations in management frames to detect misconfigurations.

Key Concept

802.1Q Native VLAN Operation and Mismatch Behavior
Question 156Question →

A network engineer is designing an infrastructure layout for a smart factory floor containing eight automated assembly stations. To guarantee that failure of a single cable run does not isolate critical controllers while maintaining reasonable cabling costs, the engineer connects all stations to a central distribution switch and adds direct redundant interconnects only between the three highest-priority controllers. Which of the following network topologies is being implemented?

Show answer & explanation

Answer: Hybrid (Star-Partial Mesh) topology

Answer

The network layout represents a Hybrid (Star-Partial Mesh) topology because it combines a central switch layout (star) with selective point-to-point redundant connections (partial mesh) between critical controllers.
The design combines two distinct topological concepts: a central switch to which all endpoints connect (star) and selective direct links between high-priority nodes (partial mesh). Combining these two structures results in a hybrid star-partial mesh topology, balancing fault tolerance for key nodes with overall deployment cost.

Step-by-Step Solution

1
Analyze the primary physical connectivity layout.
Connecting all eight stations to a central distribution switch forms a physical star topology.
Centralized switch connections characterize star network architectures.
2
Analyze the redundant interconnections.
Adding direct point-to-point links specifically between selected high-priority controllers creates a partial mesh overlay.
Selective redundancy without linking every node pair is the defining trait of partial mesh.
3
Combine the structural classifications to determine the overall topology.
The combination of star and partial mesh design elements yields a hybrid topology.
Networks incorporating multiple distinct topological structures are classified as hybrid topologies.

Key Concept

Hybrid Network Topologies
Question 157Question →

A network administrator is establishing central IP addressing services for remote branch offices across a routed WAN. Workstations residing on remote subnets must obtain IPv4 leases automatically and have their hostname-to-IP address mappings updated dynamically in corporate DNS. Which TWO of the following configurations must be implemented to fulfill these requirements?

Select all that apply

Show answer & explanation

Answer: Configure a DHCP relay agent (IP helper) on the branch router interface pointing to the central DHCP server IP address.; Enable Dynamic DNS (DDNS) integration on the DHCP server so it sends automated A and PTR record updates to the authoritative DNS server upon issuing leases.

Answer

The required configurations are placing a DHCP relay agent on the remote router interface pointing to the DHCP server, and enabling Dynamic DNS (DDNS) integration on the DHCP server to automatically update A and PTR records.
Implementing a DHCP relay agent (IP helper address) on the remote router interface allows DHCP broadcast requests to reach the centralized DHCP server across Layer 3 boundaries. Enabling Dynamic DNS (DDNS) on the DHCP server allows it to automatically register forward (A) and reverse (PTR) DNS records on behalf of client hosts when leases are issued.

Step-by-Step Solution

1
Identify the requirement for cross-subnet DHCP leasing.
Recognize that DHCP broadcast requests from remote subnets cannot cross routers without a DHCP relay agent (IP helper) converting them to unicast traffic.
Routers drop Layer 2 broadcast packets by default.
2
Identify the requirement for automatic hostname and IP registration in DNS.
Select Dynamic DNS (DDNS) updates configured on the DHCP server to update A and PTR records automatically.
DDNS automates DNS record updates whenever leases are assigned or renewed by DHCP.

Key Concept

DHCP Relay and Dynamic DNS (DDNS) Integration
Estimated Time:1m 30s
Question 158Question →

A network administrator is configuring traffic control policies across multiple subnets in an enterprise network. Which of the following functions and operations occur specifically at Layer 3 (Network Layer) of the OSI model? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: Encapsulating segments into packets containing logical source and destination IP addresses; Determining optimal path selection using dynamic routing protocols and routing tables

Answer

The correct functions operating at Layer 3 (Network Layer) are encapsulating upper-layer data into IP packets with logical addressing and determining optimal path selection using routing tables.
Layer 3 (Network Layer) is responsible for logical addressing (IP addresses), packet encapsulation, and path selection (routing) between different networks. Therefore, encapsulating segments into packets with IP addresses and determining path selection via routing tables are correct Layer 3 functions.

Step-by-Step Solution

1
Identify the primary responsibilities of Layer 3 (Network Layer) in the OSI reference model.
Layer 3 handles logical IP addressing, packet creation, and path selection (routing) across subnets.
Layer 3 enables end-to-end host communication across disparate network segments.
2
Evaluate each provided option against Layer 3 operational boundaries.
Packet encapsulation with IP addresses and routing path determination belong to Layer 3. FCS error checking belongs to Layer 2, while port-based multiplexing belongs to Layer 4.
Distinguishing protocol data units (PDUs) and header responsibilities across OSI layers ensures accurate protocol stack analysis.

Key Concept

OSI Model Layer 3 (Network Layer) Functions
Question 159Question →

A network technician is configuring an IP addressing plan for an IoT security segment that requires static IP assignments for 6262 security cameras and 22 default gateway interfaces. Which CIDR prefix length represents the most space-efficient IPv4 subnet mask that satisfies this requirement?

Show answer & explanation

Answer: /25

Answer

/25
The total number of IP addresses required is 62+2=6462 + 2 = 64. Using the usable host formula 2n−22^n - 2, a subnet with n=6n = 6 host bits provides 26−2=622^6 - 2 = 62 usable addresses, which is not enough. Increasing to n=7n = 7 host bits yields 27−2=1262^7 - 2 = 126 usable host addresses. The resulting prefix length is 32−7=2532 - 7 = 25, making the /25 prefix length the most space-efficient valid choice.

Step-by-Step Solution

1
Calculate total usable host addresses required
62 cameras + 2 gateway interfaces = 64 usable host addresses
Both endpoint devices and gateway interfaces require individual usable IP addresses within the subnet.
2
Apply the usable host formula (2n−2≥642^n - 2 \ge 64)
For n=6n = 6 host bits, 26−2=622^6 - 2 = 62 usable hosts (insufficient). For n=7n = 7 host bits, 27−2=1262^7 - 2 = 126 usable hosts (sufficient).
Two IP addresses in every IPv4 subnet are reserved for the network ID and the broadcast address, so 2 must be subtracted from total addresses.
3
Determine the CIDR prefix length
32−7=2532 - 7 = 25, corresponding to /25 prefix length
Subtracting the required host bits from 32 total IPv4 bits gives the prefix length.

Key Concept

IPv4 Subnetting and Usable Host Calculation
Question 160Question →

A network administrator needs to implement an interior gateway protocol (IGP) that uses a link-state routing algorithm to maintain a complete map of the network topology. Which of the following protocols should the administrator deploy?

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Answer: Open Shortest Path First (OSPF)

Answer

Open Shortest Path First (OSPF)
Open Shortest Path First (OSPF) is an interior gateway link-state routing protocol. Each router running OSPF maintains a Link-State Database (LSDB) containing a complete topology map of the network area to compute the shortest path to every destination subnet.

Step-by-Step Solution

1
Identify the protocol class specified in the requirement
The scenario requires an interior gateway protocol operating on a link-state algorithm to maintain a complete topology map.
Link-state protocols construct a complete view of the network topology to compute shortest paths.
2
Evaluate the candidate options against link-state protocol characteristics
OSPF uses Dijkstra's Shortest Path First algorithm and Link-State Advertisements to map the entire network area.
OSPF is the standard link-state dynamic routing protocol among the choices provided.

Key Concept

Link-State Routing Protocols vs. Distance-Vector Protocols
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