Examkin

All practice questions

2237 questions

All types2,237Multiple choice1,150Multiple select488Numeric30Matching350Ordering201Short answer18
Question 361Question →

A core router learns two separate routes to the destination network 192.168.50.0/24192.168.50.0/24 from two different dynamic routing protocols. The first route is learned via Internal EIGRP with an Administrative Distance (AD) of 9090 and a metric cost of 2,176,0002,176,000. The second route is learned via OSPF with an Administrative Distance (AD) of 110110 and a cost metric of 2020. Assuming default settings, which route will the router install into its Routing Information Base (RIB) for traffic destined to 192.168.50.0/24192.168.50.0/24?

Show answer & explanation

Answer: The EIGRP route, because lower Administrative Distance takes precedence over protocol metrics when comparing routes from different routing sources.

Answer

The EIGRP route is installed because lower Administrative Distance takes precedence over protocol metrics when evaluating routes from different sources.
When a router receives multiple routes for the exact same network prefix from different routing protocols, it uses Administrative Distance (AD) to measure administrative trustworthiness. Internal EIGRP defaults to an AD of 90, whereas OSPF defaults to an AD of 110. Because 90 is lower than 110, the router selects the EIGRP path to populate the routing table. Metric comparisons are only performed between paths learned from the exact same dynamic routing protocol.

Step-by-Step Solution

1
Evaluate destination prefix matching.
Both routes target the exact same prefix length (192.168.50.0/24192.168.50.0/24).
Longest prefix match rule applies first, but since the subnet masks are identical, the router proceeds to compare routing sources.
2
Compare Administrative Distance (AD) values between EIGRP and OSPF.
Internal EIGRP has an AD of 9090, while OSPF has an AD of 110110.
Administrative Distance indicates the trustworthiness of a routing source. Lower values indicate higher believability.
3
Select the route for RIB installation.
The route with AD 9090 (EIGRP) is selected and installed into the routing table.
Because EIGRP has a lower AD than OSPF, the OSPF route is discarded from RIB consideration, making protocol metrics irrelevant during inter-protocol path selection.

Key Concept

Administrative Distance vs. Routing Metric in Path Selection
Question 362Question →

A network technician needs to separate broadcast traffic between two departments connected to the same physical switch. Which technology should the technician implement on the switch to isolate the broadcast domains?

Show answer & explanation

Answer: Virtual Local Area Networks (VLANs)

Answer

Virtual Local Area Networks (VLANs) logically segment a physical Layer 2 switch into distinct broadcast domains.
Virtual Local Area Networks (VLANs) allow network administrators to partition a single physical switch into multiple logical broadcast domains, keeping broadcast traffic isolated within each designated group of ports.

Step-by-Step Solution

1
Analyze the network requirement
The requirement calls for isolating broadcast traffic between different host groups connected to the same physical switch.
By default, all ports on an unconfigured Layer 2 switch belong to a single shared broadcast domain.
2
Identify the appropriate Layer 2 feature
Configuring Virtual Local Area Networks (VLANs) groups switch ports into logical sub-networks.
VLANs restrict broadcast frames to only those switch ports assigned to the same VLAN ID.

Key Concept

VLAN Broadcast Domain Isolation
Estimated Time:45s
Question 363Question →

A network engineer is allocated the IPv4 address block 10.240.16.0/2010.240.16.0/20 to create subnets for distinct department segments. Each segment requires a minimum of 120120 usable host IP addresses. What is the maximum number of subnets of this minimum required size that can be created from the given block?

Show answer & explanation

Answer: 32

Answer

The maximum number of subnets of the minimum required size that can be created is 32.
To support a minimum of 120 hosts, 7 host bits are required (27−2=126≥1202^7 - 2 = 126 \ge 120). A subnet with 7 host bits has a prefix length of /25/25 (32−7=2532 - 7 = 25). Partitioning a /20/20 block into /25/25 subnets borrows 5 bits (25−20=525 - 20 = 5), resulting in 25=322^5 = 32 subnets.

Step-by-Step Solution

1
Calculate the host bits required for the minimum host capacity requirement
7 host bits are required because 27−2=126≥1202^7 - 2 = 126 \ge 120.
Using 6 host bits only provides 26−2=622^6 - 2 = 62 usable addresses, which fails to meet the 120 host requirement.
2
Calculate the required subnet mask prefix length
The prefix length is /25/25, calculated as 32−7=2532 - 7 = 25.
Subtracting host bits from the 32 total IPv4 address bits determines the network prefix.
3
Calculate the total subnets formed by splitting a /20/20 block into /25/25 subnets
225−20=25=322^{25 - 20} = 2^5 = 32 subnets.
The number of subnets created is 2 raised to the power of the borrowed subnet bits.

Key Concept

IPv4 Subnetting and Host Capacity Calculation
Question 364Question →

A network engineer is configuring a core router that receives routes to the destination network 10.100.20.0/2410.100.20.0/24 from two distinct dynamic routing sources simultaneously:

- An OSPF advertisement with a calculated path metric of 4545
- An Internal EIGRP (iEIGRP) advertisement with a calculated path metric of 2,560,0002,560,000

Which path will the router install into its active IPv4 routing table for the 10.100.20.0/2410.100.20.0/24 network, and what is the primary reason for this selection?

Show answer & explanation

Answer: The Internal EIGRP path, because Internal EIGRP has a lower administrative distance (9090) than OSPF (110110).

Answer

The router installs the Internal EIGRP path because Internal EIGRP has a lower Administrative Distance (90) than OSPF (110). Metric values are protocol-specific and only evaluated after administrative distance comparison.
When a router receives multiple routes for the exact same network prefix from different routing protocols, it compares their Administrative Distance (AD). Internal EIGRP has a lower default AD (90) than OSPF (110). Because lower AD values are preferred, the router selects and installs the Internal EIGRP route into the routing table. Metric values are never directly compared between different dynamic routing protocols.

Step-by-Step Solution

1
Identify the destination prefix and routing sources
Both routing protocols offer a route to the exact same prefix (10.100.20.0/2410.100.20.0/24).
When prefix lengths are equal, the router compares Administrative Distance (AD) to select the trustworthy routing source.
2
Compare Administrative Distance (AD) values of the competing protocols
Internal EIGRP has a default AD of 9090, while OSPF has a default AD of 110110.
Lower AD values indicate higher trustworthiness. Since 90<11090 < 110, Internal EIGRP is selected.
3
Determine why metric comparison is bypassed
Metric values (4545 vs 2,560,0002,560,000) are ignored because metrics cannot be compared across different routing protocols.
Different dynamic routing protocols use incompatible metric formulas (e.g., cost based on bandwidth vs. composite metric based on bandwidth and delay).

Key Concept

Administrative Distance vs. Routing Metric in Path Selection
Question 365Question →

Match each network scope classification on the left with its correct geographical footprint and primary application on the right.

Click a left item, then click its matching right item

Items

Personal Area Network (PAN)
Storage Area Network (SAN)
Metropolitan Area Network (MAN)
Campus Area Network (CAN)

Matches

Show answer & explanation

Answer

Personal Area Network (PAN) matches short-range personal device connections around a user; Storage Area Network (SAN) matches dedicated high-speed block storage access for data center servers; Metropolitan Area Network (MAN) matches city-wide municipal coverage; Campus Area Network (CAN) matches interconnecting LANs across adjacent buildings on a shared site.
Each network type matches its unique operational scale: PANs handle personal device range (<10m), SANs handle dedicated block storage for servers, MANs cover municipal-scale cities, and CANs connect multi-building private properties like universities.

Step-by-Step Solution

1
Determine the scale and purpose of Personal Area Network (PAN).
PAN covers small personal spaces (up to 10 meters) for peripheral pairing.
PAN technology (such as Bluetooth) is designed for short-range personal device communication.
2
Determine the scale and purpose of Storage Area Network (SAN).
SAN is a specialized data center network providing block-level access to disk arrays.
SAN offloads storage traffic from the main user LAN onto a high-performance network fabric.
3
Determine the scale and purpose of Metropolitan Area Network (MAN).
MAN spans a city-sized geographical region.
MAN connects disparate facilities situated across a municipality.
4
Determine the scale and purpose of Campus Area Network (CAN).
CAN links multiple local area networks across contiguous buildings on a shared campus.
CAN is larger than a standard LAN but smaller than a city-wide MAN, typically owned and managed by a single institution.

Key Concept

Classification of network types based on geographic coverage and architectural function.
Estimated Time:1m 0s
Question 366Question →

A network technician is using a packet capture utility to inspect outbound host traffic during an HTTPS session initiation. Arrange the following data encapsulation steps in the correct chronological sequence from the initial application request generation down to media transmission.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order follows the top-down OSI encapsulation model: Application data formatting (Upper layers), TCP segment creation with port numbers (Transport layer), IP packet creation with logical addresses (Network layer), Ethernet frame creation with MAC addresses (Data Link layer), and bit signal conversion (Physical layer).
During outbound network transmission, data is processed sequentially from Layer 7 (Application) down to Layer 1 (Physical). Application data is first generated, then encapsulated into a Transport layer segment (TCP/UDP ports), further encapsulated into a Network layer packet (IP addresses), wrapped into a Data Link layer frame (MAC addresses and FCS trailer), and finally converted into physical signals (bits) at the Physical layer.

Step-by-Step Solution

1
Identify the starting point of outbound traffic flow
Outbound transmission begins at the highest layers with data payload creation.
The client application creates the original request message before network headers are appended.
2
Determine the Layer 4 Transport encapsulation
TCP port header is added to form a segment.
Transport protocols (Layer 4) manage host-to-host session multiplexing using port numbers.
3
Determine the Layer 3 Network encapsulation
IP header is added to form a packet.
Network layer (Layer 3) handles logical host identification and path determination across network boundaries.
4
Determine the Layer 2 Data Link encapsulation
Ethernet MAC header and FCS trailer are added to form a frame.
Data Link layer (Layer 2) prepares data for physical link traversal by assigning physical hardware addresses.
5
Determine the Layer 1 Physical transmission step
Digital data is serialized and converted to signal pulses (bits).
Physical layer (Layer 1) deals directly with signal encoding across physical media.

Key Concept

OSI Model Encapsulation Sequence
Estimated Time:1m 30s
Question 367Question →

An enterprise network administrator is troubleshooting directory service integration between an internal web proxy and an Active Directory Domain Controller. Authentication using implicit SSL encryption (LDAPS) over TCP port 636 functions correctly. However, automated directory synchronization tasks configured on the proxy to use explicit TLS encryption (STARTTLS) fail to establish a secure connection. Network packet captures confirm that the proxy initiates communication over the default unencrypted LDAP port before issuing the STARTTLS upgrade command, but traffic is blocked by an intermediate firewall ACL. Which of the following transport protocols and destination port combinations must be permitted on the firewall to allow STARTTLS directory synchronization to succeed?

Show answer & explanation

Answer: TCP port 389

Answer

TCP port 389
Explicit TLS (STARTTLS) allows a client to establish an initial connection over the standard unencrypted port (TCP port 389 for LDAP) and issue a command to upgrade the active channel to TLS encryption. For STARTTLS negotiation to succeed, the firewall ACL must permit TCP port 389.

Step-by-Step Solution

1
Analyze the protocol handshake mechanism for explicit TLS (STARTTLS).
Recognize that STARTTLS is an explicit security extension that begins session negotiation over the default unencrypted service port before initiating a TLS upgrade.
Explicit TLS requires initial access to the unencrypted service port to send the protocol upgrade request.
2
Identify the standard port number associated with unencrypted directory services.
Determine that standard LDAP operates on port 389.
Lightweight Directory Access Protocol (LDAP) uses port 389 as its standard listener port.
3
Evaluate transport layer requirements for TLS session negotiation.
Select TCP over UDP for session state establishment.
TLS state handshakes and data integrity enforcement require the connection-oriented reliability offered by TCP.

Key Concept

Explicit TLS (STARTTLS) vs Implicit TLS Port Behavior
Question 368Question →

A network engineer is configuring a static route on a core router using the uncompressed IPv6 next-hop address 2001:0db8:0000:0000:0008:0000:0000:0001. According to RFC 5952 IPv6 addressing standards, which of the following represents the correct compressed form of this address?

Show answer & explanation

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

Answer

The address 2001:db8::8:0:0:1 is the correct compressed form.
The correct representation, 2001:db8::8:0:0:1, suppresses leading zeros in each hexadecimal word and compresses the first sequence of contiguous zero blocks using ::. Per RFC 5952, when two zero sequences have the same length, the first sequence must be compressed.

Step-by-Step Solution

1
Suppress leading zeros in each 16-bit hexadecimal field.
2001:0db8 becomes 2001:db8, 0008 becomes 8, and 0001 becomes 1.
Leading zeros within any 4-digit hexadecimal group are optional and omitted for brevity.
2
Identify contiguous blocks of all-zero fields.
There are two separate sequences of two consecutive zero blocks: blocks 3-4 and blocks 6-7.
IPv6 allows replacing contiguous zero blocks with a double-colon (::).
3
Apply the double-colon (::) according to RFC 5952 rules.
The first sequence of two zero blocks is replaced by ::, yielding 2001:db8::8:0:0:1.
The double-colon can only be used once per address. When zero sequences are of equal length, standard conventions require compressing the first occurrence.

Key Concept

IPv6 Address Compression Rules (RFC 5952)
Estimated Time:1m 0s
Question 369Question →

A network security analyst is monitoring how an inbound HTTPS session is processed by a host system during decapsulation. Arrange the following data processing actions in the correct sequential order as the received data moves up the OSI reference model, starting from Layer 1 (Physical) to Layer 7 (Application).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence follows the bottom-up decapsulation process of the OSI model: Physical (Layer 1) → Data Link (Layer 2) → Network (Layer 3) → Transport (Layer 4) → Presentation (Layer 6) → Application (Layer 7).
Decapsulation proceeds from Layer 1 up to Layer 7. Incoming physical signals are first converted to bits (Physical / L1), framed and verified using MAC/FCS (Data Link / L2), processed for IP routing and TTL (Network / L3), reassembled by TCP sequence and port numbers (Transport / L4), decrypted from TLS wrappers (Presentation / L6), and finally processed as an HTTP payload by the application (Application / L7).

Step-by-Step Solution

1
Identify Layer 1 operation
Converting physical signals (electrical pulses) into binary bits occurs at the Physical layer (Layer 1).
Layer 1 is responsible for the physical interface, encoding, and raw bit transmission.
2
Identify Layer 2 operation
Checking the Frame Check Sequence (FCS) and verifying the MAC address occurs at the Data Link layer (Layer 2).
Layer 2 encapsulates network packets into frames containing hardware MAC addresses and CRC error checking.
3
Identify Layer 3 operation
Inspecting IP headers, checking TTL, and handling logical address routing occurs at the Network layer (Layer 3).
Layer 3 operates on packets and processes IP addresses and routing fields.
4
Identify Layer 4 operation
TCP segment reassembly using sequence numbers and reading port numbers occurs at the Transport layer (Layer 4).
Layer 4 provides flow control, connection management, port multiplexing, and reliable segment reassembly.
5
Identify Layer 6 operation
Decrypting TLS/SSL wrappers and formatting data occurs at the Presentation layer (Layer 6).
Layer 6 manages syntax formatting, encryption/decryption, and data compression.
6
Identify Layer 7 operation
Processing the HTTP GET request by the web application service occurs at the Application layer (Layer 7).
Layer 7 directly interfaces with end-user software applications and protocol payloads.

Key Concept

OSI Model Layer Decapsulation Order and Layer Functions
Question 370Question →

Match each IPv6 base header field on the left with its correct functional description on the right.

Click a left item, then click its matching right item

Items

Traffic Class
Flow Label
Next Header
Hop Limit

Matches

Show answer & explanation

Answer

Traffic Class matches with identifying packet priority and QoS markings; Flow Label matches with maintaining the same path for a specific sequence of packets; Next Header matches with specifying upper-layer protocol or extension header; Hop Limit matches with decrementing at each hop to prevent loops.
Each fundamental IPv6 header field maps to a specific purpose: Traffic Class provides QoS markings; Flow Label maintains packet flow path consistency; Next Header identifies encapsulated protocols or extension headers; Hop Limit prevents network routing loops by decrementing per hop.

Step-by-Step Solution

1
Identify the primary role of the Traffic Class header field.
Determine that Traffic Class manages packet priority and QoS, mirroring IPv4 Differentiated Services.
Traffic Class handles packet prioritization in IPv6 header structures.
2
Identify the primary role of the Flow Label header field.
Determine that Flow Label tags specific packet streams for consistent router path delivery.
Flow Label facilitates real-time data flow handling across intermediate routers.
3
Identify the primary role of the Next Header field.
Determine that Next Header specifies payload protocols or links to optional extension headers.
Next Header replaces IPv4's Protocol field and enables modular extension header chaining.
4
Identify the primary role of the Hop Limit field.
Determine that Hop Limit decrements per layer 3 hop to mitigate routing loops.
Hop Limit replaces IPv4 TTL while serving the exact same loop-prevention mechanism.

Key Concept

IPv6 Base Header Fields and Functions
Question 371Question →

An enterprise architecture team is redesigning the physical WAN connections between 55 regional facilities. Currently, all 55 sites are interconnected using a fully meshed topology of dedicated point-to-point links to guarantee fault tolerance. To reduce recurring leased-line costs, management requests transitioning to a partial-mesh design that decommissions as many links as possible while guaranteeing that every site still retains at least two physically independent transmission paths to reach any other site. What is the maximum number of point-to-point links that can be decommissioned from the original network design?

Show answer & explanation

Answer: 5 links

Answer

5 links can be decommissioned from the original full-mesh deployment.
For a network of 55 sites (n=5n=5), a full-mesh topology requires n(n−1)2=5×42=10\frac{n(n-1)}{2} = \frac{5 \times 4}{2} = 10 links. To guarantee that every site maintains at least two independent pathways to every other site, the minimum viable partial-mesh layout is a ring topology requiring 55 links. Subtracting the 55 required links from the original 1010 links allows maximum decommissioning of 55 links.

Step-by-Step Solution

1
Calculate total point-to-point links in the existing full-mesh topology
Using the formula L=n(n−1)2L = \frac{n(n-1)}{2} where n=5n = 5: L=5×42=10L = \frac{5 \times 4}{2} = 10 physical links.
Full mesh requires a direct point-to-point cable run between every unique pair of nodes.
2
Determine the minimum number of links needed to satisfy the dual-path redundancy constraint
A minimum of 55 links configured in a closed ring (partial mesh) topology.
In a ring topology with n=5n = 5 nodes, every node has a degree of 2 (connected to two neighbors), providing two distinct physical paths (clockwise and counter-clockwise) to reach any destination node.
3
Subtract the required partial-mesh link count from the original full-mesh link count
10 links (full mesh)−5 links (ring partial mesh)=5 links decommissioned10 \text{ links (full mesh)} - 5 \text{ links (ring partial mesh)} = 5 \text{ links decommissioned}.
This yields the maximum number of links that can be removed while preserving fault tolerance constraints.

Key Concept

Full Mesh vs Partial Mesh Link Calculations
Question 372Question →

A network administrator is manually configuring a static global unicast IPv6 address on a core server interface. The administrator enters the address string `2001:db8:100:0::45::1/64` into the operating system interface configuration file. Upon applying the configuration, the system rejects the input with a syntax error. Which of the following technical explanations identifies why this IPv6 address representation is invalid?

Show answer & explanation

Answer: The double-colon shorthand notation (`::`) is included twice in a single address string, which prevents the parser from unambiguously calculating the total number of 16-bit zero blocks represented by each occurrence.

Answer

The double-colon shorthand notation (`::`) is included twice in a single address string, which prevents the parser from unambiguously calculating the total number of 16-bit zero blocks represented by each occurrence.
The double-colon (`::`) zero compression rule allows shortening contiguous 16-bit fields of zeros. However, using `::` more than once in a single address string makes it mathematically impossible to determine how many 16-bit zero blocks each `::` replaces, leading to syntax rejection by network operating systems.

Step-by-Step Solution

1
Analyze standard IPv6 address structure rules.
An IPv6 address comprises eight 16-bit hexadecimal fields (128 bits total), separated by colons.
To shorten long addresses, contiguous blocks of zero-value fields can be compressed using a double-colon (`::`).
2
Evaluate the string `2001:db8:100:0::45::1` for shorthand compliance.
The string contains two separate instances of `::`.
Multiple `::` occurrences lead to mathematical ambiguity regarding how many 16-bit zero blocks are assigned to the first versus second double-colon.
3
Determine the required syntax resolution.
The double-colon must be used only once (e.g., `2001:db8:100:0:0:45::1` or `2001:db8:100:0::45:0:1`).
This guarantees that an automated parser can reliably pad the missing zero fields to reconstruct the exact 128-bit binary address.

Key Concept

IPv6 Address Zero Compression Rules
Question 373Question →

A network administrator is configuring stateless inbound and outbound access control lists (ACLs) on an enterprise edge firewall connecting a remote branch to the corporate datacenter. Corporate security policy mandates that all administrative remote access and directory authentication services must be fully encrypted in transit, strictly prohibiting cleartext credentials or unencrypted management protocols. During post-deployment testing, packet captures show that administrative CLI management sessions and secure directory queries are being dropped by the firewall. Which of the following transport layer protocol and destination port combinations MUST be explicitly allowed in the firewall rules to restore these required secure services? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: TCP port 22; TCP port 636

Answer

The firewall must permit TCP port 22 for secure SSH terminal access and TCP port 636 for encrypted LDAPS directory service authentication.
The correct answers represent the standard transport protocol and port assignments for secure administrative access and encrypted directory services. Secure Shell (SSH) uses TCP port 22 to establish encrypted terminal sessions. Lightweight Directory Access Protocol Secure (LDAPS) uses TCP port 636 to encapsulate directory service communication within an SSL/TLS tunnel, fulfilling the corporate requirement for encrypted transit.

Step-by-Step Solution

1
Identify the protocol requirements from the scenario constraints.
The scenario requires encrypted remote CLI management and encrypted directory service authentication.
Security policy explicitly forbids cleartext protocols such as Telnet or standard LDAP.
2
Determine the transport protocol and default port number for secure remote CLI management.
SSH (Secure Shell) provides encrypted CLI administration over TCP port 22.
SSH requires connection-oriented TCP to maintain reliable session control and stream encryption.
3
Determine the transport protocol and default port number for encrypted directory authentication.
LDAPS (LDAP over TLS/SSL) operates over TCP port 636.
Standard LDAP over TCP port 389 sends credentials in cleartext; TLS wrapper on port 636 encrypts the entire LDAP session payload.

Key Concept

Identification of default port numbers and transport layer choices for secure management and directory protocols (SSH on TCP 22 vs LDAPS on TCP 636).
Question 374Question →

An network administrator is deploying Virtual Local Area Networks (VLANs) on a newly installed Layer 2 enterprise switch. Which of the following statements accurately describe key operational characteristics of VLANs? (Select TWO).

Select all that apply

Show answer & explanation

Answer: VLANs divide a single physical switch into multiple separate broadcast domains.; Hosts assigned to different VLANs on the same physical switch require a Layer 3 routing function to communicate.

Answer

VLANs divide a single physical switch into multiple separate broadcast domains, and hosts assigned to different VLANs on the same switch require a Layer 3 routing function to communicate.
VLANs logically partition a physical Layer 2 switch into independent broadcast domains. Because each VLAN is isolated at Layer 2, any communication between hosts located on different VLANs requires inter-VLAN routing performed by a Layer 3 device.

Step-by-Step Solution

1
Analyze Layer 2 segmentation properties of VLANs
Creating VLANs breaks a physical switch into distinct logical segments, where each segment acts as its own independent broadcast domain.
Broadcast traffic generated on one VLAN is restricted solely to ports assigned to that specific VLAN.
2
Determine requirements for inter-VLAN communication
Traffic destined for a host in another VLAN must pass through a Layer 3 device (such as a router or Layer 3 switch).
Layer 2 switches do not forward frames between different broadcast domains without IP routing capabilities.

Key Concept

VLAN Broadcast Isolation and Inter-VLAN Routing Requirements
Estimated Time:45s
Question 375Question →

A network administrator is mapping network application requirements to transport layer protocol features. Match each operational requirement or characteristic on the left with its corresponding protocol implementation detail on the right.

Click a left item, then click its matching right item

Items

Three-way handshake establishment
Low-latency transmission without delivery guarantees
Dynamic flow control via sliding windowing
Stateless subnet-wide broadcast support

Matches

Show answer & explanation

Answer

Three-way handshake establishment matches establishing a session using SYN, SYN-ACK, and ACK flags. Low-latency transmission matches using connectionless UDP datagrams with minimal header overhead. Dynamic flow control matches adjusting transmission rates via sliding windowing. Stateless broadcast support matches leveraging connectionless transport to distribute datagrams without tracking.
TCP is connection-oriented and utilizes sequence numbers, three-way handshakes (SYN, SYN-ACK, ACK), and sliding windows for reliability and flow control. UDP is connectionless, prioritizing low latency and low overhead while supporting broadcast modes.

Step-by-Step Solution

1
Analyze connection-oriented protocol characteristics.
Identified TCP mechanisms: SYN/SYN-ACK/ACK handshake for session initialization and sliding windowing for flow control.
TCP maintains session state and manages sender rate dynamically to guarantee reliable, ordered data delivery.
2
Analyze connectionless protocol characteristics.
Identified UDP mechanisms: lightweight 8-byte headers for time-sensitive traffic and broadcast support without connection establishment.
UDP lacks handshakes, state tracking, and flow control mechanisms, making it ideal for low-overhead or one-to-many communication.
3
Map left-side requirements to right-side implementations.
Paired left_1 to right_1, left_2 to right_2, left_3 to right_3, and left_4 to right_4.
Each feature corresponds precisely to the unique operational mechanisms of TCP or UDP.

Key Concept

Connection-Oriented (TCP) vs Connectionless (UDP) Protocol Operational Traits
Question 376Question →

An administrator is auditing a layer 2 enterprise switch topology consisting of four switches configured with IEEE 802.1D Spanning Tree Protocol (STP). The bridge priorities and base MAC addresses for the switches are as follows:

* Switch_Alpha: Priority 32768, MAC 00:0A:B8:11:22:33
* Switch_Beta: Priority 28672, MAC 00:0A:B8:AA:BB:CC
* Switch_Gamma: Priority 28672, MAC 00:0A:B8:77:88:99
* Switch_Delta: Priority 32768, MAC 00:0A:B8:00:11:22

Which switch will be elected as the STP root bridge for this broadcast domain?

Show answer & explanation

Answer: Switch_Gamma, because it has the lowest bridge priority value of 28672 tied with Switch_Beta, and a lower MAC address than Switch_Beta.

Answer

Switch_Gamma will be elected as the root bridge because it shares the lowest bridge priority (28672) and has a lower MAC address than Switch_Beta.
In 802.1D Spanning Tree Protocol, the root bridge is chosen based on the lowest Bridge ID, which consists of a 2-byte Priority and a 6-byte MAC address. Switches with priority 28672 beat switches with priority 32768. Between Switch_Beta and Switch_Gamma (both set to 28672), Switch_Gamma's MAC address (00:0A:B8:77:88:99) is lower than Switch_Beta's MAC address (00:0A:B8:AA:BB:CC), making Switch_Gamma the elected root bridge.

Step-by-Step Solution

1
Evaluate the Bridge ID components (Bridge Priority and MAC address) for all candidate switches.
Identified priority values: Alpha (32768), Beta (28672), Gamma (28672), Delta (32768).
STP root bridge selection evaluates the lowest Bridge ID, which is formed by concatenating Bridge Priority and MAC Address.
2
Filter for the switches with the lowest numerical Bridge Priority value.
Switch_Beta and Switch_Gamma both tie with the lowest priority value of 28672.
Bridge priority is evaluated first. Higher priority values (32768) are eliminated.
3
Compare the MAC addresses of the tied switches to break the tie.
Switch_Gamma MAC (00:0A:B8:77:88:99) is numerically lower than Switch_Beta MAC (00:0A:B8:AA:BB:CC).
When bridge priorities match, the switch with the lower MAC address wins the election.

Key Concept

Spanning Tree Protocol (STP) Root Bridge Election Logic
Estimated Time:1m 30s
Question 377Question →

A network administrator needs to configure a firewall rule to allow secure, encrypted command-line administration of a remote Linux server using Secure Shell (SSH). Which transport protocol and port number combination must be permitted in the firewall rule?

Show answer & explanation

Answer: TCP port 22

Answer

TCP port 22
Secure Shell (SSH) is the standard network protocol used for secure, encrypted remote command-line login and system administration. It operates over TCP port 22.

Step-by-Step Solution

1
Identify the protocol requested in the network scenario.
The requirement specifies Secure Shell (SSH) for encrypted command-line administration.
SSH provides encrypted shell access to remote servers.
2
Determine the transport protocol and default well-known port assigned to SSH.
SSH uses TCP port 22.
TCP ensures reliable connection-oriented delivery, and 22 is the standard IANA port for SSH.

Key Concept

Common Ports and Protocols - Secure Shell (SSH)
Question 378Question →

A network administrator is designing switch port profiles and configuring advanced Virtual Local Area Network (VLAN) feature sets across enterprise switches. 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

Native VLAN
PortFast
Voice VLAN
Isolated Port (Private VLAN)

Matches

Show answer & explanation

Answer

Native VLAN matches carrying untagged frames over an 802.1Q trunk; PortFast matches bypassing STP listening/learning states for immediate forwarding; Voice VLAN matches segregating IP telephony traffic and applying 802.1p CoS priority tags; Isolated Port matches restricting traffic so communication occurs strictly with promiscuous ports.
Each concept correctly maps to its core operational behavior: Native VLAN handles untagged trunk traffic, PortFast eliminates STP delays on edge ports, Voice VLAN handles prioritized telephony frames, and Isolated Ports restrict Layer 2 host-to-host connectivity strictly to promiscuous default gateways.

Step-by-Step Solution

1
Identify the operation of 802.1Q trunking regarding untagged traffic
Native VLAN handles untagged traffic across trunk links without modifying Ethernet headers.
802.1Q trunks tag frames belonging to specified VLANs but leave Native VLAN traffic untagged for backward compatibility and control plane transport.
2
Analyze Spanning Tree Protocol optimization features for access ports
PortFast transitions edge interfaces directly into the forwarding state.
End-user hosts do not create Layer 2 loops, so skipping listening and learning prevents DHCP timeout issues when devices boot up.
3
Examine traffic handling for VoIP endpoints connected to access ports
Voice VLAN isolates telephony packets and enforces 802.1p priority.
VoIP applications require minimal latency and jitter, which is achieved by separating traffic into a Voice VLAN with Quality of Service tags.
4
Evaluate Layer 2 isolation boundaries within Private VLANs (PVLANs)
An Isolated Port can only send/receive traffic to/from promiscuous ports.
Private VLAN isolated ports prevent peer-to-peer compromise by blocking frame forwarding between endpoints assigned to the same isolated VLAN segment.

Key Concept

Switching Concepts and VLAN Technologies (802.1Q Trunking, PortFast, Voice VLAN, Private VLANs)
Estimated Time:1m 30s
Question 379Question →

A network administrator is setting up a new office area where eight desktop workstations connect directly to a central network switch via individual twisted-pair Ethernet cables. If a single cable connecting one workstation is severed, only that workstation loses network connectivity, while all remaining computers continue to communicate normally. Which physical network topology is implemented in this office?

Show answer & explanation

Answer: Star topology

Answer

The network uses a physical star topology because each device connects directly to a central device (the switch), isolating individual line failures.
In a physical star topology, all endpoint devices connect directly to a centralized network device, such as a switch or hub. Because each node has a dedicated physical cable running to the central device, a cable failure only disconnects that specific host while allowing all other devices to continue transmitting data.

Step-by-Step Solution

1
Analyze the physical cabling structure described in the scenario.
Each of the eight workstations connects directly to a single central switch via individual cables.
Centralized point-to-point connections to a hub or switch are the structural hallmark of a star topology.
2
Evaluate the fault-isolation behavior of the network.
Severing one workstation cable affects only that specific host without impacting others.
Because each link is independent, single point-of-failure vulnerabilities are restricted to individual end nodes or the central switch itself.
3
Match the physical layout and failure behavior to standard network topology definitions.
Star topology is confirmed as the correct answer.
Bus and ring topologies suffer total network outages upon a single cable break, whereas a full mesh requires dedicated links between every node pair rather than central switch connections.

Key Concept

Physical Star Topology Characteristics and Resilience
Question 380Question →

A network administrator is troubleshooting an enterprise client application that fails to display incoming data correctly. Packet captures reveal that while reliable transport connections are successfully established, the application cannot decode the incoming payload due to an unnegotiated character encoding scheme and an incompatible compression format. Which OSI reference model layer is directly responsible for syntax translation, data formatting, and payload encryption?

Show answer & explanation

Answer: Presentation Layer

Answer

Presentation Layer
The Presentation Layer (Layer 6) acts as a translator for network data. It ensures that application-layer data sent by one system can be understood by the application layer of another system by handling syntax formatting, character set translation (such as ASCII or Unicode), compression, and encryption.

Step-by-Step Solution

1
Analyze the symptoms presented in the scenario.
Transport sessions are operating properly, but data decoding fails due to character encoding and payload compression mismatches.
This isolates the issue to data representation and translation rather than connection management or physical delivery.
2
Map the required functions (data formatting, character encoding, compression, encryption) to the OSI model.
These responsibilities belong to Layer 6.
The Presentation Layer (Layer 6) standardizes data formats and translates syntax between the Application Layer and lower networking layers.

Key Concept

OSI Layer 6 (Presentation Layer) Functions
Estimated Time:1m 0s
PreviousPage 19 / 112Next
All practice questions — CompTIA Network+ | Examkin