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Networking Concepts

538 questions

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Question 41Question →

A network technician is organizing documentation for network protocol operations and device functions across the Open Systems Interconnection (OSI) reference model. Match each OSI model layer on the left with its corresponding function and data structure on the right.

Click a left item, then click its matching right item

Items

Layer 2 (Data Link Layer)
Layer 3 (Network Layer)
Layer 4 (Transport Layer)
Layer 6 (Presentation Layer)

Matches

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Answer

Layer 2 matches with Frame PDU and MAC addressing; Layer 3 matches with Packet PDU and IP routing; Layer 4 matches with Segment PDU and port multiplexing; Layer 6 matches with Data formatting, syntax encoding, and encryption.
Each OSI layer serves a specific role: Layer 2 processes Frames using MAC addresses; Layer 3 routes Packets using IP addresses; Layer 4 multiplexes application streams into Segments using port numbers; Layer 6 formats, encodes, and encrypts payload Data.

Step-by-Step Solution

1
Identify the protocol data unit (PDU) and core function for Layer 2.
Layer 2 operates with Frames and hardware MAC addresses for adjacent node delivery.
Data Link controls access to the physical media and verifies frame integrity.
2
Identify the PDU and core function for Layer 3.
Layer 3 handles Packets, logical IP addresses, and routing between subnets.
Network layer provides end-to-end path determination across heterogeneous networks.
3
Identify the PDU and core function for Layer 4.
Layer 4 handles Segments/Datagrams and utilizes port numbers to direct traffic to correct process sockets.
Transport layer provides host-to-host connection management and flow control.
4
Identify the primary responsibility of Layer 6.
Layer 6 manages data representation, encoding standards, compression, and encryption.
Presentation layer ensures data sent by the application layer of one system is readable by the application layer of another.

Key Concept

OSI Model Layer Responsibilities and Protocol Data Units
Question 42Question →

A network administrator is auditing static IP assignments on a database cluster segment configured with the IPv4 subnet block 10.150.96.64/2610.150.96.64/26. Which of the following IP addresses are valid, assignable host addresses within this specific subnet range? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: 10.150.96.9510.150.96.95; 10.150.96.12610.150.96.126

Answer

The valid assignable host addresses are 10.150.96.9510.150.96.95 and 10.150.96.12610.150.96.126.
For the subnet 10.150.96.64/2610.150.96.64/26, the network ID is 10.150.96.6410.150.96.64 and the broadcast address is 10.150.96.12710.150.96.127. The range of addresses usable for host assignment is 10.150.96.6510.150.96.65 through 10.150.96.12610.150.96.126. Both 10.150.96.9510.150.96.95 and 10.150.96.12610.150.96.126 reside within this valid host range.

Step-by-Step Solution

1
Determine the subnet mask and block size from the CIDR prefix.
A prefix of /26/26 corresponds to a subnet mask of 255.255.255.192255.255.255.192. The block size is 256−192=64256 - 192 = 64.
Calculating the block size identifies the IP increment between subnets.
2
Identify the network address and broadcast address for the given block.
The network address is 10.150.96.6410.150.96.64. The next subnet boundary starts at 10.150.96.12810.150.96.128, making the broadcast address for this segment 10.150.96.12710.150.96.127.
The network address is the first address of the block, and the broadcast address is the final address before the next subnet.
3
Determine the range of usable host IP addresses.
The usable host IP range spans from 10.150.96.6510.150.96.65 (Network ID + 1) to 10.150.96.12610.150.96.126 (Broadcast ID - 1).
Network and broadcast addresses are reserved and cannot be assigned to host interfaces.
4
Evaluate the candidate options against the usable host range.
10.150.96.9510.150.96.95 and 10.150.96.12610.150.96.126 fall inside the usable range (10.150.96.6510.150.96.65 - 10.150.96.12610.150.96.126), whereas 10.150.96.6410.150.96.64 and 10.150.96.12710.150.96.127 are reserved.
Only IP addresses within the usable range can be assigned to host devices.

Key Concept

IPv4 Subnet Boundaries and Usable Host Range
Question 43Question →

During a network infrastructure audit, a system administrator observes that client endpoints in a newly provisioned subnet are failing to resolve network services by name and cannot locate the primary Domain Controller for Kerberos authentication. To resolve service locator requests for specific network protocols (such as LDAP or Kerberos) including port numbers and target hostnames, which type of DNS resource record must be configured in the domain zone?

Show answer & explanation

Answer: SRV record

Answer

The SRV (Service) record is the correct DNS resource record type required to locate servers offering specific network services such as LDAP and Kerberos.
The SRV (Service) record defines hostnames and port numbers for specific services, enabling domain clients to discover directory and authentication services such as LDAP and Kerberos.

Step-by-Step Solution

1
Identify the specific DNS requirement in the scenario.
The requirement is to map a protocol service (LDAP/Kerberos) to a destination server hostname and port number.
Client endpoints rely on service location mechanisms to discover directory and authentication services.
2
Evaluate the functions of standard DNS resource record types.
SRV records define symbolic names, target hostnames, port numbers, priority, and weight for specific services.
Other record types (A, AAAA, CNAME, PTR, MX) lack port specification fields for general network services.

Key Concept

DNS Service Location (SRV) Records
Estimated Time:1m 0s
Question 44Question →

An enterprise server receives an Ethernet frame containing an HTTP request destined for its running web service. Which of the following operations occur during the de-encapsulation process as the incoming data moves up the protocol stack at the receiving host? (Select TWO.)

Select all that apply

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Answer: The Network Interface Card validates the Frame Check Sequence in the Ethernet trailer and strips the Data Link header and trailer before passing the payload up to Layer 3.; The operating system inspects the destination port number within the Transport layer header to direct the unencapsulated payload to the specific web service process.

Answer

The correct statements are that the Network Interface Card validates the Frame Check Sequence in the Ethernet trailer and strips the Data Link header and trailer before passing the payload to Layer 3, and that the operating system inspects the destination port number within the Transport layer header to direct the unencapsulated payload to the specific web service process.
When a host receives network traffic, de-encapsulation occurs from Layer 2 up to Layer 7. The NIC validates frame integrity via the FCS and removes the Layer 2 header and trailer to pass the IP packet up. Subsequently, at Layer 4, the operating system inspects the destination port number to deliver the unencapsulated application data to the appropriate process.

Step-by-Step Solution

1
Analyze Layer 2 de-encapsulation at the receiving host
The NIC receives the physical signal, validates the Frame Check Sequence (FCS) in the Ethernet trailer for errors, and strips the Ethernet header and trailer to reveal the IP packet.
De-encapsulation unwraps data from the bottom up, removing outer Layer 2 encapsulation boundaries first.
2
Analyze Layer 3 processing
The Network layer examines the IP header, verifies the destination IP address matches the receiving host, and strips the IP header to expose the Transport layer segment.
Layer 3 processing validates addressing and identifies the encapsulated upper-layer protocol.
3
Analyze Layer 4 processing and application handoff
The Transport layer evaluates the TCP header, identifies the destination port number, and hands the clean payload to the designated application layer process.
Port numbers at Layer 4 provide socket-level multiplexing to target specific services.

Key Concept

De-encapsulation sequence and layer-specific PDU handling at the receiving host
Estimated Time:1m 30s
Question 45Question →

A network administrator is configuring an edge router connected to two upstream service providers. The router learns a default route (0.00.0/00.00.0/0) via eBGP from ISP A with an Administrative Distance of 20 and a metric of 10. Simultaneously, the administrator configures a manual static default route pointing toward ISP B using the default Administrative Distance of 1. If both links are active and healthy, which path will the router install into its Routing Information Base (RIB) for default traffic?

Show answer & explanation

Answer: The static route pointing to ISP B, because static routes have a lower Administrative Distance than eBGP routes.

Answer

The static route pointing to ISP B, because static routes have a lower Administrative Distance than eBGP routes.
When a router receives multiple candidate routes for the exact same network prefix (0.0.0.0/00.0.0.0/0) from different routing protocols or administrative sources, it compares their Administrative Distance (AD). Administrative distance rates trustworthiness, where lower numerical values indicate higher preference. Static routes default to an AD of 1, whereas external BGP (eBGP) routes have a default AD of 20. Because 1 is less than 20, the static route is selected and installed into the active routing table.

Step-by-Step Solution

1
Compare the source protocols of the identical prefix (0.0.0.0/00.0.0.0/0).
Identified two routing sources: Static Route (AD = 1) and eBGP (AD = 20).
When a router receives multiple routes for the exact same prefix length from different routing sources, it evaluates Administrative Distance (AD) first.
2
Select the route with the lower Administrative Distance.
The static route with AD = 1 is selected over the eBGP route with AD = 20.
Lower Administrative Distance values represent higher trustworthiness in routing decision logic.

Key Concept

Routing Table Path Selection & Administrative Distance Hierarchy
Question 46Question →

A network administrator is assigning static IP addresses to new database servers within an isolated VLAN allocated the block 192.168.50.128/28192.168.50.128/28. Which of the following represent valid, usable host IP addresses that can be assigned to these servers? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: 192.168.50.130192.168.50.130; 192.168.50.140192.168.50.140

Answer

The addresses 192.168.50.130192.168.50.130 and 192.168.50.140192.168.50.140 are valid usable host IP addresses within the 192.168.50.128/28192.168.50.128/28 subnetwork.
For the network block 192.168.50.128/28192.168.50.128/28, the block size is 1616 (232−282^{32-28}). The network address is 192.168.50.128192.168.50.128 and the broadcast address is 192.168.50.143192.168.50.143. Subtracting these two reserved addresses leaves the usable range of 192.168.50.129192.168.50.129 to 192.168.50.142192.168.50.142. Both 192.168.50.130192.168.50.130 and 192.168.50.140192.168.50.140 lie within this range.

Step-by-Step Solution

1
Determine the block size from the CIDR prefix length
A /28/28 prefix leaves 32−28=432 - 28 = 4 host bits, yielding a block size of 24=162^4 = 16.
Calculating the block size identifies subnet boundary increments.
2
Identify the network ID and broadcast address
The network address is 192.168.50.128192.168.50.128. Adding the block size minus 1 (128+16−1128 + 16 - 1) gives the broadcast address of 192.168.50.143192.168.50.143.
The first and last addresses in a CIDR block are reserved for network identification and subnet broadcasting.
3
Calculate the usable host address range
The usable host range is from 192.168.50.129192.168.50.129 (first usable) to 192.168.50.142192.168.50.142 (last usable).
Usable host addresses exclude the network ID and broadcast address.
4
Evaluate the candidate IP addresses against the usable range
192.168.50.130192.168.50.130 and 192.168.50.140192.168.50.140 fall within [192.168.50.129,192.168.50.142][192.168.50.129, 192.168.50.142]. 192.168.50.128192.168.50.128 is the network ID and 192.168.50.143192.168.50.143 is the broadcast address.
Only IP addresses strictly inside the usable bounds may be assigned to network hosts.

Key Concept

IPv4 Subnet Boundaries and Usable Host Allocation
Question 47Question →

An enterprise network technician is documenting baseline security parameters and service configurations across network endpoints. Match each network protocol on the left with its corresponding default transport protocol and port specification on the right.

Click a left item, then click its matching right item

Items

SNMP (Simple Network Management Protocol polling)
LDAP (Lightweight Directory Access Protocol)
POP3 (Post Office Protocol v3)
SFTP (SSH File Transfer Protocol)

Matches

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Answer

SNMP matches UDP port 161; LDAP matches TCP port 389; POP3 matches TCP port 110; SFTP matches TCP port 22.
Each listed service maps directly to its official IANA transport protocol and default port assignment: SNMP polling traffic communicates over UDP port 161; unencrypted LDAP directory services communicate over TCP port 389; unencrypted POP3 mail retrieval uses TCP port 110; and SFTP uses SSH encapsulation over TCP port 22.

Step-by-Step Solution

1
Determine transport protocol and default port for SNMP polling.
SNMP manager-to-agent queries rely on low-overhead UDP port 161.
Standard status queries operate over UDP port 161, distinct from SNMP traps which use UDP port 162.
2
Determine transport protocol and default port for LDAP.
Standard LDAP operates over connection-oriented TCP port 389.
Unencrypted directory authentication lookups use TCP 389, whereas encrypted LDAPS uses TCP 636.
3
Determine transport protocol and default port for POP3.
Standard POP3 operates over TCP port 110.
Legacy post office mail download services bind to TCP 110, whereas secure POP3S binds to TCP 995.
4
Determine transport protocol and default port for SFTP.
SFTP operates over TCP port 22.
SFTP leverages the underlying SSH protocol subsystem running on TCP port 22.

Key Concept

Standard Network Protocol and Port Assignments
Question 48Question →

A systems administrator is configuring firewall ACL rules to segregate an administrative workstation network from a management server rack. The configuration policy requires permitting remote graphical desktop administration for Windows servers and allowing network infrastructure devices to forward unencrypted system log entries to a central log collector server. Which TWO of the following port and transport protocol combinations must be allowed in the firewall rule base?

Select all that apply

Show answer & explanation

Answer: TCP port 3389; UDP port 514

Answer

TCP port 3389 and UDP port 514 must be permitted.
The scenario calls for two distinct services: Windows graphical remote desktop management and unencrypted event log collection. Remote Desktop Protocol (RDP) establishes connection-oriented management sessions on TCP port 3389. Standard Syslog event logging transmits unencrypted log entries over UDP port 514. Therefore, allowing TCP port 3389 and UDP port 514 satisfies both operational requirements.

Step-by-Step Solution

1
Identify the required network service protocols from the administrative scenario requirements.
Graphical desktop administration for Windows servers requires Remote Desktop Protocol (RDP). Unencrypted system log entry forwarding requires standard Syslog.
Mapping enterprise management functions to their standard network service protocols is necessary for firewall rule design.
2
Determine the transport layer protocol and default destination port for Remote Desktop Protocol (RDP).
RDP uses TCP port 3389.
RDP requires a reliable connection-oriented transport protocol (TCP) to manage interactive desktop display updates and keyboard/mouse inputs without packet loss.
3
Determine the transport layer protocol and default destination port for standard unencrypted Syslog.
Standard Syslog uses UDP port 514.
Native Syslog is designed as a low-overhead, connectionless log-reporting protocol that transmits event notifications over UDP.

Key Concept

Standard Port Numbers and Transport Layer Protocol Pairings
Estimated Time:1m 30s
Question 49Question →

A network administrator is assigned the IPv4 block 172.31.96.0/22172.31.96.0/22 for a data center expansion. The administrator must allocate the first subnet from this block to support a web server farm requiring at least 110110 usable host IP addresses. Which of the following correctly identifies the subnet ID and broadcast address for this web server subnet using the most space-efficient CIDR prefix?

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Answer: Subnet ID: 172.31.96.0/25172.31.96.0/25, Broadcast Address: 172.31.96.127172.31.96.127

Answer

Subnet ID: 172.31.96.0/25172.31.96.0/25, Broadcast Address: 172.31.96.127172.31.96.127
To host at least 110 devices, 7 host bits are required because 27−2=1262^7 - 2 = 126 usable hosts. A prefix length of /25/25 (32−732 - 7) creates subnets of size 128. Starting from 172.31.96.0172.31.96.0, the first /25/25 subnet covers addresses 172.31.96.0172.31.96.0 through 172.31.96.127172.31.96.127. Thus, the Subnet ID is 172.31.96.0/25172.31.96.0/25 and the Broadcast Address is 172.31.96.127172.31.96.127.

Step-by-Step Solution

1
Determine the required number of host bits (hh).
To accommodate at least 110110 usable hosts, use the formula 2h−2≥1102^h - 2 \ge 110. For h=6h=6, 26−2=622^6 - 2 = 62 (insufficient). For h=7h=7, 27−2=1262^7 - 2 = 126 usable hosts (sufficient).
Two addresses must be subtracted from the total block size for the network ID and broadcast address.
2
Calculate the CIDR prefix length.
Prefix length = 32−7=/2532 - 7 = /25.
IPv4 addresses consist of 32 bits. Subtracting 7 host bits leaves 25 network/subnet bits.
3
Determine the block size and network range for the first subnet.
Block size = 27=1282^7 = 128. Starting at 172.31.96.0172.31.96.0, the subnet spans from 172.31.96.0172.31.96.0 to 172.31.96.127172.31.96.127.
The first subnet starts at the beginning of the parent block (172.31.96.0172.31.96.0).
4
Identify the Subnet ID and Broadcast Address.
Subnet ID = 172.31.96.0/25172.31.96.0/25, Broadcast Address = 172.31.96.127172.31.96.127. Usable host range is 172.31.96.1172.31.96.1 through 172.31.96.126172.31.96.126.
The lowest address in the block is the Subnet ID, and the highest address in the block is the Broadcast Address.

Key Concept

Calculating host requirements, prefix lengths, and broadcast boundaries in IPv4 subnetting.
Question 50Question →

A network administrator is evaluating the behavior of a Layer 2 switch configured with two isolated VLANs: VLAN 10 (Sales) and VLAN 20 (Engineering). Host A, connected to a port on VLAN 10, sends a broadcast frame onto the network. Almost simultaneously, Host B, also on VLAN 10, transmits a unicast frame addressed to Host C (on VLAN 10), but Host C's MAC address is not currently listed in the switch's CAM table. Which of the following statements accurately describe how the Layer 2 switch processes these frames? (Select TWO)

Select all that apply

Show answer & explanation

Answer: The switch forwards the broadcast frame from Host A out all active ports assigned to VLAN 10 except the ingress port, but suppresses transmission to VLAN 20 ports.; The switch floods the unicast frame from Host B out all active member ports of VLAN 10 except the ingress port because the destination MAC address is unknown.

Answer

The switch forwards the broadcast frame from Host A out all active member ports of VLAN 10 except the receiving port, while isolating it from VLAN 20. In addition, because Host C's MAC address is unknown in the CAM table, the switch performs unknown unicast flooding by transmitting Host B's frame out all ports belonging to VLAN 10 except the ingress port.
VLANs create separate Layer 2 broadcast domains, meaning a broadcast frame originating on VLAN 10 will be forwarded out all active ports assigned to VLAN 10 (except the receiving port) but will never cross over to ports on VLAN 20. Furthermore, when a switch receives a unicast frame whose destination MAC address is not listed in the CAM table, it performs unknown unicast flooding, transmitting the frame to all active ports within that frame's VLAN except the ingress port.

Step-by-Step Solution

1
Analyze the broadcast domain boundaries for Host A's broadcast frame.
VLANs define logical Layer 2 broadcast domains. Broadcast traffic originating in VLAN 10 is flooded to all active ports assigned to VLAN 10 (excluding the receiving port) and blocked from crossing into VLAN 20.
Layer 2 switches do not forward broadcast traffic across different VLAN boundaries.
2
Analyze switch forwarding behavior for Host B's unicast frame with an unlearned destination MAC address.
The switch looks up Host C's destination MAC address in its CAM table. Finding no match, it performs unknown unicast flooding within VLAN 10.
Unknown unicast traffic is flooded to all ports in the source VLAN except the port of entry so the intended recipient receives the frame and can reply, allowing the switch to learn its MAC address.

Key Concept

VLAN Broadcast Domain Boundaries and Unknown Unicast Flooding Mechanics
Estimated Time:1m 30s
Question 51Question →

A biomedical technician is deploying a network of wireless wearable vital-sign sensors that continuously transmit telemetry data to a tablet located within a two-meter radius of a patient's bed. Which of the following network types best classifies this short-range wireless architecture?

Show answer & explanation

Answer: Personal Area Network (PAN)

Answer

Personal Area Network (PAN) is the correct network classification for extremely short-range communications centered around an individual or personal workspace.
A Personal Area Network (PAN) is designed for extremely short-range communication (typically under 10 meters) between personal devices or localized sensors around an individual.

Step-by-Step Solution

1
Analyze the operational radius and scope described in the scenario.
The sensors transmit data over a two-meter radius around a single patient.
Geographic scope is the primary criterion for classifying network types.
2
Match the scope to standard network classification definitions.
A network limited to a few meters focused around an individual device or patient is classified as a Personal Area Network (PAN).
PANs are specifically designed for low-power, short-range inter-device communication within immediate personal space.

Key Concept

Network Types and Geographic Scopes
Question 52Question →

Match each IP addressing service or protocol feature on the left with its core function within network operations on the right.

Click a left item, then click its matching right item

Items

Network Time Protocol (NTP)
IP Address Management (IPAM)
DHCP Snooping
Dynamic DNS (DDNS)

Matches

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Answer

Network Time Protocol (NTP) matches clock synchronization; IP Address Management (IPAM) matches centralized address space tracking; DHCP Snooping matches filtering rogue DHCP server responses; Dynamic DNS (DDNS) matches updating DNS records dynamically.
Each IP service performs a specialized function in enterprise operations: NTP handles time synchronization for log correlation; IPAM offers centralized management and tracking of address space; DHCP Snooping provides switch port security against unauthorized DHCP servers; DDNS provides automated dynamic update capabilities for DNS records.

Step-by-Step Solution

1
Identify the primary function of NTP.
NTP synchronizes time across network devices for consistent logging.
Accurate timestamps are required for system log alignment and security analysis.
2
Identify the primary function of IPAM.
IPAM plans, tracks, and manages IP address allocations across subnets.
Enterprise networks require unified management to prevent address exhaustion and IP conflicts.
3
Identify the primary security function of DHCP Snooping.
DHCP Snooping blocks rogue DHCP servers at the switch interface level.
Untrusted ports are restricted from sending DHCP Offer and ACK messages.
4
Identify the primary function of DDNS.
DDNS automatically modifies DNS records when IP addresses change dynamically.
Hostnames remain resolvable even when client IP addresses change frequently via DHCP.

Key Concept

Core functions of enterprise IP addressing and supporting network services
Question 53Question →

A network security appliance running a Syslog service generates an alert log message that must be sent to a remote monitoring server over UDP port 514. In what chronological order does the local host network stack encapsulate the log data from creation to transmission onto the wire?

Drag items to arrange them in the correct order

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Answer

The correct order of encapsulation begins with Application payload creation, followed by Transport layer UDP datagram encapsulation, Network layer IP packet creation, Data Link layer Ethernet framing with MAC addresses and FCS, and finally Physical layer bit conversion to electrical signals.
Outbound encapsulation follows the OSI model from top to bottom (Layer 7 down to Layer 1). The process begins when the Syslog process creates the application payload. Next, the Transport layer adds UDP port headers to create a datagram. Then, the Network layer appends IP addresses to form a packet. After that, the Data Link layer appends MAC addresses and an FCS trailer to construct an Ethernet frame. Finally, the Physical layer encodes the frame into physical signals to send across the network wire.

Step-by-Step Solution

1
Identify the top-layer application data creation.
The Syslog service generates the raw application data.
Encapsulation starts at the top of the OSI stack (Application layer) when user or system data is created.
2
Encapsulate at Layer 4 (Transport).
UDP headers (source and destination ports) are added to form a datagram.
The transport layer provides process-to-process addressing using port numbers.
3
Encapsulate at Layer 3 (Network).
IP headers (source and destination IP addresses) are added to form a packet.
The network layer handles logical network routing and IP addressing across subnets.
4
Encapsulate at Layer 2 (Data Link).
Ethernet header (MAC addresses) and trailer (FCS) are added to form a frame.
The data link layer provides hop-by-hop physical addressing and error detection.
5
Transmit at Layer 1 (Physical).
The binary frame is converted into electrical signals for transmission.
The physical layer transmits raw bitstreams across the physical copper cabling medium.

Key Concept

Top-down OSI model outbound data encapsulation sequence (Data -> Datagram/Segment -> Packet -> Frame -> Bits).
Question 54Question →

A network administrator is provisioning static IP addresses for a cluster of application servers located in the subnet 172.20.100.128/26172.20.100.128/26. Which of the following IPv4 addresses are valid usable host addresses within this designated subnet? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: 172.20.100.130172.20.100.130; 172.20.100.189172.20.100.189

Answer

The valid usable host addresses are 172.20.100.130172.20.100.130 and 172.20.100.189172.20.100.189.
For the subnet 172.20.100.128/26172.20.100.128/26, the block size is 64 addresses. The network address is 172.20.100.128172.20.100.128 and the broadcast address is 172.20.100.191172.20.100.191. Consequently, the assignable host addresses range from 172.20.100.129172.20.100.129 to 172.20.100.190172.20.100.190. Both 172.20.100.130172.20.100.130 and 172.20.100.189172.20.100.189 lie within this range.

Step-by-Step Solution

1
Determine the block size and network boundaries for the CIDR prefix /26/26.
A /26/26 prefix leaves 6 host bits (32−26=632 - 26 = 6). The total address block size is 26=642^6 = 64.
Calculating block size establishes the start and end boundaries of the IP subnet.
2
Identify the network ID and broadcast address for 172.20.100.128/26172.20.100.128/26.
The network ID is 172.20.100.128172.20.100.128. The broadcast address is 172.20.100.128+64−1=172.20.100.191172.20.100.128 + 64 - 1 = 172.20.100.191.
The first address in the block is reserved for the network ID, and the last address in the block is reserved for the broadcast address.
3
Calculate the usable host IP range.
The first usable IP address is 172.20.100.129172.20.100.129 and the last usable IP address is 172.20.100.190172.20.100.190.
Usable host IP addresses lie strictly between the network ID and the broadcast address.
4
Evaluate the choices against the usable host range (172.20.100.129172.20.100.129 to 172.20.100.190172.20.100.190).
172.20.100.130172.20.100.130 and 172.20.100.189172.20.100.189 are within the usable range, whereas 172.20.100.128172.20.100.128 (network ID) and 172.20.100.191172.20.100.191 (broadcast address) are unusable host addresses.
Assigning network or broadcast addresses directly to hosts results in invalid IP configurations.

Key Concept

IPv4 Subnet Boundaries and Usable Host Range Calculation
Question 55Question →

A Layer 3 router receives an IP packet and has multiple matching entries in its routing table for the destination address. Arrange the steps of the router's path selection process in the correct order from first to last.

Drag items to arrange them in the correct order

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Answer

The correct order of steps for the router path selection process is: 1) Filter candidate routes by evaluating the longest prefix match for the destination IP address, 2) Compare the administrative distance among routes with identical prefix lengths, 3) Compare protocol-specific metric values among routes with equal administrative distance, and 4) Install the optimal path into the active routing table for packet forwarding.
When an IP packet arrives, the router first identifies routes with the longest prefix match. If multiple routes share the exact same prefix length, the router compares their administrative distance to determine which protocol source is most believable. If routes share both equal prefix lengths and equal administrative distance (coming from the same protocol), the router compares protocol metrics to find the path with the lowest cost. Finally, the selected route is installed into the active forwarding table.

Step-by-Step Solution

1
Evaluate subnet mask specificity
Candidate routes are narrowed down to those matching the highest number of continuous network bits.
Longest prefix match takes precedence over all other criteria, including administrative distance.
2
Evaluate route source trustworthiness
The route originating from the source with the lower administrative distance value is preferred.
Administrative distance distinguishes between different protocol sources (e.g., EIGRP vs OSPF) when prefix specificity is equal.
3
Evaluate intra-protocol path cost
The path with the lowest metric value (e.g., OSPF cost) is selected.
Metrics are only directly comparable when candidate routes originate from the same routing protocol.
4
Commit route selection
The winning route entry is placed in the active Routing Information Base (RIB) / Forwarding Information Base (FIB).
Only the single best route (or tied equal-cost paths for load balancing) is actively used to forward traffic.

Key Concept

Router Path Selection Hierarchy (Longest Prefix Match -> Administrative Distance -> Protocol Metric)
Question 56Question →

A network administrator is deploying a new centralized AAA server to manage administrative authentication for enterprise network switches. The deployment requires using TACACS+ to separate authentication, authorization, and accounting functions while encrypting the entire payload. However, authentication requests from the switches are currently being blocked by an intervening network firewall. Which port and transport layer protocol combination must be permitted on the firewall to allow TACACS+ traffic?

Show answer & explanation

Answer: TCP port 49

Answer

TCP port 49 is required for TACACS+ communications.
TACACS+ operates over TCP port 49 by default. It utilizes TCP at the transport layer to provide reliable delivery and connection management while combining authentication, authorization, and accounting over a single port with full payload encryption.

Step-by-Step Solution

1
Identify the protocol specified in the requirement.
The scenario specifies TACACS+ (Terminal Access Controller Access-Control System Plus) for centralized network administrator AAA functionality.
Determining the target application protocol is the first step in matching port and transport requirements.
2
Determine the transport layer protocol used by TACACS+.
TACACS+ uses Transmission Control Protocol (TCP) because it requires connection-oriented reliability for AAA transactions.
Unlike RADIUS which relies on UDP, TACACS+ uses TCP at Layer 4.
3
Identify the default well-known port number for TACACS+.
The IANA standard port for TACACS+ is port 49.
Permitting TCP port 49 on the network firewall ensures TACACS+ traffic flows correctly between network switches and the AAA server.

Key Concept

Standard AAA protocols and default port mappings (TACACS+ vs RADIUS)
Estimated Time:1m 0s
Question 57Question →

A network technician connects five workstations and two servers directly to individual full-duplex ports on a managed Layer 2 switch. Three of the active ports are assigned to VLAN 15, and the remaining four active ports are assigned to VLAN 35. No Layer 3 routing device is present on the network. How many collision domains and broadcast domains exist on this switch?

Show answer & explanation

Answer: 7 collision domains and 2 broadcast domains

Answer

7 collision domains and 2 broadcast domains
Each connected port on a Layer 2 Ethernet switch operating in full-duplex mode constitutes an independent collision domain, yielding 7 collision domains for the 7 active host ports. Broadcast domains are defined by VLAN boundaries; because the switch is partitioned into VLAN 15 and VLAN 35, there are exactly 2 broadcast domains.

Step-by-Step Solution

1
Count the total number of active switch ports to determine collision domains
5 workstations + 2 servers = 7 connected switch ports, producing 7 distinct collision domains
Layer 2 switches isolate collision domains on a per-port basis when running in full-duplex mode.
2
Count the total number of configured VLANs to determine broadcast domains
VLAN 15 and VLAN 35 represent 2 distinct logical VLANs, producing 2 broadcast domains
A Virtual Local Area Network (VLAN) logically segments a physical switch into separate Layer 2 broadcast domains.

Key Concept

Collision and Broadcast Domain Segmentation in Layer 2 Switches
Question 58Question →

Match each network topology classification to its defining structural characteristic and operational failure profile.

Click a left item, then click its matching right item

Items

Bus Topology
Ring Topology
Full Mesh Topology
Star Topology

Matches

Show answer & explanation

Answer

Bus Topology matches the shared continuous trunk cable with terminators; Ring Topology matches the sequential circular token path vulnerable to single node failure; Full Mesh Topology matches complete point-to-point node connections calculated by n(n−1)/2n(n-1)/2; Star Topology matches connections radiating from a central device where single link breaks only isolate one node.
The matches accurately pair each physical topology with its physical layout and failure domain behavior. Bus topologies rely on a terminated single backbone line. Ring topologies process traffic sequentially in a circular loop. Full mesh topologies link every device pair for total redundancy. Star topologies isolate individual drop cables through a central distribution device.

Step-by-Step Solution

1
Analyze Bus Topology characteristics
Identified that bus topologies utilize a central backbone cable terminated on both ends, where a break affects all nodes.
Bus networks rely on shared physical media continuity.
2
Analyze Ring Topology characteristics
Identified that ring topologies pass data sequentially in a circle from one device to the next.
Without redundant rings, an outage at any intermediary host breaks the circuit.
3
Analyze Full Mesh Topology characteristics
Identified that full mesh requires dedicated links between every node pair, adhering to the mathematical formula L=n(n−1)2L = \frac{n(n-1)}{2}.
Full mesh offers maximum fault tolerance due to full point-to-point redundancy.
4
Analyze Star Topology characteristics
Identified that star topology connects hosts independently to a central switch or hub.
Failure of an edge cable impacts only the single connected endpoint.

Key Concept

Physical Network Topologies and Fault Tolerance Profiles
Estimated Time:1m 30s
Question 59Question →

A server receives an incoming network transmission where an HTTP request payload is encapsulated within a TCP segment, an IPv4 packet, and an Ethernet frame. As the destination host's operating system stack performs de-encapsulation at the Transport layer (Layer 4), which header field is inspected to deliver the inner payload to the correct server process?

Show answer & explanation

Answer: Destination Port Number

Answer

The destination port number in the Transport layer header is inspected by the receiving network stack to direct the payload to the appropriate listening application service.
During the de-encapsulation process, each layer of the OSI model inspects its corresponding header to determine how to route the inner Protocol Data Unit (PDU) upward. At the Transport layer (Layer 4), TCP or UDP headers contain a Destination Port field. The host OS reads this destination port to direct the segment's data payload to the exact application service bound to that port.

Step-by-Step Solution

1
Analyze the incoming data flow during de-encapsulation.
The physical network interface accepts the electrical/optical signals and strips the Layer 2 Ethernet header after checking the frame header and EtherType field.
The EtherType field notifies Layer 2 that the encapsulated payload belongs to Network layer IPv4.
2
Evaluate Network layer (Layer 3) processing.
The IP protocol stack inspects the destination IP address and the Protocol field within the IPv4 header.
The Protocol field specifies that the payload should be handed off to TCP at Layer 4.
3
Evaluate Transport layer (Layer 4) de-encapsulation.
The TCP protocol stack receives the segment and inspects the Destination Port Number in the TCP header.
The destination port number acts as the address identifier for the specific application socket listening on the server.

Key Concept

De-encapsulation and Layer 4 Port Multiplexing
Question 60Question →

A network engineer is partitioning the address block 10.45.160.0/1910.45.160.0/19 to provision a new VoIP network segment that must accommodate at least 500500 IP phones. To prevent address space depletion, the engineer allocates the smallest valid subnet starting at the beginning of the block that satisfies this host capacity requirement. What is the last usable host IP address of this newly allocated subnet?

Show answer & explanation

Answer: 10.45.161.254

Answer

10.45.161.254
To accommodate at least 500500 usable host addresses with minimal wastage, 99 host bits are needed (29−2=5102^9 - 2 = 510 usable addresses), establishing a /23/23 subnet mask. The first /23/23 subnet within 10.45.160.0/1910.45.160.0/19 ranges from 10.45.160.010.45.160.0 (Network ID) to 10.45.161.25510.45.161.255 (Broadcast Address). Subtracting 11 from the broadcast address yields 10.45.161.25410.45.161.254 as the final usable host IP address.

Step-by-Step Solution

1
Determine the required host bits and CIDR prefix length
9 host bits are required, yielding a /23 subnet prefix
The formula for usable hosts is 2h−2≥5002^h - 2 \ge 500. For h=8h = 8, 28−2=2542^8 - 2 = 254 (insufficient). For h=9h = 9, 29−2=5102^9 - 2 = 510 (sufficient). The prefix length is 32−9=2332 - 9 = 23.
2
Calculate the network block boundaries for the first /23 subnet
Network ID: 10.45.160.0/23, Broadcast Address: 10.45.161.255
A /23 subnet has a block size of 22 in the third octet (256−254=2256 - 254 = 2). Starting at 10.45.160.010.45.160.0, the network spans from 10.45.160.010.45.160.0 through 10.45.161.25510.45.161.255.
3
Identify the last usable host IP address within the subnet boundary
10.45.161.254
The last usable host IP address is always one numeric value below the broadcast address (10.45.161.255−1=10.45.161.25410.45.161.255 - 1 = 10.45.161.254).

Key Concept

Subnet Host Capacity and Subnet Boundary Calculation
Estimated Time:2m 0s
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