Networking Concepts

538 questions

Question 301Question

An enterprise network architect is designing a multi-VRF network infrastructure where clients in remote subnets must receive IP leases from a centralized DHCP server cluster while maintaining accurate physical switch-port tracking and automated domain name registration. Which TWO of the following configurations or mechanisms must be implemented to fulfill these requirements?

Select all that apply

Show answer & explanation

Answer: Enable DHCP Relay (IP Helper) on the remote subnet default gateway interfaces to append Option 82 sub-options (Circuit ID and Remote ID) before forwarding requests to the central DHCP server.; Configure Client FQDN (Option 81) or secure Dynamic DNS (DDNS) updates on the DHCP server to automatically register forward (A/AAAA) and reverse (PTR) records with authoritative DNS servers.

Answer

The two correct requirements are configuring DHCP Relay on default gateway interfaces to insert Option 82 relay information, and enabling DHCP Option 81 (Client FQDN) / secure Dynamic DNS updates on the DHCP server.
Enabling DHCP Relay with Option 82 on remote subnet gateway interfaces allows the network to relay client requests across Layer 3 boundaries while attaching switch port and circuit metadata. In addition, using Option 81 (Client FQDN) / DDNS on the DHCP server ensures seamless, automated updating of A, AAAA, and PTR records in authoritative DNS zones whenever leases are granted.

Step-by-Step Solution

1
Analyze the requirement for tracking physical location/switch-port details across Layer 3 boundaries.
Identified that DHCP Relay Agent Information (Option 82) must be inserted at the Layer 3 boundary interface (default gateway) to provide Circuit ID and Remote ID metadata to the DHCP server.
Broadcast DHCP requests cannot cross routers without a relay agent, and Option 82 supplies switch port and agent identity to the centralized server.
2
Analyze the requirement for automated domain name registration across IP addressing services.
Identified DHCP Option 81 (Client FQDN) and secure Dynamic DNS (DDNS) integration between DHCP and DNS infrastructure.
Option 81 allows the DHCP server to negotiate and execute dynamic updates for forward (A/AAAA) and reverse (PTR) lookup records upon issuing an IP address lease.
3
Evaluate and discard distractor options based on architectural error taxonomy.
Eliminated incorrect relay placement (must be on remote gateway, not central server host), incorrect protocol/port (DHCP uses UDP 67/68, not TCP 67), and invalid DNS record usage (CNAME pointing to broadcast address).
Ensures architectural compliance with CompTIA Network+ standards for IP services.

Key Concept

DHCP Relay Architecture, Option 82 (Relay Agent Information), and Option 81 / DDNS Integration
Question 302Question

A network engineer is provisioning subnets from the parent address block 192.168.64.0/20192.168.64.0/20 for a newly constructed facility. The primary VLAN requires capacity for at least 100100 usable host IP addresses while minimizing wasted address space. If the engineer allocates the first available valid subnet from this block that satisfies the host capacity requirement, what is the broadcast address of this allocated subnet?

Show answer & explanation

Answer: 192.168.64.127192.168.64.127

Answer

The broadcast address of the allocated subnet is 192.168.64.127192.168.64.127.
To host at least 100 devices with minimal address waste, 7 host bits are required (272=1262^7 - 2 = 126 usable hosts), which corresponds to a /25/25 subnet mask. The first /25/25 subnet within 192.168.64.0/20192.168.64.0/20 occupies the IP block 192.168.64.0192.168.64.0 to 192.168.64.127192.168.64.127. The final address of this block, 192.168.64.127192.168.64.127, serves as the broadcast address.

Step-by-Step Solution

1
Determine the minimum prefix length needed for 100 usable hosts.
Prefix length /25/25 (7 host bits, providing 272=1262^7 - 2 = 126 usable hosts).
A /26/26 prefix only offers 262=622^6 - 2 = 62 usable host addresses, which is insufficient. A /25/25 prefix provides 126126 usable addresses, satisfying the requirement of at least 100 hosts while minimizing wasted space.
2
Identify the network boundary of the first /25/25 subnet inside the 192.168.64.0/20192.168.64.0/20 parent block.
Network ID: 192.168.64.0/25192.168.64.0/25.
Subnetting starts sequentially from the lowest network ID of the parent block.
3
Calculate the broadcast address of the 192.168.64.0/25192.168.64.0/25 subnet.
Broadcast Address: 192.168.64.127192.168.64.127.
With a block size of 128128 IP addresses (272^7), the subnet range spans 192.168.64.0192.168.64.0 through 192.168.64.127192.168.64.127. The last IP address in the block is reserved for broadcast.

Key Concept

Subnet Masking and Broadcast Address Calculation
Question 303Question

A network administrator is verifying the automatic IPv6 interface configuration of a host. The host network adapter has a MAC address of 00:1A:2B:3C:4D:5E. Which IPv6 link-local address is generated for this interface using the Modified EUI-64 format?

Show answer & explanation

Answer: fe80::21a:2bff:fe3c:4d5e

Answer

fe80::21a:2bff:fe3c:4d5e
To form a Modified EUI-64 interface identifier from a 48-bit MAC address, the 16-bit sequence FFFE is inserted between the OUI and NIC specific bytes, and the 7th bit of the first byte is inverted (changing 00 to 02). When prefixed with fe80::/64 and formatted with standard zero suppression, the resulting address is fe80::21a:2bff:fe3c:4d5e.

Step-by-Step Solution

1
Divide the 48-bit MAC address into two 24-bit halves.
Left half: 00:1A:2B, Right half: 3C:4D:5E.
Modified EUI-64 creates a 64-bit interface identifier by inserting a fixed 16-bit value between the OUI and the device's extension identifier.
2
Insert the 16-bit hexadecimal string FFFE between the two halves.
00:1A:2B:FF:FE:3C:4D:5E.
The standard EUI-64 expansion rule requires adding FFFE into the center of a 48-bit MAC address.
3
Invert the 7th bit (universal/local bit) of the first byte.
The first byte 0x00 (00000000 in binary) becomes 0x02 (00000010 in binary).
IPv6 Modified EUI-64 rules require flipping the 7th bit of the first byte to indicate universal scope.
4
Combine the result with the link-local prefix (fe80::/64) and apply standard zero compression.
fe80::21a:2bff:fe3c:4d5e.
Leading zeros within a 16-bit hex block (021a to 21a) are omitted according to RFC 5952 formatting standard.

Key Concept

Modified EUI-64 Link-Local Address Generation
Estimated Time:1m 30s
Question 304Question

A network engineer is reviewing a protocol analyzer capture during an active client-server session. The log shows explicit control checkpoints being inserted into the data stream to allow recovery and resume capabilities, as well as managing dialog control to regulate which side can transmit. Which OSI model layer is primarily responsible for managing these session synchronization points and dialog modes?

Show answer & explanation

Answer: Session layer

Answer

Session layer
The correct answer is the Session layer because Layer 5 of the OSI model is explicitly designed to handle session setup, maintenance, tear-down, dialog control (regulating which host sends when), and insertion of synchronization checkpoints to allow recovery during long data transmissions.

Step-by-Step Solution

1
Analyze the operational requirements described in the stem.
Identified key requirements: managing dialog modes (simplex, half-duplex, full-duplex) and inserting synchronization checkpoints into communication streams.
Synchronization checkpoints and dialog control are functions dedicated to managing logical sessions between end systems.
2
Map the identified functions to the appropriate OSI model layer.
Session Layer (Layer 5) directly governs dialog control and session checkpointing.
Layer 5 establishes, manages, and terminates connections between applications at each end.

Key Concept

OSI Layer 5 (Session Layer) Functions
Estimated Time:1m 15s
Question 305Question

Match each core IP addressing service or infrastructure component on the left with its primary operational function on the right.

Click a left item, then click its matching right item

Items

DHCP Relay Agent
DNS PTR Record
IP Address Management (IPAM)
Network Time Protocol (NTP)

Matches

Show answer & explanation

Answer

DHCP Relay Agent matches with forwarding broadcast DHCP packets across routing boundaries; DNS PTR Record matches with mapping an IP address to an FQDN for reverse lookups; IPAM matches with centralizing IP space tracking and auditing; NTP matches with synchronizing system clocks for correlated logging.
Each service is correctly paired according to standard networking definitions: DHCP Relay forwards client discovery broadcasts across routers to a central server; DNS PTR records handle reverse IP-to-hostname lookups; IPAM centralizes network address management; and NTP maintains synchronized clock timing across network nodes.

Step-by-Step Solution

1
Analyze the role of DHCP Relay
DHCP Relay acts as an intermediate agent converting broadcast discovery packets into unicast traffic to traverse routers.
Routers drop layer 2 broadcast packets by default, requiring relay functionality to reach centralized DHCP services.
2
Analyze the function of DNS PTR records
PTR records map canonical IP addresses to FQDNs in reverse lookup zones.
Standard forward DNS uses A/AAAA records for domain-to-IP resolution, while reverse resolution requires PTR records.
3
Evaluate the administrative role of IPAM
IPAM integrates IP inventory tracking, subnet allocation, and capacity planning.
It prevents address collisions and simplifies management across large networks.
4
Identify the purpose of NTP
NTP synchronizes device clocks across network hosts.
Accurate timestamps are critical for logging, troubleshooting, and time-sensitive authentication mechanisms.

Key Concept

Core IP Addressing Services and Functions
Question 306Question

A network technician is configuring a local subnetwork for a department that requires a /27/27 prefix length. How many usable host IPv4 addresses can be assigned to devices within this subnetwork?

Show answer & explanation

Answer: 30

Answer

There are 30 usable host IPv4 addresses in a /27 subnet.
A /27 prefix leaves 5 host bits (3227=532 - 27 = 5). Calculating 252^5 gives 3232 total IP addresses. Subtracting 22 reserved addresses for the network ID and broadcast address results in exactly 3030 usable host IPv4 addresses.

Step-by-Step Solution

1
Determine the number of host bits remaining in a /27 CIDR prefix.
32 total bits - 27 network bits = 5 host bits.
An IPv4 address contains 32 bits divided between network bits and host bits.
2
Calculate the total capacity of IP addresses for 5 host bits.
2^5 = 32 total IP addresses.
The number of total IP addresses in a subnet is calculated using 2 raised to the power of the host bits.
3
Subtract the reserved network and broadcast addresses.
32 - 2 = 30 usable host addresses.
The lowest IP address represents the network ID and the highest IP address represents the broadcast address, neither of which can be assigned to host devices.

Key Concept

IPv4 Usable Host Address Calculation
Estimated Time:45s
Question 307Question

A network technician is performing a protocol analyzer capture to inspect traffic traversing an enterprise IPv6 router interface. Which of the following statements accurately describe the characteristics of IPv6 base headers and link-local addressing? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: The 8-bit Hop Limit field replaces the IPv4 Time to Live (TTL) field to prevent packets from circulating indefinitely in a routing loop.; Link-local addresses utilize the fe80::/10 prefix and are automatically configured for single-segment communication without being routed across subnets.

Answer

The correct statements are that the 8-bit Hop Limit field replaces the IPv4 Time to Live (TTL) field to prevent routing loops, and link-local addresses use the fe80::/10 prefix for single-segment, non-routable communication.
The Hop Limit field in the IPv6 base header serves as the direct replacement for IPv4's Time to Live (TTL) field, decrementing by one at each hop to eliminate packet loops. Furthermore, IPv6 link-local addresses fall under the fe80::/10 block (most commonly configured with fe80::/64) and are intended solely for local link communication, remaining non-routable across external networks.

Step-by-Step Solution

1
Analyze the Hop Limit field function in IPv6 headers.
Confirm that Hop Limit functions identically to IPv4 TTL, decrementing at each hop.
It prevents network packets from looping indefinitely on the network.
2
Evaluate link-local address prefixes and scope.
Identify fe80::/10 as the designated prefix scope for link-local traffic.
Link-local addresses are restricted to communication within the local broadcast/multicast domain and are discarded by routers.
3
Examine IPv6 shorthand address compression rules and Next Header functionality.
Reject invalid compression using multiple double colons and correct the field description of Next Header.
Using double colons more than once creates ambiguity, and Next Header identifies upper-layer protocols rather than QoS tags.

Key Concept

IPv6 Base Header Fields and Address Scope Characteristics
Question 308Question

A network administrator is troubleshooting an issue where new IP phones on VLAN 50 (10.50.0.0/2410.50.0.0/24) successfully obtain IPv4 leased addresses and default gateway settings from a centralized DHCP server located on VLAN 10 (10.10.0.10010.10.0.100), but repeatedly fail to locate their TFTP provisioning server (10.10.0.20010.10.0.200) and fail to sync time with the corporate NTP server (10.10.0.1010.10.0.10). The layer 3 switch interface for VLAN 50 is configured with `ip helper-address 10.10.0.100`. Packet captures on the switch show DHCP requests are forwarded, but TFTP and NTP requests sent via broadcast during initial boot are dropped. Which configuration modification on the layer 3 switch will directly resolve the provisioning and time synchronization failure for the endpoints on VLAN 50?

Show answer & explanation

Answer: Configure additional UDP forward port commands (`ip forward-protocol udp 69` and `ip forward-protocol udp 123`) or configure DHCP scope Options 66 and 42 on the central DHCP server.

Answer

Configure additional UDP forward port commands or configure DHCP scope Options 66 and 42 on the central DHCP server.
Centralized network services across routed subnets require remote hosts to receive unicast destination information or require layer 3 devices to relay specific broadcast protocols. Configuring DHCP scope Option 66 (TFTP server) and Option 42 (NTP server) delivers unicast IPv4 addresses directly to endpoints during initial DHCP address leasing. Alternatively, ensuring `ip forward-protocol udp` rules allow UDP broadcast translation for ports 69 and 123 allows legacy broadcast discovery to reach centralized servers.

Step-by-Step Solution

1
Analyze client boot sequence requirements
IP phones require IP address configuration (via DHCP), TFTP server address for boot file download (Option 66), and NTP server address for time synchronization (Option 42).
Endpoints on remote subnets cannot discover centralized services via local layer 2 broadcasts across routed boundaries.
2
Evaluate relay behavior of `ip helper-address`
The `ip helper-address` command relays broadcast UDP traffic destined for specific ports to targeted unicast destinations.
If clients broadcast to discover TFTP (port 69) or NTP (port 123) without receiving DHCP scope options, UDP broadcast forwarding must be explicitly permitted or DHCP scope options must supply unicast addresses.
3
Select the correct infrastructure configuration solution
Providing DHCP Scope Options 66 and 42 ensures clients receive unicast target addresses directly, while adjusting UDP forward rules permits necessary relay traffic.
This directly satisfies the requirement for endpoints on VLAN 50 to locate and communicate with services on VLAN 10.

Key Concept

DHCP Relay (IP Helper) Operations and Service Option Configurations
Question 309Question

A network engineer is provisioning subnets for an enterprise database cluster within the allocated block 172.22.140.0/22172.22.140.0/22. The designated subnet for this VLAN is 172.22.142.128/25172.22.142.128/25. Which of the following IPv4 addresses represent valid usable host addresses that can be statically assigned to database interfaces in this subnet? (Select ALL that apply.)

Select all that apply

Show answer & explanation

Answer: 172.22.142.129172.22.142.129; 172.22.142.200172.22.142.200

Answer

The valid usable host IPv4 addresses for the subnet are 172.22.142.129172.22.142.129 and 172.22.142.200172.22.142.200.
For the subnet 172.22.142.128/25172.22.142.128/25, the prefix length /25/25 provides 128 total IP addresses (23225=1282^{32-25} = 128). The network address is 172.22.142.128172.22.142.128 and the broadcast address is 172.22.142.255172.22.142.255. Consequently, the valid usable host IPv4 address range extends from 172.22.142.129172.22.142.129 up through 172.22.142.254172.22.142.254. Both 172.22.142.129172.22.142.129 and 172.22.142.200172.22.142.200 lie within this valid host range.

Step-by-Step Solution

1
Determine the prefix length and total block size.
A /25/25 prefix length leaves 3225=732 - 25 = 7 host bits, yielding 27=1282^7 = 128 total IP addresses per subnet.
Understanding prefix length establishes the block boundary size.
2
Identify the network ID and broadcast address for 172.22.142.128/25172.22.142.128/25.
Network ID = 172.22.142.128172.22.142.128; Directed Broadcast Address = 172.22.142.128+127=172.22.142.255172.22.142.128 + 127 = 172.22.142.255.
The network address is the starting boundary and the broadcast address is the ending boundary.
3
Calculate the usable host IP address range.
Usable host range spans from 172.22.142.129172.22.142.129 (Network ID + 1) to 172.22.142.254172.22.142.254 (Broadcast Address - 1).
Network and broadcast addresses must be excluded when reserving addresses for host interfaces.
4
Evaluate the candidate options against the usable host range.
172.22.142.129172.22.142.129 and 172.22.142.200172.22.142.200 fall strictly within the usable host range, while 172.22.142.128172.22.142.128 (Network ID) and 172.22.142.255172.22.142.255 (Broadcast) do not.
Only IP addresses strictly within the usable host range are assignable to hosts.

Key Concept

IPv4 Subnet Boundaries and Usable Host Allocation
Question 310Question

A system administrator is configuring a subnet for a new department containing 45 desktop computers and 3 network printers. Which of the following CIDR prefix lengths provides the minimum necessary capacity to accommodate all 48 devices while minimizing wasted IP addresses?

Show answer & explanation

Answer: /26

Answer

The prefix length /26 is the correct choice as it provides 62 usable host addresses, accommodating the required 48 devices with minimal address waste.
To support 48 host devices (45 PCs + 3 printers), a subnet must accommodate at least 48 usable addresses. A /26 subnet provides 262=622^6 - 2 = 62 usable IP addresses, which satisfies the host requirement while allocating the smallest possible subnet block.

Step-by-Step Solution

1
Calculate the total host IP addresses required.
45 workstations + 3 printers = 48 total host devices required.
Every active host interface requires a unique usable IP address on the subnet.
2
Determine the required host bits (h) using the formula 2h2required_hosts2^h - 2 \ge \text{required\_hosts}.
For h=5h = 5: 252=302^5 - 2 = 30 usable hosts (insufficient). For h=6h = 6: 262=622^6 - 2 = 62 usable hosts (sufficient).
Two IP addresses in every subnet are reserved for the network ID and broadcast address.
3
Convert host bits (h) to CIDR prefix length.
32 total IPv4 bits - 6 host bits = /26 prefix length.
Subtracting host bits from 32 yields the network prefix length.

Key Concept

Determining minimum CIDR prefix length based on host capacity requirements
Question 311Question

A network administrator is designing a local network topology across two switches using multiple VLANs and an 802.1Q trunk link. Which TWO of the following statements correctly describe Layer 2 switch operations and VLAN encapsulation in this environment?

Select all that apply

Show answer & explanation

Answer: VLANs partition a physical switch into multiple distinct Layer 2 broadcast domains.; The switch inserts a 4-byte 802.1Q tag into the Ethernet frame header when forwarding traffic for non-native VLANs across a trunk link.

Answer

VLANs partition a physical switch into distinct Layer 2 broadcast domains, and an 802.1Q trunk inserts a 4-byte tag header into Ethernet frames for non-native VLAN traffic.
VLANs create logically separate Layer 2 broadcast domains on a physical switch, restricting broadcast traffic and isolating subnets. Across trunk connections, IEEE 802.1Q multiplexes VLAN traffic by inserting a 4-byte tag header into standard Ethernet frames for non-native VLANs.

Step-by-Step Solution

1
Analyze the fundamental definition and purpose of VLAN segmentation at Layer 2.
Confirm that each VLAN functions as an isolated broadcast domain within the switch infrastructure.
Traffic broadcasted within one VLAN does not cross over to ports in another VLAN without Layer 3 routing.
2
Examine how 802.1Q trunking multiplexes traffic from multiple VLANs across a single physical link.
Identify that 802.1Q inserts a 4-byte VLAN tag into the frame header for non-native VLAN traffic.
Tagging allows receiving switches to correctly identify the destination VLAN of incoming frames.
3
Evaluate the incorrect statements regarding inter-VLAN routing, native VLAN frame handling, and STP priority.
Reject statements claiming direct Layer 2 inter-VLAN communication, untagged frame dropping on trunks, and high numerical STP priority root selection.
Inter-VLAN traffic requires Layer 3 routing, native VLAN handles untagged traffic, and STP prefers lower bridge priority values.

Key Concept

VLAN Broadcast Isolation and 802.1Q Trunk Tagging
Estimated Time:1m 30s
Question 312Question

A network technician is configuring an email client to retrieve user messages securely from a remote mail server using implicit TLS encryption. Which default port number should be configured for the Internet Message Access Protocol Secure (IMAPS)?

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Answer: 993

Answer

Port 993 is the default TCP port used for secure Internet Message Access Protocol (IMAPS).
Port 993 is the industry-standard well-known port designated for IMAP over SSL/TLS (IMAPS), providing encrypted email access between client and server.

Step-by-Step Solution

1
Identify the required protocol and security mode specified in the scenario.
The scenario requires configuring encrypted email retrieval using IMAP over TLS/SSL (IMAPS).
Email applications distinguish between unencrypted and encrypted retrieval protocols by using different well-known TCP ports.
2
Map IMAPS to its standard well-known port number.
IMAPS operates on TCP port 993 by default.
Standard network conventions assign TCP port 143 to unencrypted IMAP and TCP port 993 to implicit SSL/TLS encrypted IMAPS.

Key Concept

Common Ports and Protocols for Secure Mail Retrieval
Question 313Question

A network administrator needs to connect two routers across a redundant network infrastructure. The administrator wants the routers to automatically exchange reachability information and update their routing tables dynamically whenever a network path fails. Which of the following options should be configured to meet this requirement?

Show answer & explanation

Answer: Dynamic interior gateway protocol (such as OSPF)

Answer

Dynamic interior gateway protocol (such as OSPF)
Dynamic interior gateway protocols (such as OSPF, EIGRP, or RIP) allow routers to automatically discover neighboring routers, exchange network routes, and dynamically update path selections when topology changes or link failures occur.

Step-by-Step Solution

1
Analyze the operational requirements
Identified the need for automatic reachability exchange and dynamic topology convergence during link failures.
The scenario highlights that routing tables must automatically update without manual administrator intervention when network changes occur.
2
Evaluate routing mechanism options
Dynamic routing protocols dynamically adapt to link changes, whereas static routes require manual reconfiguration.
Interior Gateway Protocols (IGPs) like OSPF periodically or event-driven exchange routing messages to discover paths automatically.

Key Concept

Dynamic Routing Protocol Functionality
Question 314Question

A network engineer intends to establish a redundant path to a remote office subnet 10.45.0.0/1610.45.0.0/16. The primary path learns the route dynamically via eBGP (administrative distance of 2020). The engineer attempts to configure a backup floating static route through an IPsec tunnel interface, intended to activate only if the primary eBGP session fails. However, the engineer enters the configuration command: `ip route 10.45.0.0 255.255.0.0 172.16.1.2 15`. Under normal operating conditions with both connections active, which path will the router install into the routing table for traffic destined to 10.45.12.3010.45.12.30, and why?

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Answer: The router will install the static route via next-hop 172.16.1.2 because lower administrative distance values are prioritized over higher values, causing the static route (AD 15) to override the eBGP route (AD 20).

Answer

The router will select the static route via next-hop 172.16.1.2 because an administrative distance of 15 is numerically lower and therefore more preferred than eBGP's administrative distance of 20.
When a router receives multiple routes for the exact same destination prefix length from different routing protocols or administrative sources, it uses Administrative Distance (AD) to decide which route to populate into the routing table. Lower AD values represent higher preference. Because eBGP has a default AD of 20 and the misconfigured static route was given an AD of 15, the router prefers 15 over 20, causing the intended 'backup' static route to be active during normal operation.

Step-by-Step Solution

1
Analyze prefix lengths for incoming traffic
Both the eBGP route and the static route match the target IP address 10.45.12.30 with an identical prefix mask of /16 (10.45.0.0/16).
Longest Prefix Match (LPM) is evaluated first. Since both masks are identical (/16), the router moves to comparing Administrative Distance.
2
Compare Administrative Distance (AD) values of competing routing sources
The static route has an AD of 15 (explicitly configured), while eBGP has a default AD of 20.
Administrative distance measures route trustworthiness across different protocol sources, where lower numerical values indicate higher trustworthiness.
3
Determine the active route installed into the Routing Information Base (RIB)
Because 15 is lower than 20, the router selects the static route via next-hop 172.16.1.2 instead of the dynamic eBGP route.
To create a true floating backup static route for eBGP (AD 20), the engineer should have configured an AD greater than 20 (e.g., AD 25 or 200).

Key Concept

Administrative Distance vs Floating Static Route Configuration
Question 315Question

Match each core IP addressing service component to its corresponding primary functional role within an enterprise network architecture.

Click a left item, then click its matching right item

Items

DHCP Option 82
Dynamic DNS (DDNS / Option 81)
IP Address Management (IPAM)
Stratum 1 NTP Server

Matches

Show answer & explanation

Answer

DHCP Option 82 matches with appending switch port and device MAC metadata to lease requests. Dynamic DNS matches with automatically registering forward and reverse name resolution entries. IP Address Management (IPAM) matches with centralized tracking, scope utilization monitoring, and cross-platform synchronization of network address space. Stratum 1 NTP Server matches with synchronizing system clocks directly from an attached hardware reference source such as GPS.
Each service component is matched accurately based on its core networking role: DHCP Option 82 provides relay metadata insertion (Circuit ID/Remote ID); Dynamic DNS maintains dynamic host-to-IP resource records (A/AAAA/PTR); IPAM delivers centralized governance over address spaces and scope metrics; and Stratum 1 NTP servers link directly to hardware time sources like GPS receivers.

Step-by-Step Solution

1
Analyze DHCP Option 82 functional purpose
Identified that Option 82 is the Relay Agent Information Option used by network switches/routers to attach Circuit ID and Remote ID details.
Relay agents add geographic or topological metadata to aid DHCP server assignment policies.
2
Analyze Dynamic DNS (DDNS / Option 81) operations
Identified that DDNS updates forward (A/AAAA) and reverse (PTR) DNS records dynamically upon lease events.
Prevents stale host resolution data when IP leases are dynamically assigned via DHCP.
3
Analyze IP Address Management (IPAM) features
Identified that IPAM offers consolidated discovery, tracking, and scope utilization monitoring across infrastructure DNS/DHCP deployments.
Eliminates spreadsheet management by centralizing IP space management across active network services.
4
Analyze NTP Stratum hierarchy definitions
Identified that Stratum 1 servers connect directly to Stratum 0 time sources (GPS, atomic clocks).
Stratum level indicates distance from the reference clock, where Stratum 1 represents direct attachment.

Key Concept

IP Addressing Services and Enterprise Supporting Infrastructure
Estimated Time:2m 0s
Question 316Question

A network administrator is establishing stateful firewall rules to allow remote administrative workstations to securely manage internal Linux servers via encrypted command-line sessions and authenticate against a directory server using SSL/TLS encryption. Unencrypted management traffic must remain strictly blocked. Which TWO port and transport protocol combinations must be explicitly permitted to support these administrative requirements? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: TCP port 22; TCP port 636

Answer

TCP port 22 (for SSH) and TCP port 636 (for LDAPS) must be explicitly permitted.
Encrypted remote terminal access is provided by SSH over TCP port 22. Encrypted directory services authentication is provided by LDAPS over TCP port 636. Both protocols require connection-oriented TCP transport to establish stateful, reliable cryptographic sessions.

Step-by-Step Solution

1
Identify the protocol required for secure remote command-line administration.
Secure Shell (SSH) is required, which operates on standard TCP port 22.
SSH replaces unencrypted Telnet (TCP 23) to satisfy secure remote administration policies.
2
Identify the protocol required for encrypted directory service queries.
Lightweight Directory Access Protocol over SSL/TLS (LDAPS) is required, which operates on standard TCP port 636.
LDAPS encrypts directory authentication traffic, whereas unencrypted LDAP uses TCP port 389.
3
Verify the correct Transport Layer protocol for both services.
Both SSH and LDAPS rely on TCP for connection reliability and TLS/session handling.
UDP is connectionless and cannot support the reliable handshakes required by SSH and TLS encryption streams.

Key Concept

Standard service port numbers and transport protocol selection for secure remote access and directory services
Question 317Question

A network administrator is designing a dedicated, fault-tolerant interconnect architecture for 55 mission-critical database nodes within a single data center rack. To ensure continuous operation without a central point of failure, every node must have a direct point-to-point physical cable connection to every other node in the cluster. Which network topology is being implemented, and what is the total number of physical cabling links required for this configuration?

Show answer & explanation

Answer: Full mesh topology requiring 10 physical links

Answer

Full mesh topology requiring 10 physical links
The design specifies a fully redundant layout where every node has a direct, dedicated physical connection to every other node, which defines a full mesh topology. The total number of required links for n=5n = 5 nodes is calculated using the formula n(n1)2\frac{n(n-1)}{2}, resulting in 5×(51)2=10\frac{5 \times (5-1)}{2} = 10 physical links.

Step-by-Step Solution

1
Identify the topology based on structural requirements
Full mesh topology
Connecting every node directly to every other node without an intermediary device defines a physical full mesh topology.
2
Calculate required physical links using the mesh formula
10 physical links
The formula for calculating un-directed connections in a full mesh network with nn nodes is L=n(n1)2L = \frac{n(n-1)}{2}. Substituting n=5n = 5 yields 5×42=10\frac{5 \times 4}{2} = 10.

Key Concept

Full Mesh Topology and Link Calculation
Estimated Time:1m 30s
Question 318Question

In an enterprise network running Per-VLAN Spanning Tree Plus (PVST+), two switches (Switch A and Switch B) are interconnected. Switch A has a base MAC address of 00:11:22:33:44:55 and is configured with an STP base priority of 28672 for VLAN 50. Switch B has a base MAC address of 00:11:22:33:44:AA and currently operates with the default STP base priority of 32768 for VLAN 50. A network administrator wants Switch B to become the Root Bridge for VLAN 50. Which priority value must be configured on Switch B for VLAN 50 to guarantee it becomes the Root Bridge?

Show answer & explanation

Answer: 24576

Answer

Setting the base priority of Switch B to 24576 ensures it achieves a lower Bridge ID than Switch A and becomes the Root Bridge for VLAN 50.
In Spanning Tree Protocol, the switch with the lowest Bridge ID (BID) is elected as the Root Bridge. BID consists of Bridge Priority + System ID Extension (VLAN ID) + Base MAC Address. Configured in steps of 4096, setting Switch B's priority to 24576 results in a total priority of 24626, lower than Switch A's total priority of 28722, securing the Root Bridge election for Switch B.

Step-by-Step Solution

1
Calculate Switch A's total Bridge ID priority for VLAN 50
28672 (base priority) + 50 (System ID Extension for VLAN 50) = 28722
PVST+ combines the 4-bit priority field with the 12-bit VLAN ID extension to form the total priority portion of the Bridge ID.
2
Evaluate the outcome if Switch B matches Switch A's base priority of 28672
Total priority tie (28722 vs 28722). Switch A wins based on lower MAC address (00:11:22:33:44:55 < 00:11:22:33:44:AA).
When total priorities are equal, Spanning Tree Protocol uses the switch base MAC address as the ultimate tiebreaker, where lower numerical value wins.
3
Determine the required priority increment to guarantee Switch B wins
Select the next lower valid STP priority increment of 4096 below 28672, which is 24576.
Bridge priorities must be configured in multiples of 4096. A base priority of 24576 gives Switch B a total priority of 24626, which is lower than Switch A's 28722.

Key Concept

STP Root Bridge Selection and Priority Tie-Breaking Logic
Estimated Time:2m 0s
Question 319Question

When a router receives route advertisements from different sources for the exact same destination network, it evaluates their trustworthiness using Administrative Distance (AD). How should the following route sources be ordered from the lowest administrative distance (most trustworthy) to the highest administrative distance (least trustworthy)?

Drag items to arrange them in the correct order

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Answer

The correct sequence from lowest to highest Administrative Distance is: Directly connected interface (AD = 00), Static route (AD = 11), OSPF dynamic route (AD = 110110), and RIP dynamic route (AD = 120120).
Administrative Distance (AD) is a feature used by routers to select the best path when there are two or more different routes to the same destination from different routing protocols. The default AD values in increasing order are: Directly connected interfaces (00), Static routes (11), OSPF (110110), and RIP (120120).

Step-by-Step Solution

1
Identify the Administrative Distance (AD) value for each route source
Directly connected interface = 00, Static route = 11, OSPF = 110110, RIP = 120120.
Administrative distance measures route reliability on a scale from 00 to 255255, where lower values are preferred.
2
Arrange the sources in ascending numeric order of their AD values
00 (Directly connected) < 11 (Static) < 110110 (OSPF) < 120120 (RIP).
Lower AD values indicate higher trust and are selected first by the router.

Key Concept

Administrative Distance Hierarchy
Question 320Question

A network security team is replacing insecure legacy protocols across core infrastructure routers. The new baseline policy requires allowing only encrypted protocols for remote command-line administration and web-based graphical configuration. Which TWO of the following transport protocols and default port numbers must be permitted through the administrative management access control list (ACL) to meet these requirements?

Select all that apply

Show answer & explanation

Answer: SSH operating over TCP port 22; HTTPS operating over TCP port 443

Answer

SSH operating over TCP port 22 and HTTPS operating over TCP port 443 are the correct secure choices.
SSH on TCP port 22 provides secure, encrypted shell access for command-line management, while HTTPS on TCP port 443 provides encrypted web management using TLS. Together, these two fulfill the requirement for secure remote administration across CLI and GUI channels.

Step-by-Step Solution

1
Identify the remote management functional requirements.
Requirements specify encrypted command-line administration and secure web-based GUI configuration.
Cleartext protocols must be filtered out in favor of secure cryptographically backed alternatives.
2
Match secure command-line administration to its protocol and port.
SSH utilizes TCP port 22 to encrypt remote shell sessions.
Telnet on TCP port 23 is unencrypted and unacceptable.
3
Match secure web configuration to its protocol and port.
HTTPS utilizes TCP port 443 to encrypt HTTP traffic using TLS.
HTTP on TCP port 80 lacks payload encryption.

Key Concept

Secure remote management protocol and port pairing
Estimated Time:1m 15s
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