Network Implementation

424 questions

All types424 Multiple choice220 Multiple select80 Numeric11 Matching53 Ordering42 Short answer18

A wireless network administrator observes that roaming wireless clients in a multi-story corporate building consistently maintain connections to distant access points (APs) at very low received signal strength indicator (RSSI) values instead of roaming to nearer APs. This causes overall network degradation due to low data rates consuming excessive airtime. A physical site survey confirms that AP placement and transmit power levels are properly balanced across the floor plan. Which of the following configuration changes on the wireless LAN controller best remediates this sticky client behavior while optimizing overall airtime efficiency?

Disabling legacy data rates (such as

1\text{ Mbps}

through

11\text{ Mbps}

) and raising the minimum basic data rate threshold on the wireless network

Configuring adjacent

2.4\text{ GHz}

access points to operate on overlapping channels such as channels

1

2

, and

3

Increasing access point radio transmit power to maximum levels across all deployed

2.4\text{ GHz}

and

5\text{ GHz}

radios

Configuring

80\text{ MHz}

channel bonding across all

2.4\text{ GHz}

access point radios to increase client throughput

Show answer & explanation

Answer: Disabling legacy data rates (such as $1\text{ Mbps}$ through $11\text{ Mbps}$ ) and raising the minimum basic data rate threshold on the wireless network

Answer

Disabling legacy data rates and raising the minimum basic data rate threshold on the wireless network

Disabling low legacy data rates and increasing the minimum basic rate forces clients to disassociate and seek a stronger access point once their signal degrades below the set threshold. This prevents clients from lingering on distant access points at low transmission rates, freeing up wireless channel airtime and resolving sticky client issues.

Step-by-Step Solution

Analyze the problem symptoms and underlying cause

Identify that sticky clients remain connected to distant APs at very low RSSI, transmitting at low data rates (e.g.,

1\text{ Mbps} - 11\text{ Mbps}

) which consumes excessive channel airtime.

Client devices evaluate roaming thresholds based on signal strength and data rate availability broadcast by the access point.

Evaluate wireless LAN controller rate tuning mechanisms

Disabling mandatory support for low legacy data rates shrinks the effective management cell boundary of each AP without altering RF physical power.

When a client drops below the configured minimum basic rate (e.g.,

12\text{ Mbps}

24\text{ Mbps}

), it can no longer maintain management frames with that AP and is forced to roam to a closer AP with higher RSSI.

Verify airtime fairness impact

Eliminating low bitrates frees up valuable airtime for high-speed transmissions, improving total throughput for all connected wireless clients.

Slower bitrates require significantly longer channel occupancy times to transmit the same payload compared to higher OFDM rates.

Key Concept

Minimum Basic Data Rate Tuning and Airtime Fairness

Question 322Question →

A network administrator is configuring dual-stack services on an enterprise internal network. The administrator needs to create a DNS entry on the local domain controller so client systems can resolve the host name `app.corp.local` to its assigned IPv6 address `2001:db8:10:50::100`. Which of the following DNS record types should be created?

AAAA record

A record

CNAME record

PTR record

Show answer & explanation

Answer: AAAA record

Answer

An AAAA record must be configured on the DNS server.

The quad-A (AAAA) record is specifically designed to resolve domain names to 128-bit IPv6 addresses in DNS forward lookup zones.

Step-by-Step Solution

Identify the destination address family required for host name resolution.

The target address `2001:db8:10:50::100` is a 128-bit IPv6 address.

Selecting the appropriate DNS resource record type requires matching the record type with the IP version being resolved.

Select the corresponding forward lookup DNS record type for IPv6.

An AAAA record maps host names directly to IPv6 addresses.

Standard IPv4 host records use A records, whereas IPv6 uses AAAA (quad-A) records.

Key Concept

DNS Record Types for IPv6 Deployment

Question 323Question →

A network consultant is designing a high-density Wi-Fi deployment for a university sports arena housing thousands of concurrent wireless clients. To minimize co-channel interference (CCI), optimize spectrum efficiency, and maintain reliable roaming performance across adjacent access point (AP) cells, which of the following deployment strategies should the consultant implement? (Select TWO.)

Select all that apply

Install directional antennas focused downward onto designated seating sections to contain RF coverage cells and reduce cell overlap.

Configure wide

80\text{ MHz}

channel bonding across all

5\text{ GHz}

access point radios to maximize theoretical throughput for individual clients.

Utilize Dynamic Frequency Selection (DFS) channels in the

5\text{ GHz}

band to increase the total number of non-overlapping channels available for AP allocation.

Set

2.4\text{ GHz}

radio transmit power to maximum output across all APs to ensure deep signal penetration through concrete structural elements.

Assign contiguous channels such as Channel 2, Channel 4, and Channel 7 across adjacent

2.4\text{ GHz}

radios to distribute traffic.

Show answer & explanation

Answer: Install directional antennas focused downward onto designated seating sections to contain RF coverage cells and reduce cell overlap.; Utilize Dynamic Frequency Selection (DFS) channels in the $5\text{ GHz}$ band to increase the total number of non-overlapping channels available for AP allocation.

Answer

The consultant should install directional antennas focused downward onto designated seating sections to restrict RF coverage cell boundaries and utilize Dynamic Frequency Selection (DFS) channels in the

5\text{ GHz}

spectrum to expand available non-overlapping channels.

In high-density environments such as sports arenas, containment of RF coverage cells and maximization of available non-overlapping spectrum are critical. Installing directional antennas focuses signal propagation into targeted seating sectors, preventing RF leakage into neighboring AP coverage zones. Additionally, enabling Dynamic Frequency Selection (DFS) allows access points to operate on UNII-2 and UNII-2 Extended spectrum in the

5\text{ GHz}

band, significantly expanding the pool of available non-overlapping channels needed for dense channel reuse plans.

Step-by-Step Solution

Analyze high-density RF interference challenges

Identified co-channel interference (CCI) and channel reuse limits as the primary performance bottlenecks in a crowded stadium environment.

When hundreds of clients connect to nearby APs operating on the same spectrum, overlapping cells degrade throughput and cause contention.

Evaluate antenna pattern selection

Selected directional patch/sector antennas over omnidirectional antennas.

Directional antennas shape radiation patterns into narrow beams focused on specific seating sections, physically constraining RF propagation and preventing cell overlap.

Evaluate channel planning in the

5\text{ GHz}

spectrum

Selected DFS channel inclusion while avoiding wide channel bonding (

80\text{ MHz}

DFS unlocks additional

5\text{ GHz}

channels (UNII-2 and UNII-2 Extended), allowing a dense matrix of APs to operate on non-overlapping channels. Wide

80\text{ MHz}

channels deplete available channels, exacerbating CCI.

Key Concept

High-Density Wireless RF Cell Containment and DFS Spectrum Allocation

Estimated Time:2m 0s

Question 324Question →

Three switches—Switch A, Switch B, and Switch C—are interconnected in a redundant Layer 2 topology running standard IEEE 802.1D Spanning Tree Protocol (STP). Switch A is configured with a bridge priority of 4096, while Switch B and Switch C both retain the default bridge priority of 32768. The link between Switch A and Switch B is 1 Gbps (cost = 4), the link between Switch A and Switch C is 10 Gbps (cost = 2), and the link between Switch B and Switch C is 1 Gbps (cost = 4). Based on this topology and STP operational rules, which of the following statements correctly describe the resulting converged network state?

Select all that apply

Switch A is elected as the Root Bridge for the topology because it has the lowest bridge priority value.

Switch C's interface connected directly to Switch A is selected as Switch C's Root Port due to having the lowest path cost to the Root Bridge.

Switch B elects its interface connected to Switch C as a Designated Port because Switch B has a lower MAC address than Switch C.

Switch B can force itself to become the Root Bridge by increasing its bridge priority setting to a higher numerical value.

A native VLAN mismatch on the trunk between Switch B and Switch C will force Switch C to automatically become a secondary Root Bridge.

Show answer & explanation

Answer: Switch A is elected as the Root Bridge for the topology because it has the lowest bridge priority value.; Switch C's interface connected directly to Switch A is selected as Switch C's Root Port due to having the lowest path cost to the Root Bridge.

Answer

Switch A is elected as the Root Bridge for the topology because it has the lowest bridge priority value, and Switch C's interface connected directly to Switch A is selected as Switch C's Root Port due to having the lowest path cost to the Root Bridge.

Switch A becomes the Root Bridge because its priority of 4096 is lower than all other switches. Switch C selects its direct 10 Gbps interface (cost 2) as its Root Port because it provides the lowest cumulative path cost to the Root Bridge compared to any alternate route.

Step-by-Step Solution

Determine the Root Bridge election result by comparing Bridge IDs (Bridge Priority + MAC Address).

Switch A has priority 4096, which is lower than Switch B (32768) and Switch C (32768). Switch A becomes the Root Bridge.

STP selects the switch with the lowest numerical Bridge ID as the Root Bridge.

Determine Root Ports for non-root switches by evaluating cumulative root path costs.

Switch C's direct link to Switch A has a cost of 2 (10 Gbps link), which is lower than the indirect path through Switch B (cost 8). The direct port is chosen as Switch C's Root Port.

Each non-root switch chooses the single port with the lowest path cost back to the Root Bridge.

Evaluate the port roles on the link between Switch B and Switch C.

Switch C advertises a cost of 2 to reach the root, while Switch B advertises a cost of 4. Switch C wins designated port status on that link, placing Switch B's port on that segment in the Blocking state.

Designated Ports on shared segments are assigned to the switch with the lowest root path cost.

Key Concept

Spanning Tree Protocol Operations (Root Bridge Election, Path Cost Calculation, and Port Role Determination)

Question 325Question →

A network administrator is configuring DNS resource records for a newly deployed service hostname `app.example.com`. The service requires mapping the domain directly to an IPv6 address and publishing domain ownership verification text. A diagnostic `dig` command yields the following output:

;; QUESTION SECTION:
;app.example.com. IN ANY

;; ANSWER SECTION:
app.example.com. 3600 IN A 192.0.2.45

Which TWO of the following DNS record types must the administrator add to the authoritative zone file to satisfy these technical requirements?

Select all that apply

AAAA record

TXT record

PTR record

SRV record

CNAME record

Show answer & explanation

Answer: AAAA record; TXT record

Answer

The administrator must add an AAAA record to resolve the hostname to an IPv6 address and a TXT record to publish domain ownership verification text.

To satisfy both requirements, an AAAA record is required to map the hostname to its corresponding 128-bit IPv6 address, and a TXT record is required to hold the domain verification text string.

Step-by-Step Solution

Analyze the technical requirements for hostname address mapping

Mapping `app.example.com` to an IPv6 address requires an IPv6 host record.

An AAAA (quad-A) record specifically handles 128-bit IPv6 address resolution for domain names.

Analyze the technical requirements for verification data storage

Publishing verification strings requires an arbitrary text storage record.

TXT records allow domain owners to store free-form text strings in DNS zone files for third-party domain validation and security policies.

Key Concept

DNS Resource Record Types (AAAA and TXT)

Estimated Time:1m 30s

Question 326Question →

Four network switches (SW1, SW2, SW3, SW4) are connected in a local area network running standard IEEE 802.1D Spanning Tree Protocol (STP). The bridge priorities and base MAC addresses for the switches are configured as follows:

- SW1: Bridge Priority = $32768$ , MAC = $0200.1111.1111$
- SW2: Bridge Priority = $32768$ , MAC = $0200.2222.2222$
- SW3: Bridge Priority = $16384$ , MAC = $0200.3333.3333$
- SW4: Bridge Priority = $32768$ , MAC = $0200.0000.1111$

Which switch will be elected as the STP root bridge in this network topology?

SW3, because it possesses the lowest numerical bridge priority value.

SW4, because it possesses the lowest numerical MAC address among all switches.

SW2, because default priority ties force the election to select the highest MAC address.

SW1, because it retains the default bridge priority and lowest MAC among default-configured switches.

Show answer & explanation

Answer: SW3, because it possesses the lowest numerical bridge priority value.

Answer

SW3 is elected as the root bridge because it has the lowest Bridge Priority (16384), making its overall Bridge ID the lowest in the network.

In 802.1D Spanning Tree Protocol, the switch with the lowest Bridge Identifier (BID) becomes the root bridge. The BID is composed of the Bridge Priority and the MAC address. Bridge Priority is compared first; the lower numerical value wins. Since SW3 has a Bridge Priority of 16384 while SW1, SW2, and SW4 all have the default priority of 32768, SW3 wins the root bridge election automatically without needing a MAC address tie-breaker.

Step-by-Step Solution

Understand the Spanning Tree Protocol (STP) Bridge ID structure.

Bridge ID (BID) is calculated as:

\text{BID} = \text{Bridge Priority} + \text{MAC Address}

STP elects the switch with the lowest numerical BID as the root bridge.

Compare the Bridge Priority values of all participating switches.

SW1 = 32768, SW2 = 32768, SW3 = 16384, SW4 = 32768.

Bridge Priority is evaluated before MAC addresses.

Identify the switch with the lowest priority.

SW3 has a priority of 16384, which is lower than 32768.

Since SW3 has the lowest priority, MAC address tie-breaking is not needed for SW3.

Key Concept

STP Root Bridge Election Process

Estimated Time:1m 0s

Question 327Question →

Match each IPv6 address prefix or designated address on the left with its corresponding network deployment scope or functional purpose on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

fe80::/10 matches the link-local prefix; ff02::1 matches the all-nodes link-local multicast address; 2001:db8::/32 matches the documentation prefix; fc00::/7 matches the Unique Local Address (ULA) prefix.

Each IPv6 prefix serves a distinct functional scope in network deployment: fe80::/10 provides link-local unicast connectivity required on every active interface; ff02::1 targets all local nodes via link-scope multicast; 2001:db8::/32 is designated by RFC 3849 for technical documentation examples; and fc00::/7 provides private, non-globally routable enterprise addressing (Unique Local Address).

Step-by-Step Solution

Identify link-local unicast prefix

fe80::/10 is reserved for link-local communication within a single layer 2 domain.

Link-local addresses always start with fe80::/10 and are non-routable across routers.

Identify IPv6 multicast addresses

ff02::1 belongs to the ff00::/8 multicast block and targets all nodes on the link.

The ff02 scope indicates link-local multicast, and group ID 1 represents all nodes.

Identify documentation and private address spaces

2001:db8::/32 is for documentation, while fc00::/7 defines Unique Local Addresses.

RFC standards explicitly reserve 2001:db8::/32 for lab manuals/textbooks and fc00::/7 for private site-local style deployments.

Key Concept

IPv6 Address Scopes and Functional Block Classifications

Question 328Question →

A network administrator connects several new switches in a redundant topology to ensure high availability. Place the steps of the initial IEEE 802.1D Spanning Tree Protocol (STP) convergence process in the correct chronological order from link startup to topology stabilization.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct chronological order of IEEE 802.1D STP convergence is: 1) Switches originate BPDUs claiming to be the Root Bridge, 2) The switch with the lowest Bridge ID is elected as the Root Bridge, 3) Non-root switches elect their Root Ports, 4) Segment-level Designated Ports are elected, and 5) Remaining redundant ports enter the Blocking state.

The standard IEEE 802.1D STP algorithm operates sequentially: switches announce themselves as root via initial BPDUs, elect the true Root Bridge based on the lowest Bridge ID, select Root Ports on non-root switches, designate single forwarding ports per segment, and block remaining redundant links to eliminate loops.

Step-by-Step Solution

Identify initial BPDU transmission upon boot.

Switches send initial BPDUs asserting Root Bridge status.

Before exchanging information, switches lack topology visibility and default to claiming root status.

Complete Root Bridge election process.

The switch with the lowest Bridge Priority (or lowest MAC address as tiebreaker) becomes the agreed Root Bridge.

BPDUs with superior (lower) Bridge IDs override inferior claims across the broadcast domain.

Select non-root switch Root Ports.

Each non-root switch assigns its port with the lowest Root Path Cost as its Root Port.

Non-root switches require exactly one active path to send traffic toward the Root Bridge.

Elect segment-level Designated Ports.

The switch port on each segment with the lowest cost to the Root Bridge becomes the Designated Port.

Each LAN segment must have one designated forwarding interface to prevent loop formation.

Place unassigned redundant ports into Blocking state.

All non-root, non-designated ports transition to Blocking.

Ports not assigned active forwarding roles must block data traffic to eliminate physical switching loops.

Key Concept

Spanning Tree Protocol Convergence Sequence

Question 329Question →

Place the standard IEEE 802.1D Spanning Tree Protocol (STP) port states in chronological sequence from initial link activation on an unconfigured switch interface to full user data transmission capability.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct chronological sequence of IEEE 802.1D STP port states is Blocking State, Listening State, Learning State, and Forwarding State.

Under standard IEEE 802.1D STP operations, an activated interface progresses through four distinct operational states in order: Blocking (prevents loops by receiving BPDUs only), Listening (exchanges BPDUs to determine root bridge and port roles), Learning (builds MAC address tables without forwarding user traffic), and Forwarding (fully processes user data frames and BPDUs).

Step-by-Step Solution

Identify the initial state immediately following physical link establishment.

The port starts in the Blocking state to discard frames and listen for incoming BPDUs.

Immediate data forwarding upon link activation could cause bridging loops before topology information is evaluated.

Determine the second state after the initial Forward Delay timer expires.

The port transitions to the Listening state.

During Listening, the port actively transmits and receives BPDUs to participate in root bridge election and determine port roles without learning MAC addresses.

Determine the third state after the second Forward Delay timer expires.

The port transitions to the Learning state.

In the Learning state, the switch populates its MAC address table from incoming frame source addresses while continuing to hold user data frame forwarding.

Identify the final operational state.

The port enters the Forwarding state.

Once MAC table entry acquisition completes without loop detection, the port safely transitions into full user data frame processing.

Key Concept

Chronological IEEE 802.1D STP Port State Transitions

Question 330Question →

An enterprise organization is assigned the IPv6 global routing prefix $2001:\text{db8}:\text{abc}::/48$ . A network engineer is configuring a dedicated $/64$ subnet for VLAN $30$ (represented as hexadecimal $0030$ in the subnet field) and needs to assign the first usable host IPv6 address in that subnet to the default gateway interface. Which of the following represents the correctly compressed IPv6 address with its prefix length for the default gateway interface?

2001:db8:abc:30::1/64

2001:db8:abc::30::1/64

2001:db8:abc:0030:0:0:0:1/64

fe80:db8:abc:30::1/64

Show answer & explanation

Answer: 2001:db8:abc:30::1/64

Answer

2001:db8:abc:30::1/64

The correct answer properly applies RFC 5952 IPv6 address shorthand rules. The 16-bit subnet hextet 0030 suppresses leading zeros to become 30, and the contiguous run of zero hextets in the interface ID (0000:0000:0000:0001) is compressed into a single double colon (::), yielding 2001:db8:abc:30::1/64.

Step-by-Step Solution

Determine the uncompressed 128-bit IPv6 address structure.

Combining the global prefix (2001:0db8:0abc), the 16-bit subnet ID for VLAN 30 (0030), and the first host interface ID (0000:0000:0000:0001) yields 2001:0db8:0abc:0030:0000:0000:0000:0001.

A /48 global prefix allocates bits 49–64 for the subnet ID and bits 65–128 for the interface ID.

Suppress leading zeros in each 16-bit hextet.

0db8 becomes db8, 0abc becomes abc, 0030 becomes 30, and 0001 becomes 1. The address becomes 2001:db8:abc:30:0:0:0:1.

RFC 5952 mandates the suppression of leading zeros in every hextet.

Compress contiguous zero hextets.

The four consecutive zero hextets (:0:0:0:) are replaced by a single double colon (::), producing 2001:db8:abc:30::1/64.

RFC 5952 specifies replacing the longest run of consecutive zero hextets with a single '::' symbol.

Key Concept

IPv6 Address Compression Rules (RFC 5952)

Question 331Question →

A client workstation on a corporate network initiates a web connection to `service.partner.org`. Assuming no DNS records are cached on the client or intermediate servers, place the following DNS resolution steps in the correct chronological order from the initial request to the final address delivery.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct chronological sequence begins with the client stub resolver querying the local recursive resolver. The local resolver then performs iterative lookups by querying the Root Name Server, followed by the `.org` TLD Name Server, and finally the Authoritative Name Server for `partner.org`. Lastly, the local resolver caches the record and returns the IP address to the client.

In standard DNS architecture, name resolution follows a hierarchical top-down lookup pattern when un-cached. The client host sends a recursive query to its local DNS server. The local DNS server then performs iterative queries starting at the Root servers (top level), proceeding to the TLD servers (`.org`), and ending at the domain's Authoritative servers (`partner.org`). Finally, the local server caches the answer and forwards it back to the client host.

Step-by-Step Solution

Identify the initiating query

The client's stub resolver sends a recursive query to its configured local recursive resolver.

Clients do not traverse the DNS hierarchy directly; they rely on recursive resolvers.

Query the top of the DNS hierarchy

The local resolver queries a Root Name Server to find the TLD server for `.org`.

Root servers direct resolvers to the appropriate Top-Level Domain name servers.

Query the TLD server

The local resolver queries the `.org` TLD server to obtain the authoritative name servers for `partner.org`.

TLD servers maintain delegation records (NS records) for domains registered under their extension.

Query the Authoritative server

The local resolver queries the authoritative server for `partner.org` and retrieves the A/AAAA record for `service.partner.org`.

Authoritative servers store the actual DNS resource records for the targeted domain.

Deliver and cache the final result

The local resolver caches the answer and returns the IP address to the originating client host.

Caching speeds up future queries, and returning the answer allows the client host to establish a TCP/IP connection.

Key Concept

Recursive and Iterative DNS Resolution Flow

Question 332Question →

A network engineer is analyzing poor throughput and high packet loss reported by mobile client devices roaming through a multi-floor facility. A wireless spectrum analyzer indicates that three neighboring access points providing coverage on the same floor are configured to broadcast on 2.4 GHz Channels 1, 3, and 5, respectively. Which of the following best explains the primary cause of the performance degradation and the required resolution?

The channel selection causes severe adjacent-channel interference due to overlapping 22 MHz channel spectra; the access points should be reconfigured to non-overlapping Channels 1, 6, and 11.

The channel selection results in co-channel interference because odd-numbered channels share identical center frequencies; the access points should be reconfigured to operate on Channels 2, 4, and 6.

The channel selection triggers mandatory Dynamic Frequency Selection (DFS) radar avoidance mechanisms; the access points should be configured to disable DFS on 2.4 GHz and enable Transmit Power Control (TPC).

The channel selection causes multipath distortion from 40 MHz channel bonding; the access points should be restricted to Channels 1, 4, and 8 to prevent phase cancellation.

Show answer & explanation

Answer: The channel selection causes severe adjacent-channel interference due to overlapping 22 MHz channel spectra; the access points should be reconfigured to non-overlapping Channels 1, 6, and 11.

Answer

The channel selection causes severe adjacent-channel interference due to overlapping 22 MHz channel spectra; the access points should be reconfigured to non-overlapping Channels 1, 6, and 11.

In 2.4 GHz Wi-Fi networks, channels are 20-22 MHz wide but spaced only 5 MHz apart. Assigning adjacent access points to Channels 1, 3, and 5 causes severe spectral overlap (adjacent-channel interference), which corrupts packets and prevents clear channel assessment. Reconfiguring the deployment to use non-overlapping Channels 1, 6, and 11 ensures sufficient frequency separation to eliminate adjacent-channel interference.

Step-by-Step Solution

Analyze the frequency allocation in the 2.4 GHz ISM band.

In standard wireless networking, 2.4 GHz channels are spaced 5 MHz apart, while each channel requires a signal bandwidth of 20 to 22 MHz.

Understanding channel width and spacing is necessary to evaluate spectral overlap.

Evaluate the impact of deploying adjacent APs on Channels 1, 3, and 5.

Channel 1 spans 2.401-2.423 GHz, Channel 3 spans 2.411-2.433 GHz, and Channel 5 spans 2.421-2.443 GHz. Their frequencies significantly overlap with each other, leading to uncoordinated adjacent-channel interference (ACI).

ACI degrades performance because overlapping transmissions create RF noise rather than manageable CSMA/CA contention.

Determine the proper non-overlapping channel scheme.

Channels 1, 6, and 11 have center frequencies of 2.412 GHz, 2.437 GHz, and 2.462 GHz (25 MHz spacing), providing complete isolation without spectral overlap.

Using non-overlapping channels allows adjacent access points to operate concurrently without interfering with each other's transmissions.

Key Concept

2.4 GHz Non-Overlapping Channel Allocation

Estimated Time:1m 30s

Question 333Question →

A network administrator is deploying IPv6 on an enterprise subnet. The design requires host devices to automatically generate their own global unicast IP addresses using prefix information advertised by the local router, while simultaneously obtaining DNS server IP addresses from a central DHCPv6 server. Which of the following configurations and address assignments correctly support this deployment strategy? (Select TWO.)

Select all that apply

The local router must transmit Router Advertisement (RA) frames with the Managed Address Configuration (M) flag set to 0 and the Other Configuration (O) flag set to 1.

The enterprise DNS server must be configured with IPv4 A records to resolve hostnames to the newly assigned IPv6 global unicast addresses.

Host devices dynamically formulate their 128-bit global unicast address by combining the advertised /64 network prefix with an interface identifier derived via SLAAC.

The default gateway on host interfaces must be manually configured using the compressed address 2001:db8::1::10/64.

Show answer & explanation

Answer: The local router must transmit Router Advertisement (RA) frames with the Managed Address Configuration (M) flag set to 0 and the Other Configuration (O) flag set to 1.; Host devices dynamically formulate their 128-bit global unicast address by combining the advertised /64 network prefix with an interface identifier derived via SLAAC.

Answer

The correct configurations are setting the Router Advertisement flags to M=0 and O=1 (stateless DHCPv6 mode) and allowing host devices to combine the advertised /64 network prefix with a SLAAC-generated interface identifier.

Stateless DHCPv6 mode allows clients to derive their global unicast IPv6 address using SLAAC (combining the router's /64 prefix advertisement with a 64-bit host interface ID) while querying a DHCPv6 server for DNS details triggered by the RA flag configuration M=0 and O=1.

Step-by-Step Solution

Analyze the autoconfiguration requirement for host address generation.

Hosts generating their own global unicast addresses from advertised prefixes require SLAAC, which uses RA prefix information combined with a 64-bit interface identifier.

SLAAC eliminates the need for a stateful DHCPv6 server to assign IP addresses.

Determine the required Router Advertisement (RA) flag settings for Stateless DHCPv6.

The Managed Address Configuration flag must be M=0 (no stateful IP assignment) and the Other Configuration flag must be O=1 (obtain additional settings like DNS via DHCPv6).

This combination signals to clients that IP addressing comes from SLAAC while DNS server information comes from a stateless DHCPv6 server.

Evaluate record types and address compression syntax.

DNS requires AAAA records for IPv6 mapping, and valid IPv6 address notation permits only one double-colon (::) per address.

A records only support 32-bit IPv4 addresses, and multiple double-colons create unresolvable ambiguity in 128-bit address expansion.

Key Concept

Stateless Address Autoconfiguration (SLAAC) with Stateless DHCPv6

Question 334Question →

A network systems engineer is configuring name resolution for a newly deployed dual-stack web server. Workstations on the corporate network can successfully connect to the server over IPv4 using the hostname `app.finance.net`. However, IPv6 connectivity attempts fail because the hostname does not resolve to an IPv6 address.

To troubleshoot, the engineer executes the following diagnostic query from a workstation:

text
$ dig AAAA app.finance.net +short

The command produces no output, whereas running `dig A app.finance.net +short` returns `172.16.40.25`.

Which resource record type must be configured in the authoritative DNS zone file to enable host name resolution to the server's IPv6 address?

An AAAA record mapping app.finance.net to the server's 128-bit IPv6 address

A PTR record mapping the IPv4 address 172.16.40.25 to the server's IPv6 address

A CNAME record mapping app.finance.net to port 53 on the IPv6 host

A TXT record defining the socket transport protocol for IPv6 DNS queries

Show answer & explanation

Answer: An AAAA record mapping app.finance.net to the server's 128-bit IPv6 address

Answer

An AAAA record mapping the hostname directly to the server's 128-bit IPv6 address must be added to the zone file.

In DNS infrastructure, forward name resolution for IPv6 endpoints requires an AAAA (Quad-A) record, which maps a fully qualified domain name (FQDN) to a 128-bit IPv6 address. Because `dig AAAA app.finance.net` returned no answer, creating an AAAA record in the authoritative zone file directly resolves the issue.

Step-by-Step Solution

Analyze the diagnostic command output

The dig AAAA query returned an empty response, confirming no Quad-A record exists for app.finance.net.

When a DNS server has an A record but lacks an AAAA record for a hostname, IPv6-capable clients cannot resolve the IPv6 endpoint address.

Determine the appropriate DNS record type required for IPv6 forward lookup

An AAAA (quad-A) record is required to map a domain hostname to a 128-bit IPv6 address.

Standard IPv4 forward resolution uses A records (32-bit), whereas IPv6 forward resolution requires AAAA records (128-bit).

Key Concept

DNS Record Types for Dual-Stack (IPv4/IPv6) Name Resolution

Estimated Time:1m 15s

Question 335Question →

A network systems engineer is designing an IPv6 addressing scheme for an internal enterprise network containing multiple subnets connected via internal routers. The design requirement dictates that internal systems must be routable across all corporate locations without being globally reachable on the public Internet. Which of the following IPv6 address types should be assigned to these hosts?

Unique Local addresses starting within the fd00::/8 prefix block

Link-Local addresses starting within the fe80::/10 prefix block

Global Unicast addresses starting within the 2000::/3 prefix block

Multicast addresses starting within the ff02::/16 prefix block

Show answer & explanation

Answer: Unique Local addresses starting within the fd00::/8 prefix block

Answer

Unique Local addresses starting within the fd00::/8 prefix block should be assigned.

Unique Local Addresses (ULA) using the fd00::/8 block (within fc00::/7) are designed specifically for internal site communication across routed networks. They permit multi-subnet internal routing while remaining non-routable on the public IPv6 Internet.

Step-by-Step Solution

Analyze the routability and accessibility requirements for the network scenario

Internal devices must communicate across routed subnets internally, but must remain unreachable from the public IPv6 Internet.

This requirement calls for an internally routable, non-globally scoped IPv6 unicast prefix.

Evaluate the candidate IPv6 address scopes against the criteria

Unique Local Addresses (ULA, prefix fc00::/7, commonly fd00::/8) serve as the IPv6 equivalent of IPv4 private addresses (RFC 1918). They route freely across internal routers while ISP routers discard them by default.

Link-Local addresses cannot cross routers, and Global Unicast addresses are publicly routable.

Key Concept

IPv6 Unique Local Address Scope & Deployment

Estimated Time:1m 0s

Question 336Question →

A network administrator is documenting the configuration of a enterprise web server. The server interface is assigned the uncompressed IPv6 address $2001:0\text{db}8:0000:0001:0000:0000:0000:0020$ . What is the correctly formatted, compressed IPv6 address according to RFC 5952 standard rules?

Show answer & explanation

Answer: 2001:db8:0:1::20; 2001:DB8:0:1::20

Answer

2001:db8:0:1::20

According to RFC 5952 standard rules for IPv6 address representation: 1) Leading zeros within each 16-bit field must be removed (`0db8` -> `db8`, `0001` -> `1`, `0020` -> `20`). 2) The longest contiguous sequence of all-zero 16-bit fields must be replaced with `::`. The sequence of three zero fields at hextets 5, 6, and 7 is longer than the single zero at hextet 3. Therefore, `0000:0000:0000` is compressed to `::`, while hextet 3 is represented as a single `0`. The resulting canonical address is `2001:db8:0:1::20`.

Step-by-Step Solution

Suppress leading zeros in each 16-bit hexadecimal field.

The fields `0db8`, `0000`, `0001`, and `0020` become `db8`, `0`, `1`, and `20` respectively.

RFC 5952 requires that leading zeros in any hextet must be omitted (e.g., `0db8` -> `db8`, `0001` -> `1`).

Identify all zero fields and locate the longest consecutive run of all-zero fields.

There is a single zero field at hextet 3 (`0`) and a run of three consecutive zero fields at hextets 5, 6, and 7 (`0000:0000:0000`).

RFC 5952 states that the double colon `::` must replace the single longest contiguous sequence of all-zero fields.

Apply double-colon `::` compression to the longest zero run.

Hextets 5, 6, and 7 are replaced by `::`, while hextet 3 remains as a single `0`.

The single zero at hextet 3 cannot be replaced with `::` because a double colon can only appear once in an address, and the run of three zeros takes precedence.

Combine the fields into the finalized compressed address string.

`2001:db8:0:1::20`

This represents the strict RFC 5952 canonical representation of the IPv6 address.

Key Concept

RFC 5952 IPv6 Address Text Representation Rules

Question 337Question →

An IT technician is auditing a wireless network setup in an enterprise office operating on the $2.4\text{ GHz}$ band. Users report frequent packet loss and degraded network performance. The audit reveals that adjacent wireless access points are currently assigned to channels 1, 3, 5, and 7. Which channel re-assignment strategy should the technician implement to eliminate adjacent-channel interference (ACI)?

Reconfigure the access points to use non-overlapping channels 1, 6, and 11.

Reconfigure the access points to use channels 2, 4, 6, and 8 to evenly space out the frequency ranges.

Enable 40 MHz channel bonding on all access points to override overlapping channel frequencies.

Increase the transmit power on channels 3 and 5 to overpower interference from channels 1 and 7.

Show answer & explanation

Answer: Reconfigure the access points to use non-overlapping channels 1, 6, and 11.

Answer

Reconfigure the access points to use non-overlapping channels 1, 6, and 11.

In 802.11 wireless deployment on the

2.4\text{ GHz}

band, each channel spans

22\text{ MHz}

while channel center frequencies are spaced

5\text{ MHz}

apart. To prevent adjacent-channel interference (ACI), access points must be configured using channels that do not share frequency space. Channels 1, 6, and 11 are the standard non-overlapping channels that eliminate spectral overlap.

Step-by-Step Solution

Analyze the frequency structure of the 2.4 GHz Wi-Fi spectrum.

Each 2.4 GHz channel has a bandwidth of

22\text{ MHz}

, but channels are spaced only

5\text{ MHz}

apart.

Channels closer than 5 channel numbers apart (such as 1, 3, 5, 7) overlap significantly in frequency, creating adjacent-channel interference (ACI).

Identify the standard non-overlapping channel scheme for 2.4 GHz wireless deployments.

Channels 1, 6, and 11 have center frequencies of

2412\text{ MHz}

2437\text{ MHz}

, and

2462\text{ MHz}

, providing sufficient separation so their frequency envelopes do not collide.

Using 1, 6, and 11 ensures zero spectral overlap between neighboring access points.

Select the correct deployment reconfiguration.

Assigning channels 1, 6, and 11 across adjacent access points eliminates adjacent-channel interference.

This reconfiguration adheres to standard 802.11 wireless deployment practices for the 2.4 GHz band.

Key Concept

2.4 GHz Non-Overlapping Channel Assignment

Estimated Time:1m 15s

Question 338Question →

A network administrator is organizing corporate DNS infrastructure records to ensure proper service resolution and security compliance. Match each DNS resource record type on the left with its correct functional description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

SRV Record matches the definition of host and port location for services; PTR Record matches reverse DNS resolution from IP to domain; CNAME Record matches domain aliasing to another domain; TXT Record matches storing arbitrary text strings for email authentication like SPF and DKIM.

Each DNS record type has a dedicated function: SRV locates service protocols and port numbers; PTR enables reverse resolution of IP addresses to hostnames; CNAME acts as a domain alias pointing to another hostname; and TXT stores text metadata such as SPF and DKIM authentication signatures.

Step-by-Step Solution

Analyze SRV Record functionality

Identifies that SRV specifies hostnames along with port numbers and service protocols (e.g., SIP, LDAP).

SRV records are unique among standard records because they include protocol and port metadata.

Analyze PTR Record functionality

Identifies that PTR performs reverse mapping of IP addresses to canonical domain names.

PTR records populate reverse lookup zones.

Analyze CNAME Record functionality

Identifies that CNAME points an alias hostname to another canonical DNS name.

CNAME allows multiple subdomains to share the target canonical host resolution.

Analyze TXT Record functionality

Identifies that TXT contains freeform text strings used by protocols like SPF and DKIM for security verification.

TXT records provide human and machine readable metadata without altering routing.

Key Concept

DNS Resource Record Functions and Applications

Question 339Question →

A network engineer deploys four switches in a redundant enterprise topology running standard IEEE 802.1D Spanning Tree Protocol. The switches are configured with the following bridge priorities and MAC addresses:

• Switch-Alpha: Priority 4096, MAC 00:11:22:33:44:55
• Switch-Beta: Priority 8192, MAC 00:11:22:33:44:AA
• Switch-Gamma: Priority 4096, MAC 00:11:22:33:44:11
• Switch-Delta: Priority 32768, MAC 00:11:22:33:44:00

Which switch will be elected as the Root Bridge for this STP domain?

Switch-Gamma

Switch-Alpha

Switch-Delta

Switch-Beta

Show answer & explanation

Answer: Switch-Gamma

Answer

Switch-Gamma will be elected as the Root Bridge because it shares the lowest bridge priority value (4096) and has the lowest MAC address among the tied switches.

In Spanning Tree Protocol (STP), the Root Bridge election evaluates the Bridge Identifier (BID), which consists of a 2-byte Bridge Priority and a 6-byte MAC address. The switch with the lowest numerical BID becomes the Root Bridge. Priority is evaluated first; Switch-Alpha and Switch-Gamma tie with the lowest priority of 4096. Evaluating the MAC address as the tie-breaker reveals that Switch-Gamma's MAC address (00:11:22:33:44:11) is lower than Switch-Alpha's MAC address (00:11:22:33:44:55), making Switch-Gamma the Root Bridge.

Step-by-Step Solution

Evaluate the Bridge Priority values of all switches in the STP domain.

Switch-Alpha (4096) and Switch-Gamma (4096) have the lowest priority values, while Switch-Beta (8192) and Switch-Delta (32768) are eliminated.

STP elects the switch with the lowest numerical Bridge ID (Priority + MAC address). Priority is evaluated first.

Compare the MAC addresses of the switches tied for the lowest priority value.

Switch-Gamma's MAC address (00:11:22:33:44:11) is lower than Switch-Alpha's MAC address (00:11:22:33:44:55).

When priority values are identical, the switch with the lowest MAC address serves as the tie-breaker to determine the Root Bridge.

Key Concept

STP Root Bridge Election Mechanics

Question 340Question →

Match each wireless propagation concept or deployment parameter on the left with its corresponding description or operational impact on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Fresnel zone clearance matches maintaining an unobstructed elliptical RF boundary between wireless bridge endpoints to avoid signal degradation. Antenna gain matches increasing signal strength in a focused direction without raising total transmitter RF power output. Polarization alignment matches ensuring maximum signal transfer by orienting transmitting and receiving antennas along the same spatial plane. RF absorption matches attenuating RF signal amplitude as wave energy passes through dense physical obstacles like reinforced concrete.

Fresnel zone clearance involves maintaining an unobstructed elliptical path between point-to-point wireless nodes to avoid signal degradation from phase cancellation. Antenna gain describes focusing RF energy into specific beam patterns without altering actual transmitter power output. Polarization alignment requires aligning transmitting and receiving antenna vectors along the same axis to maximize received signal quality. RF absorption describes the attenuation of radio waves as they pass through dense physical obstructions such as concrete walls.

Step-by-Step Solution

Identify the definition and requirement for Fresnel zone clearance in wireless deployments.

Matches maintaining an unobstructed elliptical boundary between wireless bridge endpoints.

Physical obstacles inside the Fresnel zone cause reflection and phase cancellation that weaken signal strength.

Identify the physical behavior of antenna gain.

Matches increasing signal strength in a focused direction without raising transmitter output power.

Antennas achieve gain passively by directing radiated energy into narrowed beams rather than generating additional electrical energy.

Determine the principle of polarization alignment.

Matches ensuring maximum signal transfer by orienting transmitting and receiving antennas along the same spatial plane.

Mismatched antenna polarization (e.g., vertical vs horizontal) causes significant attenuation at the receiver.

Identify the mechanism of RF absorption in indoor/outdoor environments.

Matches attenuating RF signal amplitude as wave energy passes through dense physical obstacles like reinforced concrete.

Dense structural materials absorb RF energy, converting it into heat and attenuating signal propagation.

Key Concept

Wireless RF propagation characteristics and antenna deployment principles

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