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A network technician receives a ticket stating that remote users at a new satellite office cannot reach the company's internal intranet application. The technician interviews affected users, duplicates the failure on a test host, and receives a 'Destination Host Unreachable' message. The technician then formulates a hypothesis that an incorrect VLAN tagging policy on the WAN router interface is stripping traffic. To verify this, the technician runs a packet capture on the router interface and confirms that frames are arriving with an unexpected VLAN ID tag, validating the root cause. According to the CompTIA troubleshooting methodology, which of the following steps should the technician execute NEXT?

Establish a plan of action to resolve the issue and identify potential effects.

Perform complete end-to-end connectivity tests across all satellite office workstations.

Update the network infrastructure ticketing system with the final root cause analysis and resolution steps.

Question the obvious by verifying physical Layer 1 cable continuity and link status indicators on the switch.

Show answer & explanation

Answer: Establish a plan of action to resolve the issue and identify potential effects.

Answer

Establish a plan of action to resolve the issue and identify potential effects.

The scenario describes a technician who has already identified the problem, established a theory of probable cause, and tested the theory to confirm the root cause via packet capture analysis. According to the CompTIA troubleshooting methodology, once the theory is confirmed, the immediate next step is to establish a plan of action to resolve the problem and identify potential effects.

Step-by-Step Solution

Identify the current phase completed in the scenario.

The technician interviewed users, reproduced the issue (Step 1: Identify the problem), established a hypothesis (Step 2: Establish a theory), and confirmed the root cause via packet capture (Step 3: Test the theory to determine cause).

Determining the completed step is necessary to find the next sequential phase in the CompTIA methodology.

Identify the mandatory next step in the CompTIA 6-step troubleshooting methodology.

Step 4 is 'Establish a plan of action to resolve the problem and identify potential effects'.

After a theory is proven true, a plan must be formulated before implementing fixes or verifying system functionality.

Key Concept

CompTIA Troubleshooting Methodology Steps

Question 642Question →

A network router receives two identical candidate routes to destination network $192.168.50.0/24$ : one route was learned via OSPF (default administrative distance of $110$ ) and the other route was learned via RIP (default administrative distance of $120$ ). Assuming default protocol configurations are used, which route will the router install into its active routing table?

The OSPF route, because it has a lower administrative distance value than RIP

The RIP route, because distance-vector protocols bypass administrative distance checks at Layer 2

The RIP route, because lower numerical administrative distance values indicate lower protocol preference

The OSPF route, because OSPF automatically applies Port Address Translation (PAT) to override hop counts

Show answer & explanation

Answer: The OSPF route, because it has a lower administrative distance value than RIP

Answer

The OSPF route will be installed into the routing table because its default administrative distance of 110 is lower and therefore more trustworthy than RIP's default administrative distance of 120.

When a router receives multiple paths to the exact same destination network prefix from different routing protocols, it compares their Administrative Distance (AD). AD represents the trustworthiness of the source. Lower AD values are preferred over higher AD values. Because OSPF has a default AD of 110 and RIP has a default AD of 120, the router selects the OSPF route for installation into the routing table.

Step-by-Step Solution

Identify the administrative distance (AD) values for both competing routing sources.

OSPF has a default AD of 110; RIP has a default AD of 120.

Administrative distance measures the trustworthiness of a routing information source when multiple protocols learn the same prefix.

Compare the AD values to determine path selection precedence.

110 is less than 120.

Lower administrative distance values are always preferred by the router selection algorithm.

Select the winning route for entry into the active routing table.

The router installs the OSPF route into its active routing table.

The candidate route with the lowest administrative distance is selected.

Key Concept

Administrative Distance Precedence in Route Selection

Question 643Question →

A network administrator attempts to add VLAN 30 to an active 802.1Q trunk line between two distribution switches that currently carries VLAN 10 and VLAN 20. The administrator enters the following command on interface GigabitEthernet1/0/1:

`SW-Dist1(config-if)# switchport trunk allowed vlan 30`

Immediately after executing this command, users on VLAN 10 and VLAN 20 report loss of connectivity across the trunk link. Which of the following best explains why this connectivity issue occurred?

The command replaced the existing allowed VLAN list with only VLAN 30 rather than appending it to the list.

VLAN 30 became the new native VLAN on the interface, causing all untagged traffic to be dropped automatically.

VLAN 10 and VLAN 20 were converted from isolated broadcast domains into access mode ports on GigabitEthernet1/0/1.

The trunk link disabled IEEE 802.1Q frame tagging and reverted to unencapsulated Ethernet frame transmission.

Show answer & explanation

Answer: The command replaced the existing allowed VLAN list with only VLAN 30 rather than appending it to the list.

Answer

The command executed overwrote the entire allowed VLAN list on the trunk interface with VLAN 30, thereby pruning VLAN 10 and VLAN 20 from traversing the link.

When executing `switchport trunk allowed vlan <vlan-list>` without using the `add` keyword, the switch replaces the entire set of allowed VLANs on the trunk with only the specified VLAN(s). As a result, VLAN 10 and VLAN 20 were pruned from GigabitEthernet1/0/1, causing cross-switch traffic for those VLANs to be dropped.

Step-by-Step Solution

Analyze the CLI command executed on the switch interface.

Identified the command as `switchport trunk allowed vlan 30` without the `add` parameter.

In Cisco IOS and standard network switch syntax, `switchport trunk allowed vlan <vlan-id>` sets an explicit allowed list, replacing any previously permitted VLANs.

Evaluate the impact on active VLAN traffic traversing the trunk.

VLAN 10 and VLAN 20 are immediately removed from the trunk's allowed list.

Frames belonging to VLAN 10 and VLAN 20 are dropped at egress/ingress on GigabitEthernet1/0/1 because those VLAN IDs are no longer permitted on the trunk interface.

Determine the proper command syntax to prevent this issue.

The administrator should have executed `switchport trunk allowed vlan add 30`.

Using the `add` parameter appends the specified VLAN ID to the existing list of permitted VLANs without clearing current entries.

Key Concept

802.1Q Trunk Allowed VLAN List Overwrite vs. Append

Estimated Time:1m 30s

Question 644Question →

A network technician is performing diagnostic testing on several physical cabling installations across a enterprise facility. Match each physical layer cabling issue on the left with its underlying root cause or correct remediation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The physical layer symptoms correspond to their underlying causes as follows: optical attenuation/reflection without break indicates fiber end-face contamination; 100% DC continuity with high NEXT indicates a split pair; intermittent errors during motor startup indicate EMI on UTP runs; and total lack of fiber link light indicates a Tx/Rx polarity reversal.

Each cabling symptom matches standard physical layer troubleshooting principles: end-face dirt causes insertion/return loss; split pairs maintain continuity while generating NEXT; motor operation induces EMI in UTP; and duplex fiber requires proper Tx/Rx crossover alignment.

Step-by-Step Solution

Analyze the fiber attenuation and reflection symptom.

High insertion loss accompanied by back-reflection at a junction points to connector contamination, which scatters optical power.

Microscopic contaminants on fiber end-faces create air gaps and scattering surfaces that degrade light transmission.

Evaluate the UTP continuity versus crosstalk conflict.

A basic wiremapper only validates electrical continuity per pin. Preserving pin continuity while swapping conductors between pairs creates a split pair, destroying crosstalk cancellation.

Differential signaling relies on balanced twisted pairs; splitting pairs exposes conductors to high near-end crosstalk.

Correlate environmental machinery cycles with copper link drops.

Signal drops occurring during motor startup indicate electromagnetic induction on unshielded copper cabling.

Unshielded twisted-pair cabling lacks metallic shielding against high-voltage electromagnetic radiation.

Diagnose the missing link light on the duplex fiber connection.

Duplex fiber requires a crossover of Tx to Rx. If Tx connects to Tx, light fails to hit the receiving photodiode.

Optical transceivers send light from Tx and expect light into Rx; reversing polarity restores the signal path.

Key Concept

Troubleshooting Physical Cabling and Connectors

Question 645Question →

A network administrator needs to restore a corrupted domain controller on Thursday morning. The network backup strategy consists of a full backup executed every Sunday at 00:00, followed by daily incremental backups executed Monday through Wednesday at 23:00. Arrange the restoration steps in the correct chronological order required to recover the server to its most recent state prior to the failure.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order to restore the system using incremental backups is: 1) Restore the full baseline backup from Sunday, 2) Apply the incremental backup from Monday, 3) Apply the incremental backup from Tuesday, and 4) Apply the incremental backup from Wednesday.

Incremental backups capture only data that has changed since the last full or incremental backup and clear the archive bit. Therefore, to fully recover data to the most recent state (Wednesday night), you must first restore the full Sunday baseline backup, followed by every individual incremental backup in exact chronological sequence (Monday, then Tuesday, then Wednesday).

Step-by-Step Solution

Identify the backup architecture type

Recognize that incremental backups capture only changes made since the previous backup of any type (full or incremental).

Understanding the backup type determines whether all intermediary backups are required or if only the latest differential backup is needed.

Determine the baseline restoration point

Select and restore the Sunday full backup first.

All backup restoration processes require restoring the foundational baseline image (full backup) before any delta changes can be applied.

Apply subsequent incremental backups in chronological sequence

Apply Monday's incremental backup, followed by Tuesday's incremental backup, and finally Wednesday's incremental backup.

Because each incremental backup clears the archive bit and depends on the state established by the immediately preceding incremental backup, skipping any incremental set will result in data loss or corrupted file states.

Key Concept

Incremental Backup Restoration Sequence

Question 646Question →

An enterprise network administrator configures a Layer 3 access switch as a DHCP relay agent for VLAN 55 ( $10.55.0.0/24$ ), designating a remote server at $192.168.1.100$ via an `ip helper-address` command on SVI 55. Downstream Layer 2 edge switches append DHCP Option 82 (Relay Agent Information) metadata to client requests before forwarding them to the Layer 3 switch. However, client hosts on VLAN 55 fail to obtain IP addresses, and packet drops are logged on the Layer 3 switch for incoming DHCPDISCOVER packets containing Option 82 with a gateway IP address (`giaddr`) of $0.0.0.0$ . Which configuration change on the Layer 3 switch will resolve this issue?

Enable DHCP Option 82 trust on the relay agent to allow untrusted ingress packets containing Option 82 data

Move the `ip helper-address` configuration from SVI 55 to the routed uplink interface connected to the DHCP server

Reconfigure the helper address mechanism to encapsulate forwarded client requests in TCP port 67 instead of UDP port 67

Configure a static route on the switch to forward VLAN 55 Layer 2 broadcast traffic directly across subnets to the DHCP server

Show answer & explanation

Answer: Enable DHCP Option 82 trust on the relay agent to allow untrusted ingress packets containing Option 82 data

Answer

The correct action is to enable DHCP Option 82 trust on the Layer 3 relay switch so it accepts and processes incoming client requests pre-tagged with Option 82 information.

When downstream Layer 2 switches append Option 82 information to a client DHCP request before it reaches the Layer 3 relay agent, the request arrives on an untrusted port with a gateway IP address (`giaddr`) of

0.0.0.0

. By default security design, many relay agents discard such packets to prevent Option 82 injection attacks. Enabling Option 82 trust allows the Layer 3 switch to accept these packets, insert its SVI IP into `giaddr`, and forward the unicast packet to the DHCP server.

Step-by-Step Solution

Analyze the reported symptom and packet behavior

Client broadcast packets contain DHCP Option 82 inserted by downstream Layer 2 switches, but arrive at the Layer 3 relay agent with `giaddr = 0.0.0.0`.

The Layer 2 edge switch appends Option 82 metadata without modifying IP header fields such as `giaddr`.

Identify default security rules for DHCP relay agents

Standard DHCP relay security policy dictates dropping packets containing Option 82 if received on untrusted interfaces when `giaddr` is set to

0.0.0.0

This behavior prevents Option 82 spoofing attacks from untrusted edge ports.

Determine the necessary configuration fix

Configuring `ip dhcp relay information trust-all` or enabling interface-level Option 82 trust instructs the relay switch to process pre-existing Option 82 fields and forward the unicast request to

192.168.1.100

Explicitly trusting edge Option 82 insertion enables valid cross-VLAN relay functionality in tiered switch architectures.

Key Concept

DHCP Option 82 Untrusted Ingress Processing & Relay Agent Trust Policies

Question 647Question →

A network administrator receives reports that a workstation at IP address 192.168.10.45 periodically loses connectivity to the default gateway. The administrator runs the `arp -a` command on the workstation and views the following output:

Interface: 192.168.10.45 --- 0x2
Internet Address Physical Address Type
192.168.10.1 00-11-22-33-44-55 dynamic
192.168.10.12 00-11-22-33-44-55 dynamic

Which of the following conditions is most directly indicated by this terminal output?

A duplicate MAC address collision exists where another local IP address shares the gateway's physical address.

The local DNS server failed to answer host resolution requests for the default gateway.

The workstation host is attempting to connect to the gateway across an mismatched TCP port number.

The workstation requires an AAAA resource record to populate dynamic entries in the IP cache.

Show answer & explanation

Answer: A duplicate MAC address collision exists where another local IP address shares the gateway's physical address.

Answer

A duplicate MAC address collision exists where another local IP address shares the gateway's physical address.

The `arp -a` utility displays the host's Address Resolution Protocol cache. Seeing two separate IP addresses (the gateway at 192.168.10.1 and a host at 192.168.10.12) bound to the identical physical MAC address (00-11-22-33-44-55) indicates a MAC collision, incorrect static assignment, or ARP spoofing, which leads to intermittent reachability issues to the default gateway.

Step-by-Step Solution

Analyze the output of the `arp -a` command.

Observe that IP address 192.168.10.1 (default gateway) and IP address 192.168.10.12 both resolve to MAC address 00-11-22-33-44-55.

The ARP table maintains mappings between Layer 3 IPv4 addresses and Layer 2 MAC addresses on the local network.

Evaluate the functional impact of identical physical addresses assigned to different logical IP addresses.

Frames destined for either IP address will be delivered to the same physical network interface card.

Unless running specialized multi-homing or virtual gateway clustering protocols, two distinct active IPv4 addresses should not dynamically resolve to the exact same MAC address on a local broadcast domain without causing frame misdirection or MAC address duplication conflicts.

Key Concept

Inspecting ARP Cache Entries to Identify Layer 2 / Layer 3 Addressing Conflicts

Estimated Time:1m 15s

Question 648Question →

During a routine security audit of host communications, a technician inspects raw protocol headers passing through a socket interface. The payload header contains source and destination port multiplexing fields along with window size parameters for flow control, but contains no IP routing headers or hardware physical addresses. At which layer of the OSI model does this data unit function?

Transport Layer

Network Layer

Data Link Layer

Session Layer

Show answer & explanation

Answer: Transport Layer

Answer

Transport Layer

The Transport Layer (Layer 4) of the OSI model manages end-to-end transport services. Key functions at this layer include service multiplexing using port numbers (e.g., TCP/UDP ports), segmentation, error control via checksums, and flow control using mechanisms like sliding window size.

Step-by-Step Solution

Analyze the header fields described in the scenario.

Identified port numbers (source/destination) and window size parameters (flow control).

Port numbers provide service multiplexing, and window sizes regulate data transmission rates.

Map identified protocol functions to the OSI model layers.

Port assignment and flow control are core duties of Layer 4.

The Transport Layer manages end-to-end communication control between end hosts.

Key Concept

OSI Layer 4 (Transport Layer) Functions and Protocol Data Units

Question 649Question →

A core border router receives an IP packet destined for host address $10.20.105.42$ . The router's active routing table contains four operational candidate entries that match this target network space:

1. Static route to $10.20.96.0/19$ via next-hop $172.16.1.1$ (Administrative Distance: $1$ )
2. OSPF route to $10.20.104.0/21$ via next-hop $172.16.2.1$ (Administrative Distance: $110$ , Metric: $20$ )
3. Internal EIGRP route to $10.20.104.0/22$ via next-hop $172.16.3.1$ (Administrative Distance: $90$ , Metric: $15$ )
4. External BGP route to $10.20.0.0/16$ via next-hop $172.16.4.1$ (Administrative Distance: $20$ , Metric: $100$ )

Which next-hop IP address will the router select to forward this packet?

172.16.1.1, because static routes possess the lowest administrative distance among all matching table entries

172.16.3.1, because the /22 subnet mask provides the longest matching prefix for the destination address

172.16.4.1, because BGP exterior gateway routes take precedence over interior routing protocols when crossing autonomous system boundaries

172.16.2.1, because OSPF link-state metrics are preferred over distance-vector summary calculations

Show answer & explanation

Answer: 172.16.3.1, because the /22 subnet mask provides the longest matching prefix for the destination address

Answer

172.16.3.1, because the /22 subnet mask provides the longest matching prefix for the destination address.

The fundamental decision rule for IPv4/IPv6 packet forwarding is Longest Prefix Match (LPM). When a router consults its forwarding table, it selects the route with the most specific (longest) subnet mask that covers the target IP address. Here, host address

10.20.105.42

falls within all four candidate ranges, but

/22

(22 matching network bits) is more specific than

/21

/19

, or

/16

. Consequently, the router routes to next-hop

172.16.3.1

without evaluating Administrative Distance or protocol metrics.

Step-by-Step Solution

Determine which routing entries encompass the destination IP address 10.20.105.42

All four entries match: 10.20.0.0/16 (10.20.0.0 - 10.20.255.255), 10.20.96.0/19 (10.20.96.0 - 10.20.127.255), 10.20.104.0/21 (10.20.104.0 - 10.20.111.255), and 10.20.104.0/22 (10.20.104.0 - 10.20.107.255).

Before comparing routing metrics, all matching candidate prefixes must be identified.

Apply the Longest Prefix Match (LPM) rule across all valid matches

The prefix length of /22 (22 matching network bits) is the most specific subnet mask compared to /21, /19, and /16.

Routers always evaluate prefix specificity first; Administrative Distance (AD) and metric are only compared when two identical prefix lengths exist from different sources.

Select the next-hop associated with the longest prefix entry

The entry 10.20.104.0/22 points to next-hop 172.16.3.1.

Because /22 has the highest number of matching bits, the router immediately forwards the packet to 172.16.3.1.

Key Concept

Longest Prefix Match (LPM) Routing Precedence

Question 650Question →

A network administrator needs to segment the IPv4 network block $10.200.16.0/20$ to accommodate multiple remote office LANs, with the largest subnet requiring support for up to 500 usable host IP addresses. Which of the following statements regarding the subnet design and IP address allocation are correct? (Select TWO.)

Select all that apply

A /23 prefix (subnet mask 255.255.254.0) is the smallest subnet mask that accommodates 500 usable host IP addresses.

The first usable host IP address in the second /23 subnet carved from this block is 10.200.18.1.

A /24 prefix (subnet mask 255.255.255.0) can be assigned to accommodate 500 usable hosts by using variable-length subnet masking.

The broadcast address for the first /23 subnet allocated from this block is 10.200.16.255.

Show answer & explanation

Answer: A /23 prefix (subnet mask 255.255.254.0) is the smallest subnet mask that accommodates 500 usable host IP addresses.; The first usable host IP address in the second /23 subnet carved from this block is 10.200.18.1.

Answer

The correct statements are that a /23 prefix (255.255.254.0) is the smallest prefix capable of supporting 500 usable hosts, and the first usable host IP address in the second /23 subnet is 10.200.18.1.

To support 500 hosts, 9 host bits are necessary (

2^9 - 2 = 510

), resulting in a /23 mask (

255.255.254.0

). The first /23 subnet ranges from

10.200.16.0

10.200.17.255

with broadcast address

10.200.17.255

. The second /23 subnet starts at

10.200.18.0

, making

10.200.18.1

its first usable host address.

Step-by-Step Solution

Determine the required host bits and prefix length for 500 hosts

9 host bits are needed because

2^8 - 2 = 254

(insufficient) and

2^9 - 2 = 510

usable hosts (sufficient). The prefix length is

32 - 9 = /23

(subnet mask

255.255.254.0

Calculating usable hosts requires subtracting 2 (network and broadcast addresses) from

2^h

Calculate subnet boundaries for consecutive /23 subnets starting from 10.200.16.0/20

Subnet 1:

10.200.16.0/23

(Network ID:

10.200.16.0

, First Usable:

10.200.16.1

, Last Usable:

10.200.17.254

, Broadcast:

10.200.17.255

). Subnet 2:

10.200.18.0/23

(Network ID:

10.200.18.0

, First Usable:

10.200.18.1

, Last Usable:

10.200.19.254

, Broadcast:

10.200.19.255

A /23 subnet increments by 2 in the third octet.

Evaluate the choices against the calculated results

The statement specifying /23 as the minimum mask for 500 hosts is correct. The statement identifying 10.200.18.1 as the first usable IP of the second subnet is correct. Statements proposing /24 or 10.200.16.255 as the first subnet broadcast address are incorrect.

Comparing calculated addressing parameters against each statement verifies validity.

Key Concept

IPv4 Subnetting and Host Capacity Calculation

Question 651Question →

A network administrator is configuring a GigabitEthernet 0/1 trunk link on switch SW-BuildingA to connect to SW-Core. The link is intended to carry traffic for VLAN 10 (Data), VLAN 20 (Voice), and VLAN 99 (Management). VLAN 99 is configured as the native VLAN on both switches. The administrator executes the command `switchport trunk allowed vlan 10,20` on interface GigabitEthernet 0/1. Immediately after executing the command, remote management access to SW-BuildingA via VLAN 99 drops. Which of the following best explains why management traffic is no longer traversing the trunk link?

The explicit allowed VLAN list replaces the default allowed range and excludes VLAN 99, preventing its frames from traversing the trunk regardless of native VLAN designation.

Native VLAN traffic is automatically tagged with 802.1Q headers when allowed VLAN lists are modified, causing a tag mismatch at the receiving switch.

Untagged native VLAN frames cannot cross an 802.1Q trunk unless inter-VLAN routing is active on the physical interface.

Applying an allowed VLAN list automatically reverts the switchport operational mode from trunk to access, disabling multi-VLAN support.

Show answer & explanation

Answer: The explicit allowed VLAN list replaces the default allowed range and excludes VLAN 99, preventing its frames from traversing the trunk regardless of native VLAN designation.

Answer

The explicit allowed VLAN list replaces the default allowed range and excludes VLAN 99, preventing its frames from traversing the trunk regardless of native VLAN designation.

On an 802.1Q trunk, the allowed VLAN list controls which VLANs can send and receive frames across the link. When `switchport trunk allowed vlan 10,20` is executed without the `add` keyword, it replaces the entire allowed list with only VLANs 10 and 20. Because VLAN 99 is excluded from this list, all frames associated with VLAN 99 are blocked, causing loss of management access.

Step-by-Step Solution

Analyze the command executed on the trunk interface.

The command `switchport trunk allowed vlan 10,20` sets the active allowed list on the interface strictly to VLANs 10 and 20.

By default, trunks allow all VLANs (1-4094). Specifying a list parameter without the `add` keyword overwrites the existing allowed list.

Evaluate the interaction between native VLAN configuration and allowed VLAN lists.

VLAN 99 is designated as the native VLAN, but it is not present in the allowed list (10, 20).

An 802.1Q trunk will only forward frames for VLANs that are explicitly present in its allowed VLAN list, even if that VLAN is configured as the native VLAN.

Determine the impact on management traffic.

Management traffic belonging to VLAN 99 is pruned and dropped at interface GigabitEthernet 0/1.

Because VLAN 99 is excluded from the allowed list, remote management traffic sent over VLAN 99 cannot traverse the link.

Key Concept

802.1Q Trunk Allowed VLAN Filtering

Question 652Question →

A network administrator is provisioning a dedicated management VLAN to connect 14 top-of-rack switch interfaces and 1 central default gateway interface. To minimize wasted IP address space, which of the following CIDR prefix lengths represents the smallest subnet mask that can accommodate all 15 required interfaces?

27

28

26

29

Show answer & explanation

Answer: / $27$

Answer

The prefix length /

27

is the smallest subnet that provides enough usable IP addresses for 15 host interfaces.

The prefix length /

27

allocates 5 host bits (

32 - 27 = 5

), providing

2^5 - 2 = 30

usable IP addresses. This is the smallest valid subnet mask that accommodates all 15 required host interfaces (14 switches + 1 default gateway).

Step-by-Step Solution

Determine the total number of required usable host IP addresses.

14 switch interfaces + 1 default gateway interface = 15 usable IP addresses required.

Every active device interface on the subnet requires a unique usable IPv4 address.

Calculate usable host capacity using the IPv4 host capacity formula (

2^n - 2

For

n=4

host bits (/

28

), capacity is

2^4 - 2 = 14

usable hosts. For

n=5

host bits (/

27

), capacity is

2^5 - 2 = 30

usable hosts.

Two IP addresses in every IPv4 subnet are reserved for the network ID and broadcast address and cannot be assigned to hosts.

Select the smallest prefix length that meets or exceeds the required 15 hosts.

The prefix length /

27

provides 30 usable host IP addresses, satisfying the 15 host requirement with minimal address waste.

A /

28

subnet only provides 14 usable addresses, falling short by 1 IP address.

Key Concept

IPv4 Subnet Mask Selection and Usable Host Calculation

Question 653Question →

An administrator configures two separate VLANs—VLAN $10$ (Sales) and VLAN $20$ (Marketing)—on a single Layer 2 switch. Host computers in both departments are connected to switch access ports assigned to their respective VLAN IDs. Although physical links are up and all devices are correctly assigned IP addresses within their subnets, hosts in VLAN $10$ cannot communicate with hosts in VLAN $20$ . Which of the following best explains why these hosts cannot communicate, and what is required to enable connectivity between them?

VLANs create separate Layer 2 broadcast domains; therefore, a Layer 3 routing device, such as a router or multi-layer switch, is required to forward traffic between them.

Access ports block inter-subnet traffic by default; converting all host access ports into 802.1Q trunk ports will allow untagged communication between VLAN

10

and VLAN

20

The switch drops inter-VLAN frames because the native VLAN on the switch has not been set to VLAN

10

to match the Sales department subnet.

Spanning Tree Protocol (STP) has placed the switch ports into a blocking state due to a mismatch, requiring an adjustment to the STP root bridge priority.

Show answer & explanation

Answer: VLANs create separate Layer 2 broadcast domains; therefore, a Layer 3 routing device, such as a router or multi-layer switch, is required to forward traffic between them.

Answer

VLANs create separate Layer 2 broadcast domains; therefore, a Layer 3 routing device, such as a router or multi-layer switch, is required to forward traffic between them.

Each VLAN configured on a switch acts as a separate Layer 2 broadcast domain. Because Layer 2 switches forward Ethernet frames strictly within the origin VLAN, traffic destined for another VLAN must be forwarded to a Layer 3 device (such as a router or Layer 3 switch) capable of routing packets between IP subnets.

Step-by-Step Solution

Identify the Layer 2 boundary created by Virtual Local Area Networks (VLANs).

Recognize that VLAN

10

and VLAN

20

exist in completely separate broadcast domains on the switch.

By design, Layer 2 switches do not forward frames between different VLANs.

Determine the necessary networking layer component required to bridge distinct broadcast domains.

A Layer 3 routing function (such as a router-on-a-stick or Switched Virtual Interfaces on a Layer 3 switch) must evaluate IP packet headers.

Inter-VLAN communication requires IP routing because Layer 2 switching logic isolates VLAN traffic.

Key Concept

VLAN Isolation and Inter-VLAN Routing Requirements

Question 654Question →

A network administrator is troubleshooting connectivity on an enterprise server host configured with the IP address $10.205.75.142/21$ . To establish proper firewall routing rules, the administrator needs to identify the complete range of usable host IP addresses within this specific subnet. Which of the following represents the correct usable host IP address range for this subnetwork segment?

10.205.72.1

10.205.79.254

10.205.72.0

10.205.79.255

10.205.75.1

10.205.75.254

10.205.64.1

10.205.79.254

Show answer & explanation

Answer: $10.205.72.1$ to $10.205.79.254$

Answer

The usable host range for the

10.205.75.142/21

subnet is

10.205.72.1

10.205.79.254

The CIDR notation

/21

indicates that the first 21 bits represent the network prefix, corresponding to a subnet mask of

255.255.248.0

. In the third octet, subnets increment by multiples of

8

(

256 - 248 = 8

). For an IP address with

75

in the third octet, the network boundary starts at

72

(

9 \times 8 = 72

), making the network address

10.205.72.0

and the broadcast address

10.205.79.255

. Excluding the network and broadcast addresses yields the usable range of

10.205.72.1

through

10.205.79.254

Step-by-Step Solution

Determine the subnet mask and third-octet block size for a /21 CIDR prefix.

A /21 prefix leaves 11 host bits (

32 - 21 = 11

). The subnet mask is

255.255.248.0

. The third octet increment (block size) is calculated as

256 - 248 = 8

The CIDR prefix defines the boundary between the network portion and host portion of the IP address.

Find the network address by calculating the nearest lower multiple of the block size (8) for the third octet (75).

Multiples of 8 in the third octet are

0, 8, 16, 24, 32, 40, 48, 56, 64, 72, 80

. Since 75 falls between 72 and 80, the network ID is

10.205.72.0

The network ID is the first IP address of the subnet segment.

Determine the broadcast address for the subnet segment.

The next subnet begins at

10.205.80.0

, making the broadcast address of the current segment

10.205.79.255

The broadcast address is the last address in the subnet block immediately preceding the start of the next block.

Calculate the usable host IP address range by subtracting network and broadcast addresses.

First usable host IP:

10.205.72.1

. Last usable host IP:

10.205.79.254

Usable host addresses consist of all IP addresses between the network address and the broadcast address.

Key Concept

Calculating CIDR Subnet Boundaries and Usable Host Ranges for Non-Byte-Aligned Prefixes

Question 655Question →

A network technician is setting up a router for a newly expanded office. The organization requires that if an active link goes down, the router should automatically discover alternate paths and update its routing table without requiring manual administrator intervention. Which of the following network technologies should be implemented to fulfill this requirement?

Dynamic routing protocols

Port Address Translation (PAT)

VLAN broadcast domain isolation

OSI Layer 2 frame encapsulation

Show answer & explanation

Answer: Dynamic routing protocols

Answer

Dynamic routing protocols provide automatic network discovery and dynamic route recalculation upon link changes.

Dynamic routing protocols enable routers to dynamically advertise reachability, discover topology changes, and update routing tables automatically when network links go down.

Step-by-Step Solution

Identify the core network requirement in the scenario.

The router must dynamically recalculate paths and update routing tables upon link failure without manual configuration.

The scenario highlights the need for continuous automated reachability updates.

Evaluate the function of dynamic routing protocols.

Dynamic protocols (such as OSPF or RIP) automatically learn remote networks and adjust active paths when topology shifts happen.

Dynamic routing automates path determination and routing table updates.

Key Concept

Dynamic Routing Protocols and Route Convergence

Question 656Question →

An enterprise edge router receives an IP packet destined for host address $172.16.42.100$ . The router's active routing table contains four candidate routes:

Protocol Source	Destination Subnet	Administrative Distance	Metric	Next Hop
Static	$172.16.0.0/16$	1	0	Null0
eBGP	$172.16.40.0/21$	20	100	$192.168.1.1$
OSPF	$172.16.42.0/24$	110	20	$10.0.0.2$
RIPv2	$172.16.42.96/28$	120	2	$10.0.0.6$

Which route will the router select to forward the packet, and why?

The RIPv2 route (

172.16.42.96/28

), because Longest Prefix Match takes precedence over Administrative Distance.

The Static route (

172.16.0.0/16

), because static routes have the lowest Administrative Distance.

The eBGP route (

172.16.40.0/21

), because exterior gateway protocols are prioritized over interior gateway protocols.

The OSPF route (

172.16.42.0/24

), because it offers the optimal balance between prefix specificity and Administrative Distance.

Show answer & explanation

Answer: The RIPv2 route ( $172.16.42.96/28$ ), because Longest Prefix Match takes precedence over Administrative Distance.

Answer

The router will select the RIPv2 route (

172.16.42.96/28

) because Longest Prefix Match (LPM) is the absolute first criterion used in IP routing decisions, taking precedence over Administrative Distance and metric.

When a router receives a packet, it evaluates all matching routing table entries using the Longest Prefix Match (LPM) rule first. Because

172.16.42.96/28

has the longest subnet mask (/28 = 28 matching bits) among all matching candidate routes, it is chosen immediately. Administrative Distance (AD) and metrics are only evaluated to select a winner when two or more routes advertise identical destination prefix lengths.

Step-by-Step Solution

Determine which routing table entries match the target IP address (

172.16.42.100

All four entries match:

172.16.0.0/16

(range

172.16.0.0 - 172.16.255.255

172.16.40.0/21

(range

172.16.40.0 - 172.16.47.255

172.16.42.0/24

(range

172.16.42.0 - 172.16.42.255

), and

172.16.42.96/28

(range

172.16.42.96 - 172.16.42.111

The router must filter candidate routes to only those subnets that encompass the destination address.

Compare the prefix lengths (subnet masks) of all matching candidate routes.

Prefix lengths are /16, /21, /24, and /28. The longest prefix match is /28.

In IP routing, the most specific route (highest prefix length) is always selected before Administrative Distance or metric are evaluated.

Identify the protocol associated with the longest prefix match (/28).

The /28 route corresponds to the RIPv2 learned entry via next-hop

10.0.0.6

Administrative Distance (120 vs 1 vs 20 vs 110) is ignored because the prefix lengths are not equal.

Key Concept

Longest Prefix Match (LPM) Rule

Question 657Question →

A network technician is configuring a local area network interface using the IPv4 network block $10.20.30.0/26$ . Which of the following statements regarding this subnet configuration are correct? (Select TWO.)

Select all that apply

The subnet mask in dotted-decimal notation is

255.255.255.192

The total number of usable host IP addresses in this subnet is

62

The total number of usable host IP addresses in this subnet is

64

The IP address

10.20.30.63

can be assigned directly to a host workstation interface.

Show answer & explanation

Answer: The subnet mask in dotted-decimal notation is $255.255.255.192$ .; The total number of usable host IP addresses in this subnet is $62$ .

Answer

The correct statements are that the subnet mask in dotted-decimal notation is

255.255.255.192

and the subnet contains

62

usable host IP addresses.

A CIDR prefix of

/26

uses 26 bits for the network portion and 6 bits for the host portion (

32 - 26 = 6

). In dotted-decimal notation, 26 subnet bits translate to

255.255.255.192

. The total number of IP addresses in a

/26

subnet is

2^6 = 64

. Subtracting 2 (one for the network address

10.20.30.0

and one for the broadcast address

10.20.30.63

) leaves exactly 62 usable host IP addresses.

Step-by-Step Solution

Calculate the subnet mask from CIDR notation /26

26 binary 1s followed by 6 binary 0s gives 11111111.11111111.11111111.11000000, which converts to

255.255.255.192

Converting CIDR prefix to dotted-decimal notation is necessary to determine the subnet boundary.

Determine host bits and total IP address capacity

Remaining host bits =

32 - 26 = 6

. Total addresses =

2^6 = 64

Host bits determine the size of the IP block.

Calculate usable host IP addresses

Usable host addresses =

64 - 2 = 62

. Network ID is

10.20.30.0

and Broadcast address is

10.20.30.63

The first address (network ID) and last address (broadcast address) are reserved and cannot be assigned to end devices.

Key Concept

IPv4 Subnet Masking and Usable Host Calculation

Question 658Question →

During a network security audit, an incident response analyst discovers a compromised host with the IPv4 address $172.16.89.138/22$ . What is the broadcast address for the subnet to which this host belongs?

Show answer & explanation

Answer: 172.16.91.255

Answer

The broadcast address for the subnet is 172.16.91.255.

For an IP address of

172.16.89.138/22

, the subnet mask is 255.255.252.0. The third octet increments in multiples of 4 (

256 - 252 = 4

). The network containing the third octet value of 89 starts at 88 (

172.16.88.0

). The subnet spans up to

172.16.91.255

, making

172.16.91.255

the broadcast address where all 10 host bits are set to 1.

Step-by-Step Solution

Determine the subnet mask and block size from the CIDR prefix /22.

A /22 prefix corresponds to a subnet mask of 255.255.252.0. The block size in the third octet is

256 - 252 = 4

The prefix /22 leaves 10 host bits across the third and fourth octets, giving

2^{10} = 1024

total IP addresses (

4

blocks of 256).

Identify the network address for the host IP address 172.16.89.138.

The third octet boundaries increment by 4 (

0, 4, 8, \dots, 84, 88, 92

). Since 89 lies between 88 and 91, the network address is 172.16.88.0.

The network ID uses the lowest address in the block formed by the multiple of the block size (

88

Calculate the broadcast address of the subnet.

The broadcast address is 172.16.91.255.

The broadcast address has all 10 host bits set to 1, which corresponds to the last IP address in the

172.16.88.0 - 172.16.91.255

range.

Key Concept

IPv4 Non-Octet Boundary Subnetting and Broadcast Address Calculation

Estimated Time:2m 0s

Question 659Question →

A network technician is provisioning an isolated subnet for a small cluster of administrative servers using the $/29$ IPv4 prefix. What is the maximum number of usable host IP addresses available on this subnet?

Show answer & explanation

Answer: 6

Answer

The maximum number of usable host IP addresses available on a

/29

subnet is 6.

For a

/29

prefix,

32 - 29 = 3

bits remain for host addressing. The total number of IP addresses in the block is

2^3 = 8

. Because 1 address is reserved for the network ID and 1 address is reserved for the directed broadcast ID, the number of usable host IP addresses is

8 - 2 = 6

Step-by-Step Solution

Calculate the number of host bits available in the CIDR prefix.

32 total IPv4 bits - 29 network bits = 3 host bits.

IPv4 addresses consist of 32 bits total.

Calculate total IP addresses supported by 3 host bits.

2^3 = 8 total IP addresses.

Each host bit doubles the number of possible IP addresses.

Subtract reserved network and broadcast addresses.

8 total IPs - 2 = 6 usable host addresses.

The network address (all host bits 0) and broadcast address (all host bits 1) cannot be assigned to host interfaces.

Key Concept

Calculating usable host IP addresses in IPv4 subnetting by subtracting network and broadcast addresses.

Estimated Time:45s

Question 660Question →

A network administrator is replacing static routes with a dynamic routing protocol across an enterprise network. Which TWO of the following are key operational features provided by dynamic routing protocols? (Select TWO.)

Select all that apply

Automatically discover reachable remote subnets through neighbor information exchanges

Automatically recalculate alternate paths when a active network link fails

Eliminate the requirement for Layer 3 network layer addresses on routing interfaces

Translate multiple private internal IP addresses into a single public IP address

Show answer & explanation

Answer: Automatically discover reachable remote subnets through neighbor information exchanges; Automatically recalculate alternate paths when a active network link fails

Answer

Dynamic routing protocols automatically discover remote networks by communicating with neighboring routers and recalculate valid backup paths when a link failure occurs.

Dynamic routing protocols automatically discover remote subnets by exchanging network state and path metrics with neighboring routers. In addition, when network topology changes or a link fails, dynamic routing protocols recalculate paths and update routing tables without manual intervention.

Step-by-Step Solution

Identify the primary purpose of dynamic routing protocols

Dynamic routing protocols automate route discovery and topology maintenance across routers.

Static routing requires manual route entry, while dynamic protocols continuously update routing tables based on network state.

Evaluate the choices against dynamic routing protocol functionality

Discovering remote subnets from neighbor routers and dynamically re-routing traffic around failed links are core features of dynamic routing protocols.

Protocols like OSPF, EIGRP, and RIP continuously share state and metric information to update network topology maps.

Distinguish routing from addressing and NAT services

Layer 3 addressing remains essential for routing interfaces, and address translation is handled by PAT/NAT rather than dynamic routing protocols.

Routing protocols direct traffic between subnets using Layer 3 IP headers; they do not remove IP addressing requirements or perform NAT overload functions.

Key Concept

Dynamic Routing Protocol Features and Operations

Estimated Time:1m 0s

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