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Networking Concepts

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Question 261Question →

A network administrator connects two Layer 2 switches using an 802.1Q trunk link. Switch A has native VLAN 10 configured on its trunk port, while Switch B has native VLAN 20 configured on its trunk port. Host A, attached to an access port in VLAN 10 on Switch A, sends a broadcast frame. Switch A transmits this frame untagged across the trunk link. Which of the following best describes how Switch B processes this incoming untagged frame?

Show answer & explanation

Answer: Switch B assigns the incoming untagged frame to its local native VLAN (VLAN 20) and forwards it to all active access ports in VLAN 20.

Answer

Switch B assigns the incoming untagged frame to its local native VLAN (VLAN 20) and forwards it to all active access ports in VLAN 20.
Under 802.1Q trunking specifications, frames belonging to the native VLAN are sent across the trunk untagged. When an untagged frame arrives at a receiving switch interface, that switch implicitly tags or assigns the frame to its own locally configured native VLAN. Because Switch A sends VLAN 10 traffic untagged and Switch B has native VLAN 20 configured, Switch B maps the incoming untagged frame into VLAN 20, causing cross-VLAN traffic leakage.

Step-by-Step Solution

1
Analyze how Switch A transmits VLAN 10 traffic across the trunk link.
Because VLAN 10 is configured as the native VLAN on Switch A's trunk port, Switch A strips the 802.1Q tag and sends the frame untagged across the link.
802.1Q trunking rules dictate that frames belonging to the port's native VLAN are sent without an explicit 802.1Q VLAN header.
2
Determine how Switch B interprets untagged ingress frames on its trunk port.
Switch B receives the untagged frame and associates it with its own configured native VLAN, which is VLAN 20.
A switch interface assigned an 802.1Q native VLAN treats any incoming frame without a VLAN tag as member traffic of that local native VLAN.
3
Evaluate the forwarding destination of the broadcast frame on Switch B.
Switch B floods the broadcast frame out all active interfaces belonging to VLAN 20.
Broadcast frames are flooded throughout their assigned Layer 2 broadcast domain.

Key Concept

802.1Q Native VLAN Operation and Mismatch Behavior
Question 262Question →

A network administrator is configuring Layer 2 switching features and security controls across an enterprise switch environment. Match each switching concept or operational mechanism on the left to its corresponding functional behavior on the right.

Click a left item, then click its matching right item

Items

802.1Q Native VLAN
STP PortFast
BPDU Guard
Dynamic MAC Aging

Matches

Show answer & explanation

Answer

802.1Q Native VLAN matches untagged traffic transport across trunk links; STP PortFast matches direct transition to forwarding state; BPDU Guard matches err-disabling ports upon BPDU reception; Dynamic MAC Aging matches flushing idle CAM table entries.
Each feature is correctly mapped to its core Layer 2 switching mechanism: the 802.1Q native VLAN handles untagged trunk traffic, PortFast accelerates edge port transition to forwarding, BPDU Guard protects edge ports by err-disabling them when BPDUs are detected, and dynamic MAC aging clears inactive CAM table entries.

Step-by-Step Solution

1
Analyze the role of the 802.1Q Native VLAN.
Identified that native VLANs handle traffic that does not contain an 802.1Q tag across a trunk link.
By default in 802.1Q trunking, frames belonging to the native VLAN are sent untagged.
2
Analyze the function of STP PortFast.
Identified immediate forwarding state transition on end-user access ports.
PortFast is intended for edge ports so end-user clients do not experience DHCP or connection timeouts while waiting for STP listening/learning phases.
3
Analyze the operation of BPDU Guard.
Identified err-disable port state reaction upon receiving BPDUs.
Edge ports should not receive BPDUs from other switches; receiving one indicates a potential rogue switch or loop risk.
4
Analyze the function of Dynamic MAC Aging.
Identified purging of idle MAC addresses from the switch's CAM table.
Switches use an aging timer to clear inactive MAC addresses to ensure frame forwarding remains efficient and accurate.

Key Concept

Switching Concepts and VLANs
Question 263Question →

A security log server receives an incoming transmission containing a syslog message sent from a remote network gateway. As the destination host network stack processes this data unit through de-encapsulation, which transformation occurs specifically at Layer 3 immediately before payload processing is handed off to Layer 4?

Show answer & explanation

Answer: The IPv4 header is inspected and stripped, converting the Layer 3 packet into a Layer 4 UDP datagram containing source and destination port parameters.

Answer

The IPv4 header is inspected and stripped, converting the Layer 3 packet into a Layer 4 UDP datagram containing source and destination port parameters.
During inbound de-encapsulation on a receiving host, each layer inspects its corresponding control header, performs integrity and address checks, and strips its header before passing the encapsulated payload upward. At Layer 3 (Network Layer), the host inspects the IP header, verifies the IP destination address, and removes the IP header. The remaining payload is a Layer 4 PDU (a UDP datagram containing source and destination port numbers), which is passed to Layer 4 for transport processing.

Step-by-Step Solution

1
Identify the OSI layer processing boundary described in the scenario
The scenario focuses on the specific de-encapsulation step occurring at Layer 3 (Network Layer) moving toward Layer 4 (Transport Layer).
De-encapsulation unwraps data headers sequentially from Layer 1 up to Layer 7.
2
Determine the PDU state at Layer 3
Layer 3 receives an IP packet from Layer 2 after the Ethernet header and trailer have been removed.
The Data Link layer strips Layer 2 framing before passing the enclosed packet to the Network layer.
3
Analyze Layer 3 processing actions
Layer 3 validates fields in the IP header (such as destination IP and Protocol identifier 17 for UDP), strips the IP header, and forwards the resulting Layer 4 segment/datagram to the Transport layer.
Stripping the Layer 3 header reveals the Layer 4 PDU (UDP datagram).

Key Concept

De-encapsulation PDU Transformations across OSI Layers
Estimated Time:2m 0s
Question 264Question →

Match each OSI model layer on the left with its primary operational function and corresponding Protocol Data Unit (PDU) characteristics on the right.

Click a left item, then click its matching right item

Items

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

Matches

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Answer

The Data Link Layer (Layer 2) pairs with node-to-node frame delivery via MAC addresses; the Network Layer (Layer 3) pairs with packet routing via IP addresses; the Transport Layer (Layer 4) pairs with end-to-end process multiplexing via segments and ports; and the Presentation Layer (Layer 6) pairs with syntax translation, compression, and encryption.
Each OSI layer serves a specific purpose in network communications: Layer 2 uses Frames and MAC addresses for local transmission; Layer 3 uses Packets and IP addresses for routing; Layer 4 uses Segments and port numbers for transport multiplexing; and Layer 6 provides data translation, compression, and encryption.

Step-by-Step Solution

1
Analyze Layer 2 (Data Link Layer) characteristics.
Layer 2 builds Frames using MAC addressing for physical network segment communication.
Hardware addressing and framing are defined at the Data Link Layer.
2
Analyze Layer 3 (Network Layer) characteristics.
Layer 3 manages Packets and IP routing across subnets.
Path selection and logical addressing occur at the Network Layer.
3
Analyze Layer 4 (Transport Layer) characteristics.
Layer 4 uses Segments and port numbers for end-to-end communication.
Process-to-process communication and multiplexing are handled by TCP/UDP at the Transport Layer.
4
Analyze Layer 6 (Presentation Layer) characteristics.
Layer 6 performs encryption, compression, and format conversion.
Data syntax standardization between sender and receiver occurs at the Presentation Layer.

Key Concept

OSI Layer Functions and PDU Mapping
Question 265Question →

A network administrator is implementing a Variable Length Subnet Masking (VLSM) plan for a company using the parent address space 172.16.32.0/20172.16.32.0/20. Subnets are allocated sequentially starting from the lowest available network address. First, the Engineering department is allocated a subnet accommodating at least 100100 usable host interfaces. Immediately following the Engineering subnet, the Sales department is allocated a subnet to accommodate at least 2828 usable host interfaces. Which of the following IPv4 addresses represent valid, usable host IP addresses within the Sales department subnet? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: 172.16.32.130172.16.32.130; 172.16.32.158172.16.32.158

Answer

The addresses 172.16.32.130172.16.32.130 and 172.16.32.158172.16.32.158 are the valid usable host IP addresses within the Sales department subnet.
The Sales department requires a subnet for at least 2828 host interfaces. A /27/27 subnet provides 3232 total addresses (3030 usable hosts). Since the Engineering department uses 172.16.32.0/25172.16.32.0/25 (address range 172.16.32.0172.16.32.0 - 172.16.32.127172.16.32.127), the Sales subnet begins at 172.16.32.128/27172.16.32.128/27. Its network address is 172.16.32.128172.16.32.128, its broadcast address is 172.16.32.159172.16.32.159, and its usable host range spans 172.16.32.129172.16.32.129 through 172.16.32.158172.16.32.158. Therefore, 172.16.32.130172.16.32.130 and 172.16.32.158172.16.32.158 are both valid, assignable host addresses within the Sales subnet.

Step-by-Step Solution

1
Calculate the Engineering department subnet size and prefix length
Engineering requires 100100 usable host IPs. 27−2=126≥1002^7 - 2 = 126 \ge 100, so 77 host bits are needed, giving a prefix length of /25/25 (32−7=2532 - 7 = 25).
Determining host bit requirements establishes the boundary for sequential allocation.
2
Determine the network block allocated to the Engineering department
The Engineering subnet is 172.16.32.0/25172.16.32.0/25. Network ID: 172.16.32.0172.16.32.0, Usable range: 172.16.32.1172.16.32.1 to 172.16.32.126172.16.32.126, Broadcast address: 172.16.32.127172.16.32.127.
Sequential allocation requires knowing where the Engineering block finishes so the next subnet can start.
3
Calculate the Sales department subnet size and prefix length
Sales requires 2828 usable host IPs. 25−2=30≥282^5 - 2 = 30 \ge 28, so 55 host bits are needed, giving a prefix length of /27/27 (32−5=2732 - 5 = 27).
Establishes the block size for the Sales department subnet.
4
Determine the address boundaries and usable host range for the Sales department subnet
The Sales subnet starts immediately at 172.16.32.128/27172.16.32.128/27. Network ID: 172.16.32.128172.16.32.128, Usable host range: 172.16.32.129172.16.32.129 to 172.16.32.158172.16.32.158, Broadcast address: 172.16.32.159172.16.32.159.
Allows identification of valid usable host IPs versus network/broadcast or out-of-range IPs.
5
Evaluate the candidate options against the calculated usable host range
172.16.32.130172.16.32.130 and 172.16.32.158172.16.32.158 fall strictly within the inclusive range 172.16.32.129172.16.32.129 through 172.16.32.158172.16.32.158.
Matching options to valid host address criteria.

Key Concept

VLSM Subnet Allocation & Host Range Determination
Estimated Time:3m 0s
Question 266Question →

A system administrator is configuring core IP services across multiple subnets in an enterprise network. Which TWO of the following configurations are required to allow hosts on remote subnets to automatically acquire IPv4 addresses and resolve domain names to IPv4 addresses? (Select TWO)

Select all that apply

Show answer & explanation

Answer: Configuring a DHCP relay agent (IP helper) on the default gateway interface of each client subnet; Creating DNS A records on the domain name server for internal hostnames

Answer

The necessary configurations are deploying a DHCP relay agent (IP helper) on local router interfaces to forward DHCP broadcasts across subnets, and creating DNS A records to resolve hostnames to IPv4 addresses.
A DHCP relay agent (IP helper) allows clients on remote subnets to reach a central DHCP server by converting local broadcast requests into unicast traffic. Additionally, DNS A records perform forward address resolution from domain names to IPv4 addresses.

Step-by-Step Solution

1
Determine the service required to pass DHCP requests across router boundaries.
Identify that routers block broadcast packets, requiring a DHCP relay agent (IP helper) on the gateway interface to forward client lease requests to a remote DHCP server.
DHCP discover messages are sent as Layer 2 broadcasts, which cannot traverse subnets without a relay.
2
Identify the appropriate DNS record type for resolving hostnames to IPv4 addresses.
Select DNS A records as the proper mechanism for forward IPv4 resolution.
A records specifically map fully qualified domain names (FQDNs) to 32-bit IPv4 addresses.

Key Concept

IP Addressing Services (DHCP Relay and DNS Record Types)
Question 267Question →

Workstations in a newly configured branch office network (VLAN 20) are unable to obtain IPv4 addresses dynamically. The network administrator verifies that the centralized DHCP server is online in the datacenter (VLAN 100) and that IP routing between the subnets is functioning properly. Packet captures at the datacenter reveal that broadcast DHCP DISCOVER messages originated by branch clients never arrive at the server. Which of the following configuration steps must be performed on the branch router to resolve this issue?

Show answer & explanation

Answer: Configure an IP helper address pointing to the DHCP server IP on the client VLAN 20 gateway interface.

Answer

An IP helper address must be configured on the client VLAN 20 gateway interface of the branch router, pointing directly to the centralized DHCP server's IP address.
The correct action is to configure an IP helper address on the client VLAN 20 gateway interface of the branch router. By default, Layer 3 devices drop Layer 2 broadcast frames, preventing DHCP DISCOVER messages from reaching servers outside the local broadcast domain. The IP helper command enables a DHCP relay agent on the client-facing gateway interface, transforming broadcast traffic into unicast packets routed across the network to the central DHCP server.

Step-by-Step Solution

1
Analyze client traffic behavior across Layer 3 boundaries
Identify that Layer 2 broadcast messages (such as DHCP DISCOVER packets) are dropped by routers by default.
Routers establish broadcast domain boundaries and do not forward L2 broadcast packets across subnets without explicit relay configuration.
2
Determine the required IP service mechanism
Recognize the requirement for a DHCP Relay Agent (IP helper address).
A DHCP relay agent intercepts local client broadcasts and encapsulates them into unicast IP packets directed to the designated DHCP server.
3
Identify the correct interface placement for the relay agent
Apply the IP helper configuration on the inbound interface receiving the client broadcast traffic (VLAN 20 gateway interface).
The relay agent must listen on the specific local interface where client broadcast requests originate to successfully relay them to another subnet.

Key Concept

DHCP Relay Agent and IP Helper Address Placement
Estimated Time:1m 15s
Question 268Question →

Match each dynamic routing protocol on the left with its defining metric or structural classification on the right.

Click a left item, then click its matching right item

Items

RIP (Routing Information Protocol)
OSPF (Open Shortest Path First)
BGP (Border Gateway Protocol)

Matches

Show answer & explanation

Answer

RIP matches with the distance-vector protocol using hop count (max 15 hops); OSPF matches with the link-state protocol using bandwidth cost; BGP matches with the path-vector exterior gateway protocol connecting autonomous systems.
Each protocol is accurately matched to its operational category and primary metric calculation method. RIP uses distance-vector logic and hop counts, OSPF uses link-state logic and bandwidth-based cost, and BGP serves as the path-vector exterior gateway protocol for routing between autonomous systems.

Step-by-Step Solution

1
Identify the protocol characteristics of RIP.
RIP relies on distance-vector algorithms and measures distance by hop count up to 15.
Hop count is the defining metric for legacy RIP implementations.
2
Identify the protocol characteristics of OSPF.
OSPF is a link-state protocol using interface bandwidth to determine link cost.
Link-state protocols build map topologies and use bandwidth cost for shortest path selection.
3
Identify the protocol characteristics of BGP.
BGP is an exterior gateway protocol using path-vector logic for inter-AS routing.
BGP handles routing between distinct autonomous systems across the Internet backbone.

Key Concept

Routing Concepts and Dynamic Routing Protocols
Question 269Question →

A network administrator is deploying a multi-VLAN enterprise network architecture. Workstations connected to VLAN 40 (172.16.40.0/24172.16.40.0/24) fail to receive dynamic IP address configurations and revert to Automatic Private IP Addressing (169.254.x.x169.254.x.x). The centralized DHCP server is located on VLAN 10 (172.16.10.50172.16.10.50). Analysis indicates that client broadcast DHCPDISCOVER messages are arriving at the default gateway interface but are dropped at the Layer 3 boundary. Which of the following commands must the administrator configure on the VLAN 40 router interface to resolve this issue?

Show answer & explanation

Answer: ip helper-address 172.16.10.50

Answer

ip helper-address 172.16.10.50
Applying the `ip helper-address 172.16.10.50` command configures the router interface as a DHCP Relay Agent. It intercepts local broadcast DHCPDISCOVER traffic from VLAN 40 clients and forwards it as unicast UDP traffic to the target DHCP server IP address on VLAN 10.

Step-by-Step Solution

1
Identify the cause of client IP configuration failure across subnets
DHCPDISCOVER broadcast packets cannot traverse Layer 3 boundaries without explicit relay intervention.
Routers drop Layer 2/3 broadcast frames by default to prevent broadcast storms.
2
Determine the correct IP addressing service mechanism required
A DHCP Relay Agent command (`ip helper-address`) must be applied to the client subnet's ingress router interface.
The relay agent intercepts local broadcast DHCPDISCOVER messages and encapsulates them into unicast UDP packets destined for the centralized DHCP server address.

Key Concept

DHCP Relay Agent and IP Helper Configuration
Question 270Question →

A network technician is configuring Quality of Service (QoS) for real-time VoIP traffic across an IPv6 enterprise network. The technician needs to tag IPv6 packet headers at Layer 3 so that intermediate routers can classify and prioritize voice packets. Which IPv6 header field is specifically used to carry Differentiated Services Code Point (DSCP) and Explicit Congestion Notification (ECN) information?

Show answer & explanation

Answer: Traffic Class

Answer

The Traffic Class field is the IPv6 header component designed to carry DSCP and ECN values for packet classification and QoS enforcement.
The Traffic Class field in the fixed 40-byte IPv6 header is an 8-bit field designed specifically for packet classification and traffic prioritization. It directly corresponds to the IPv4 Type of Service (ToS) / Differentiated Services (DiffServ) byte, carrying 6 bits of DSCP data and 2 bits of ECN data.

Step-by-Step Solution

1
Identify the QoS requirement in the scenario
The requirement is to mark Layer 3 IPv6 headers with DSCP and ECN values for traffic prioritization.
Quality of Service relies on standardized header fields to inform intermediate network devices how to queue and handle traffic.
2
Evaluate the function of IPv6 header fields
The 8-bit Traffic Class field stores 6 bits for DSCP (Differentiated Services Code Point) and 2 bits for ECN (Explicit Congestion Notification).
This field mirrors the IPv4 DiffServ/ToS byte structure, enabling compatible QoS policy enforcement across IPv6 routers.

Key Concept

IPv6 Base Header Structure and Traffic Class Field
Question 271Question →

A system administrator needs to configure firewall access for a new web server that will support secure command-line administration and encrypted web traffic. Which of the following transport protocols and default port numbers must be permitted through the firewall? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: TCP port 22; TCP port 443

Answer

TCP port 22 and TCP port 443
SSH uses TCP port 22 to enable encrypted administrative terminal sessions, and HTTPS uses TCP port 443 to secure web communications with TLS/SSL encryption.

Step-by-Step Solution

1
Determine the protocol and transport port required for secure command-line management.
SSH uses TCP port 22.
SSH provides encrypted shell access and requires TCP for reliable, connection-oriented data delivery.
2
Determine the protocol and transport port required for encrypted web traffic.
HTTPS uses TCP port 443.
HTTPS encrypts HTTP communication using TLS over connection-oriented TCP port 443.

Key Concept

Default transport protocol and port mappings for secure administration (SSH on TCP port 22) and secure web communications (HTTPS on TCP port 443).
Question 272Question →

A network technician is troubleshooting an enterprise Network Management System (NMS). The NMS can successfully poll managed routers and switches for status metrics, but it fails to receive automated event notifications, such as link-down alerts, generated directly by the network devices across an internal firewall. Network documentation confirms that outbound traffic from the NMS to managed devices on UDP port 161 is permitted. Which of the following firewall rule modifications will resolve the issue?

Show answer & explanation

Answer: Permit inbound traffic from the managed devices to the NMS on UDP port 162.

Answer

Permit inbound traffic from the managed devices to the NMS on UDP port 162.
Simple Network Management Protocol (SNMP) uses two separate port numbers: UDP port 161 for agent polling (GET/SET requests initiated by the NMS) and UDP port 162 for trap notifications (spontaneous alerts initiated by managed agent devices to the NMS). Because the NMS can poll but cannot receive alerts, the firewall is blocking agent-initiated traffic destined for UDP port 162 on the management server.

Step-by-Step Solution

1
Analyze the operational traffic flow described in the scenario.
The NMS polls devices successfully using SNMP GET/SET requests, but spontaneous alert messages (traps/informs) originating from the devices are blocked.
SNMP operates using two distinct communication channels and default port assignments depending on whether the manager or agent initiates communication.
2
Identify the standard protocol and port assigned to SNMP manager polling versus agent traps.
Manager-to-agent polling uses UDP port 161, whereas agent-to-manager asynchronous trap notifications use UDP port 162.
Firewalls must be configured to permit traffic initiated by remote agents destined for the monitoring server's trap receiver service.
3
Select the access list modification that permits asynchronous alerts.
Permitting UDP port 162 inbound from managed network equipment to the management server restores trap logging.
UDP port 162 specifically handles incoming SNMP alert notifications.

Key Concept

SNMP Ports (UDP 161 for Polling/Agent vs UDP 162 for Traps/NMS Manager)
Estimated Time:2m 0s
Question 273Question →

A network technician is configuring and auditing Layer 2 features on an enterprise switch stack. Match each Layer 2 switching concept on the left to its correct operational description on the right.

Click a left item, then click its matching right item

Items

802.1Q Encapsulation
Native VLAN
Port Security
Spanning Tree Protocol

Matches

Show answer & explanation

Answer

802.1Q Encapsulation matches with inserting a 4-byte VLAN field into the Ethernet header. Native VLAN matches with processing untagged frames on trunk ports. Port Security matches with restricting access based on source MAC addresses. Spanning Tree Protocol matches with placing redundant ports in a non-forwarding state to prevent loops.
Each Layer 2 switching feature directly matches its operational behavior: 802.1Q inserts a 4-byte header for trunk tagging; Native VLAN handles untagged frames on trunk links; Port Security restricts ingress based on source MAC address rules; and Spanning Tree Protocol blocks redundant ports to avoid Layer 2 loops.

Step-by-Step Solution

1
Identify the primary frame tagging mechanism used across Layer 2 trunk interfaces.
802.1Q Encapsulation modifies standard Ethernet frames by adding a 4-byte header tag containing the VLAN ID.
Trunk links require frame tagging to distinguish frames belonging to different logical networks.
2
Determine how unencapsulated (untagged) frames are processed on an 802.1Q trunk port.
The Native VLAN is assigned to process any untagged frames sent or received across an 802.1Q trunk link.
Backward compatibility and explicit handling of control traffic require a designated default untagged path.
3
Analyze Layer 2 access control mechanisms on individual switch interfaces.
Port Security enforces MAC address limits on switch ports to restrict unauthorized network devices.
Evaluating ingress source MAC addresses prevents unauthorized physical network additions.
4
Examine loop suppression protocols in redundant switch topologies.
Spanning Tree Protocol blocks specific redundant switch ports to maintain a loop-free logical forwarding topology.
Broadcast storms caused by physical redundant paths must be mitigated at Layer 2.

Key Concept

Core Layer 2 Switching Concepts and VLAN Operation Mechanics
Estimated Time:1m 30s
Question 274Question →

A technician is assigning IPv4 addresses to host interfaces within the 10.0.1.0/2910.0.1.0/29 subnet. Which of the following are valid, usable host IP addresses for this subnet? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: 10.0.1.410.0.1.4; 10.0.1.610.0.1.6

Answer

The valid usable host IP addresses are 10.0.1.410.0.1.4 and 10.0.1.610.0.1.6.
For the 10.0.1.0/2910.0.1.0/29 network block, a /29/29 mask provides 23=82^3 = 8 total IP addresses. The first address (10.0.1.010.0.1.0) serves as the network ID and the final address (10.0.1.710.0.1.7) serves as the broadcast address. The usable host range includes 10.0.1.110.0.1.1 through 10.0.1.610.0.1.6. Therefore, 10.0.1.410.0.1.4 and 10.0.1.610.0.1.6 are valid usable host addresses.

Step-by-Step Solution

1
Determine the block size from the prefix length
A /29/29 prefix leaves 3 host bits (32−29=332 - 29 = 3), giving a block size of 23=82^3 = 8 IP addresses.
Calculating the block size identifies the total boundaries of the subnet.
2
Identify the network and broadcast addresses
The network address is 10.0.1.010.0.1.0 and the broadcast address is 10.0.1.710.0.1.7.
The first address in the block is reserved for the network ID and the last address is reserved for broadcast.
3
Determine the usable host IP range
The usable range spans from 10.0.1.110.0.1.1 to 10.0.1.610.0.1.6.
Usable host IPs consist of all addresses strictly between the network ID and the broadcast address.
4
Select options that fall within the usable range
Both 10.0.1.410.0.1.4 and 10.0.1.610.0.1.6 lie within 10.0.1.110.0.1.1 to 10.0.1.610.0.1.6.
Any address inside this range can be assigned to a network interface.

Key Concept

Usable IPv4 Host Range Calculation
Question 275Question →

A network administrator is allocating subnets sequentially from the 10.10.16.0/2010.10.16.0/20 parent block using Variable Length Subnet Masking (VLSM), assigning the largest required blocks first. The administrator has already completed subnets for three departments:

- Department A requires 500500 usable hosts
- Department B requires 250250 usable hosts
- Department C requires 120120 usable hosts

Immediately following these allocations, the administrator provisions a new subnet for Department D, which requires at least 6060 usable hosts. What is the broadcast address of Department D's newly provisioned subnet?

Show answer & explanation

Answer: 10.10.19.19110.10.19.191

Answer

The broadcast address of Department D's newly provisioned subnet is 10.10.19.19110.10.19.191.
To find the broadcast address of Department D's subnet, subnets must be allocated sequentially in order of decreasing size starting from 10.10.16.0/2010.10.16.0/20:
1. Department A (500500 hosts) requires a /23/23 block (512512 addresses), covering 10.10.16.010.10.16.0 – 10.10.17.25510.10.17.255.
2. Department B (250250 hosts) requires a /24/24 block (256256 addresses), covering 10.10.18.010.10.18.0 – 10.10.18.25510.10.18.255.
3. Department C (120120 hosts) requires a /25/25 block (128128 addresses), covering 10.10.19.010.10.19.0 – 10.10.19.12710.10.19.127.
4. Department D (6060 hosts) requires a /26/26 block (6464 addresses), starting at 10.10.19.12810.10.19.128 and ending at 10.10.19.19110.10.19.191. Thus, the broadcast address for Department D is 10.10.19.19110.10.19.191.

Step-by-Step Solution

1
Determine prefix length and address range for Department A (500 usable hosts)
Prefix length: /23/23 (29−2=510≥5002^9 - 2 = 510 \ge 500). Range: 10.10.16.010.10.16.0 to 10.10.17.25510.10.17.255. Next available IP: 10.10.18.010.10.18.0.
A /23/23 subnet provides 512 total addresses (510 usable), satisfying the 500 host requirement.
2
Determine prefix length and address range for Department B (250 usable hosts)
Prefix length: /24/24 (28−2=254≥2502^8 - 2 = 254 \ge 250). Range: 10.10.18.010.10.18.0 to 10.10.18.25510.10.18.255. Next available IP: 10.10.19.010.10.19.0.
A /24/24 subnet provides 256 total addresses (254 usable), satisfying the 250 host requirement.
3
Determine prefix length and address range for Department C (120 usable hosts)
Prefix length: /25/25 (27−2=126≥1202^7 - 2 = 126 \ge 120). Range: 10.10.19.010.10.19.0 to 10.10.19.12710.10.19.127. Next available IP: 10.10.19.12810.10.19.128.
A /25/25 subnet provides 128 total addresses (126 usable), satisfying the 120 host requirement.
4
Calculate the subnet boundaries and broadcast address for Department D (60 usable hosts)
Subnet block: 10.10.19.128/2610.10.19.128/26. Network ID: 10.10.19.12810.10.19.128, Usable host range: 10.10.19.12910.10.19.129 – 10.10.19.19010.10.19.190, Broadcast address: 10.10.19.19110.10.19.191.
Department D requires 6060 hosts, so 26−2=62≥602^6 - 2 = 62 \ge 60 requires a /26/26 prefix (block size of 64). Adding 64 to the starting address 10.10.19.12810.10.19.128 gives the next network boundary 10.10.19.19210.10.19.192, making 10.10.19.19110.10.19.191 the broadcast address.

Key Concept

Variable Length Subnet Masking (VLSM) Allocation and Boundary Calculation
Estimated Time:3m 0s
Question 276Question →

A network administrator needs to enable automatic network route updates between routers within a single autonomous system. Which type of routing protocol should the administrator deploy to meet this requirement?

Show answer & explanation

Answer: Interior Gateway Protocol (IGP)

Answer

Interior Gateway Protocol (IGP) is the correct choice because IGPs are designed for routing within a single autonomous system.
Interior Gateway Protocols (IGPs) are built specifically to exchange routing information automatically among routers inside a single autonomous system or internal network administration boundary.

Step-by-Step Solution

1
Identify the boundary requirement in the scenario.
The network updates must occur within a single autonomous system.
Routing protocols are classified primarily by their operational scope: interior versus exterior.
2
Match the requirement to the protocol scope classification.
Interior Gateway Protocols (IGPs) operate within a single domain/autonomous system.
Protocols such as OSPF, EIGRP, and RIP are IGPs built for intra-domain dynamic routing.

Key Concept

Interior Gateway Protocol vs. Exterior Gateway Protocol Scope
Question 277Question →

A network engineer is tracing traffic from a host on Subnet A to an application server on Subnet B. The traffic traverses a Layer 2 switch, a Layer 3 default gateway router, and a second Layer 2 switch before reaching the destination server. Assuming Network Address Translation (NAT) is not configured, which of the following statements accurately describe how headers are modified or preserved during encapsulation and de-encapsulation across these network devices? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: The Layer 3 source and destination IP addresses encapsulated by the originating host remain unchanged throughout the path across both subnets.; The default gateway router strips the incoming Layer 2 frame header and encapsulates the IP packet into a new Layer 2 frame before egressing onto the next hop.

Answer

The correct statements are that the Layer 3 source and destination IP addresses remain unchanged across both subnets (when NAT is not used), and the default gateway router strips the incoming Layer 2 frame header before encapsulating the IP packet in a new Layer 2 frame for the destination subnet.
Layer 3 IP headers provide end-to-end addressing and remain intact across routed hops unless NAT is specifically configured. Conversely, Layer 2 frames are link-local; when a router forwards traffic between subnets, it de-encapsulates the incoming frame (stripping the Layer 2 header and trailer) and re-encapsulates the payload packet into a brand new Layer 2 frame appropriate for the next hop.

Step-by-Step Solution

1
Analyze Layer 2 switch operations during transit.
Layer 2 switches evaluate incoming frames at the Data Link layer using MAC address tables and forward the frame intact without stripping or modifying Layer 2, Layer 3, or Layer 4 headers.
Switches do not perform de-encapsulation above Layer 2.
2
Analyze Layer 3 router operations during hop-by-hop forwarding.
The router receives the frame on its ingress interface, performs de-encapsulation by removing the Layer 2 header/trailer, inspects the Layer 3 IP destination address to make a forwarding decision, and then encapsulates the IP packet into a new Layer 2 frame with updated source and destination MAC addresses.
Data Link layer encapsulation is local to each physical network segment/hop.
3
Evaluate end-to-end protocol headers (Layer 3 and Layer 4).
In the absence of NAT or proxying, Layer 3 IP addresses and Layer 4 TCP/UDP headers pass through intermediate routers untouched.
Layer 3 provides logical end-to-end addressing while Layer 4 manages end-to-end communication sessions between hosts.

Key Concept

Hop-by-hop Layer 2 re-framing vs. end-to-end Layer 3/4 header preservation
Question 278Question →

An enterprise network engineer is deploying Voice over IP (VoIP) telephony across multiple remote subnets. Devices on VLAN 10 receive IP addresses and function normally. However, newly deployed IP phones residing on VLAN 30 successfully acquire IPv4 addresses, default gateway settings, and DNS server addresses from the central DHCP server, but remain stuck in a boot loop reporting 'Server Unreachable' when attempting to download their configuration files. Analysis confirms that IP routing between VLAN 30 and the TFTP/provisioning server is fully functional. Which of the following DHCP service configurations on the central server is missing or misconfigured for the VLAN 30 scope?

Show answer & explanation

Answer: The custom vendor-specific DHCP option (Option 66 or Option 150) specifying the IP address or hostname of the provisioning server is missing from the DHCP scope for VLAN 30.

Answer

The custom vendor-specific DHCP option (Option 66 or Option 150) specifying the TFTP/provisioning server IP address or hostname is missing from the VLAN 30 DHCP scope.
When VoIP endpoints boot up, basic IP address allocation is insufficient for full operation. IP phones rely on specific DHCP scope options, such as Option 66 (TFTP Server Name) or Option 150 (Cisco proprietary option for TFTP server IP addresses), to locate the server containing their operational firmware and configuration files. Because the phones acquired basic IP details but fail specifically at file download, the missing component is the DHCP option pointing to the provisioning server.

Step-by-Step Solution

1
Analyze the symptoms described in the scenario.
IP phones receive standard IP parameters (IP address, gateway, DNS) via DHCP and IP connectivity to the TFTP server exists, but phones cannot locate the configuration server.
Standard DHCP parameters (Options 3 and 6) do not automatically provide vendor-specific provisioning server locations to VoIP devices.
2
Identify the required IP addressing service configuration for automated VoIP provisioning.
DHCP Option 66 (TFTP Server Name) or Option 150 (TFTP Server IP Address) must be defined within the specific DHCP scope.
IP phones require these options to discover where to download firmware and initial configuration files upon completing the basic DHCP lease process.

Key Concept

DHCP Options and Vendor-Specific Boot Configuration
Question 279Question →

A network technician is documenting IPv6 protocol standards for a network operations team. Match each IPv6 header field or address scope prefix on the left with its correct functional definition on the right.

Click a left item, then click its matching right item

Items

ff00::/8
fe80::/10
Next Header
Hop Limit

Matches

Show answer & explanation

Answer

ff00::/8 matches Multicast address scope; fe80::/10 matches Link-Local unicast address scope; Next Header matches Payload protocol identification/extension header; Hop Limit matches Hop-by-hop decrement counter replacing IPv4 TTL.
Each IPv6 prefix and header field is matched according to standard RFC specifications: ff00::/8 defines multicast, fe80::/10 defines link-local unicast, Next Header designates encapsulated protocol or extension headers, and Hop Limit prevents layer 3 routing loops.

Step-by-Step Solution

1
Analyze IPv6 address prefixes
Identify ff00::/8 as the dedicated prefix for multicast traffic and fe80::/10 as the dedicated prefix for link-local unicast traffic.
IPv6 uses distinct leading hex digits to define address scope and delivery methods.
2
Analyze IPv6 fixed header fields
Identify Next Header as the field pointing to payload protocols/extension headers and Hop Limit as the field replacing the IPv4 TTL.
The 40-byte fixed IPv6 header streamlined several IPv4 fields for faster router processing.

Key Concept

IPv6 Addressing Scopes and Base Header Field Functions
Question 280Question →

A network technician is configuring a local DNS server to support host resolution for IPv6 network devices. Which DNS record type must be created to resolve a hostname directly to an IPv6 address?

Show answer & explanation

Answer: AAAA record

Answer

The AAAA record is used to resolve a domain hostname to an IPv6 address.
The AAAA (quad-A) record is specifically designed to map hostnames to 128-bit IPv6 addresses within Domain Name System (DNS) servers.

Step-by-Step Solution

1
Identify the requested service and address family.
The scenario asks for name resolution mapping hostnames to 128-bit IPv6 addresses.
Different address families require distinct resource record types within DNS infrastructure.
2
Select the corresponding DNS resource record type.
The AAAA (quad-A) record stores 128-bit IPv6 addresses.
Standard A records only support 32-bit IPv4 addresses.

Key Concept

DNS Resource Record Types for IP Addressing Services
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