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Question 661Question →

An IT technician is deploying a network infrastructure for a regional facility consisting of four distinct buildings situated on a single contiguous corporate park. The internal networks of each building are connected together using private, high-speed optical fiber cabling owned and managed internally by the organization. Which of the following network scope classifications best describes this setup?

Show answer & explanation

Answer: Campus Area Network (CAN)

Answer

Campus Area Network (CAN)
A Campus Area Network (CAN) is a network connecting multiple local area networks (LANs) within a contiguous geographical area, such as a corporate office park or university campus, utilizing privately owned cabling and switches.

Step-by-Step Solution

1
Analyze the physical scope and infrastructure ownership
The network spans multiple buildings within a single corporate park using privately owned infrastructure.
Geographical boundary and ownership differentiate network types.
2
Map the characteristics to standardized network classifications
A network connecting multiple LANs across a single contiguous property or facility complex is defined as a Campus Area Network (CAN).
CAN is smaller than a MAN/WAN and relies on private infrastructure interconnecting localized LANs.

Key Concept

Campus Area Network (CAN) Characteristics
Estimated Time:1m 0s
Question 662Question →

A network administrator is designing and hardening VLAN trunking parameters across enterprise switches to ensure proper frame encapsulation, DTP negotiation, and VLAN prune/allow list configurations. Match each VLAN configuration parameter or frame header attribute on the left with its corresponding operational behavior on the right.

Click a left item, then click its matching right item

Items

switchport trunk native vlan 99
switchport trunk allowed vlan add 10,20
switchport mode dynamic auto
IEEE 802.1Q Tag Structure

Matches

Show answer & explanation

Answer

Each configuration command or 802.1Q frame attribute corresponds to a specific operational mechanism: native VLAN designation handles untagged frame egress, the allowed list 'add' modifier preserves existing allowed VLAN definitions while appending new ones, dynamic auto modes passively negotiate trunking via DTP, and 802.1Q tagging places a 4-byte header containing VLAN ID and CoS fields immediately following the Ethernet Source MAC address.
The matching pairs correctly connect switch configuration commands and frame attributes to their exact operational outcomes: 'switchport trunk native vlan 99' specifies untagged frame handling for VLAN 99; 'switchport trunk allowed vlan add 10,20' preserves existing allowed VLANs while appending VLANs 10 and 20; 'switchport mode dynamic auto' passively waits for DTP requests; and the IEEE 802.1Q Tag Structure inserts a 4-byte header (TPID 0x8100 and TCI) between the Source MAC address and the EtherType field.

Step-by-Step Solution

1
Analyze native VLAN behavior under IEEE 802.1Q standard.
Identified that native VLAN traffic bypasses 802.1Q encapsulation and egresses the trunk port untagged.
802.1Q trunk ports send and expect native VLAN frames without a tag header for backward compatibility with unmanaged switches.
2
Evaluate CLI syntax for trunk allowed VLAN modification.
Determined that using 'switchport trunk allowed vlan add' appends entries without clearing existing allowed VLANs.
Omitting the 'add' parameter overwrites the entire permitted VLAN list, which is a common misconfiguration error.
3
Evaluate Dynamic Trunking Protocol (DTP) operational modes.
Matched dynamic auto to passive trunk negotiation.
Ports in dynamic auto mode respond to incoming DTP request frames from dynamic desirable or trunk ports but will not actively initiate negotiation.
4
Analyze Ethernet frame fields for IEEE 802.1Q encapsulation.
Matched 802.1Q header details to its 4-byte structure containing TPID 0x8100, PCP, DEI, and 12-bit VLAN ID.
The 802.1Q tag is inserted between the Source MAC address and the original EtherType/Length field in Ethernet frames.

Key Concept

VLAN Configuration, DTP Operational Modes, and 802.1Q Frame Header Tagging Mechanics
Estimated Time:2m 30s
Question 663Question →

A network administrator is configuring IP addressing services for a newly created client subnet (VLAN 30) across a routed enterprise network. The centralized DHCP server is located on a different subnet (VLAN 10). Users on VLAN 30 report that their workstations fail to acquire dynamic IP addresses and instead receive APIPA addresses in the range 169.254.0.0/16169.254.0.0/16. Which configuration change must be implemented on the default gateway router interface for VLAN 30 to resolve this issue?

Show answer & explanation

Answer: Configure an IP helper address (DHCP relay agent) on the VLAN 30 gateway interface pointing to the DHCP server IP address

Answer

Configure an IP helper address (DHCP relay agent) on the VLAN 30 gateway interface pointing to the DHCP server IP address
DHCP clients discover servers using local broadcasts. Because Layer 3 routers drop broadcast traffic by default, a DHCP relay agent (configured via an IP helper address on the router interface local to the client subnet) is required to translate client broadcast requests into unicast packets sent directly to the central DHCP server.

Step-by-Step Solution

1
Analyze client failure symptoms
Workstations receive APIPA addresses (169.254.x.x169.254.x.x), indicating they cannot reach a DHCP server to obtain an IP lease.
Routers do not forward Layer 2 broadcast packets (such as DHCPDISCOVER messages) across subnet boundaries by default.
2
Identify the required network service component
A DHCP relay agent (IP helper address) is needed on the client-facing gateway interface.
The relay agent intercepts client broadcast messages on VLAN 30 and encapsulates them into unicast packets directed to the DHCP server on VLAN 10.

Key Concept

DHCP Relay Agent and IP Helper Functionality across Layer 3 Subnet Boundaries
Question 664Question →

A network architect is designing a VLSM address allocation plan starting at the beginning of the 192.168.100.0/23192.168.100.0/23 address block. Subnets must be provisioned sequentially without wasting address space in the following order:
1. Subnet A: Requires support for 120120 usable host IP addresses.
2. Subnet B: Requires support for 5050 usable host IP addresses.
3. Subnet C: Requires a point-to-point link supporting 22 router interfaces.
4. Subnet D: Requires support for 2525 usable host IP addresses.

Assuming all subnets are properly aligned to their required CIDR boundary offsets, what is the broadcast IP address of Subnet D?

Show answer & explanation

Answer: 192.168.100.255192.168.100.255

Answer

The broadcast IP address of Subnet D is 192.168.100.255192.168.100.255.
Subnet A requires 120 hosts, fitting into a /25 block (192.168.100.0−192.168.100.127192.168.100.0 - 192.168.100.127). Subnet B requires 50 hosts, fitting into a /26 block (192.168.100.128−192.168.100.191192.168.100.128 - 192.168.100.191). Subnet C requires 2 hosts, fitting into a /30 block (192.168.100.192−192.168.100.195192.168.100.192 - 192.168.100.195). The next available IP address is 192.168.100.196192.168.100.196. Subnet D requires 25 hosts, requiring a /27 subnet (block size 32). Because /27/27 subnets must start on a boundary divisible by 32 (0,32,64,96,128,160,192,2240, 32, 64, 96, 128, 160, 192, 224), Subnet D must start at 192.168.100.224/27192.168.100.224/27. The range for Subnet D is 192.168.100.224192.168.100.224 through 192.168.100.255192.168.100.255, making 192.168.100.255192.168.100.255 the correct broadcast address.

Step-by-Step Solution

1
Calculate prefix size and address range for Subnet A
Subnet A needs 120120 usable hosts (27−2=126≥1202^7 - 2 = 126 \ge 120), requiring a /25/25 mask (block size 128128). Range: 192.168.100.0/25192.168.100.0/25 (192.168.100.0192.168.100.0 to 192.168.100.127192.168.100.127).
Determines the ending address of Subnet A.
2
Calculate prefix size and address range for Subnet B
Subnet B starts at 192.168.100.128192.168.100.128. It needs 5050 usable hosts (26−2=62≥502^6 - 2 = 62 \ge 50), requiring a /26/26 mask (block size 6464). Range: 192.168.100.128/26192.168.100.128/26 (192.168.100.128192.168.100.128 to 192.168.100.191192.168.100.191).
Determines the ending address of Subnet B.
3
Calculate prefix size and address range for Subnet C
Subnet C starts at 192.168.100.192192.168.100.192. It needs 22 usable hosts (22−2=22^2 - 2 = 2), requiring a /30/30 mask (block size 44). Range: 192.168.100.192/30192.168.100.192/30 (192.168.100.192192.168.100.192 to 192.168.100.195192.168.100.195).
Determines the next available unaligned address (192.168.100.196192.168.100.196).
4
Determine prefix size and boundary alignment for Subnet D
Subnet D needs 2525 usable hosts (25−2=30≥252^5 - 2 = 30 \ge 25), requiring a /27/27 mask (block size 3232). The next unaligned IP is 192.168.100.196192.168.100.196. Since a /27/27 network address must be divisible by 3232, the next valid network boundary is 192.168.100.224192.168.100.224.
Enforces natural CIDR boundary alignment rules for variable-length subnets.
5
Calculate the broadcast address of Subnet D
Network ID: 192.168.100.224/27192.168.100.224/27. Range: 192.168.100.224192.168.100.224 to 192.168.100.255192.168.100.255. Broadcast address = 192.168.100.255192.168.100.255.
Adding the block size minus one (32−1=3132 - 1 = 31) to the network address gives the broadcast address.

Key Concept

VLSM Subnet Sizing and Bit-Boundary Alignment
Estimated Time:3m 0s
Question 665Question →

A network administrator configures a DHCP relay agent on a Layer 3 router interface to forward IP address requests from client hosts to a centralized DHCP server located on a remote subnet. Which TWO actions are performed by the DHCP relay agent during this process? (Select TWO).

Select all that apply

Show answer & explanation

Answer: It converts incoming client Layer 2 DHCP broadcast packets into unicast IP packets destined for the central DHCP server IP address.; It populates the Gateway IP Address (giaddr) field in the DHCP header with its own interface IP address.

Answer

The DHCP relay agent converts local broadcast DHCP client messages into unicast IP packets targeted at the remote DHCP server, and populates the Gateway IP Address (giaddr) field in the DHCP packet header with its own interface IP address.
When client hosts generate broadcast DHCP requests, the relay agent on their local gateway interface captures the broadcast, places its own interface IP into the Gateway IP Address (giaddr) field, and forwards the request as a unicast packet to the configured DHCP server IP. These two functions ensure cross-subnet routing and correct scope selection.

Step-by-Step Solution

1
Analyze how broadcast packets cross Layer 3 boundaries
Layer 3 routers block broadcast traffic by default, requiring the relay agent to convert the client broadcast into a unicast packet.
This allows the DHCPDISCOVER/DHCPREQUEST packets to be routed across subnets to the central server IP address.
2
Determine how the remote DHCP server identifies the correct IP pool
The relay agent sets the Gateway IP Address (giaddr) field to its receiving interface IP address.
The central server evaluates the giaddr value to select the matching address pool for the requesting client subnet.

Key Concept

DHCP Relay Agent Broadcast-to-Unicast Forwarding and GIADDR Field Tagging
Estimated Time:1m 0s
Question 666Question →

An edge router at a regional office receives an IPv4 packet destined for 172.28.64.15172.28.64.15. The active routing table contains multiple valid routes from OSPF, BGP, and static configurations that match the destination network. Which TWO of the following statements correctly describe the decision process the router uses to select the forwarding path?

Select all that apply

Show answer & explanation

Answer: The router evaluates and selects the route with the longest prefix match (most specific subnet mask) prior to considering administrative distance or metric.; When multiple candidate routes have identical prefix lengths, the router selects the path associated with the lowest administrative distance value.

Answer

The router first selects the route with the longest prefix match (most specific subnet mask). If candidate routes have identical prefix lengths, the router then selects the route with the lowest administrative distance.
When forwarding an IP packet, a router follows a strict decision hierarchy. First, it identifies the route with the longest prefix match (the most specific network mask). Second, if multiple routes from different sources have identical prefix lengths, the router compares their administrative distance (AD) and selects the route with the lowest AD value.

Step-by-Step Solution

1
Evaluate subnet mask specificity across candidate routes.
Identify the route with the longest prefix match.
The router always prioritizes the most specific destination path regardless of protocol metric or administrative distance.
2
Compare administrative distance values if prefix lengths are equal.
Select the route learned from the source with the lowest administrative distance.
Administrative distance measures the trustworthiness of the routing source (e.g., connected, static, OSPF, BGP).

Key Concept

Routing Decision Hierarchy: Longest Prefix Match and Administrative Distance
Question 667Question →

A network infrastructure engineer is dividing the IPv4 address space 172.28.144.0/20172.28.144.0/20 using Variable Length Subnet Masking (VLSM). The subnets must be allocated contiguously starting from the lowest available network address in descending order of host requirements without wasting address blocks between subnets:

1. Subnet A: Requires 450450 usable host IP addresses.
2. Subnet B: Requires 190190 usable host IP addresses.
3. Subnet C: Requires 6060 usable host IP addresses.

What is the broadcast IP address of Subnet C?

Show answer & explanation

Answer: 172.28.147.63

Answer

172.28.147.63
Subnet C begins at 172.28.147.0172.28.147.0 after allocating a /23/23 (512512 addresses) for Subnet A and a /24/24 (256256 addresses) for Subnet B. A /26/26 prefix accommodates 6060 hosts (6262 usable, block size of 6464), making the address range 172.28.147.0172.28.147.0 through 172.28.147.63172.28.147.63. The last address in this subnet block (172.28.147.63172.28.147.63) serves as the broadcast IP address.

Step-by-Step Solution

1
Determine prefix length and block size for Subnet A (450 usable hosts)
Subnet A requires 99 host bits (29−2=5102^9 - 2 = 510 usable IPs), giving a CIDR mask of /23/23 (32−932 - 9) and a block size of 512512 addresses (292^9). Allocated range: 172.28.144.0172.28.144.0 to 172.28.145.255172.28.145.255.
Subnets must be allocated in descending order of required hosts.
2
Determine prefix length, block size, and range for Subnet B (190 usable hosts)
Next available start IP is 172.28.146.0172.28.146.0. Subnet B requires 88 host bits (28−2=2542^8 - 2 = 254 usable IPs), giving a CIDR mask of /24/24 (32−832 - 8) and a block size of 256256 addresses (282^8). Allocated range: 172.28.146.0172.28.146.0 to 172.28.146.255172.28.146.255.
Subnet B must immediately follow Subnet A contiguously.
3
Determine prefix length, block size, and broadcast address for Subnet C (60 usable hosts)
Next available start IP is 172.28.147.0172.28.147.0. Subnet C requires 66 host bits (26−2=622^6 - 2 = 62 usable IPs), giving a CIDR mask of /26/26 (32−632 - 6) and a block size of 6464 addresses (262^6). Network ID: 172.28.147.0/26172.28.147.0/26. Broadcast address: 172.28.147.63172.28.147.63.
The broadcast address is the last IP in the /26/26 block (172.28.147.0+63172.28.147.0 + 63).

Key Concept

Variable Length Subnet Masking (VLSM) contiguous allocation
Question 668Question →

A network administrator is setting up a new client subnet on VLAN 10. The centralized enterprise DHCP server resides on a separate subnet across a router on VLAN 20. When client workstations on VLAN 10 boot up, they fail to obtain an IP address because DHCP Discover broadcast messages are stopped by the router. Which command or feature should be configured on the router's VLAN 10 interface to forward these requests to the DHCP server?

Show answer & explanation

Answer: DHCP Relay Agent (ip helper-address) specifying the IP address of the DHCP server

Answer

Configure a DHCP Relay Agent (or ip helper-address) on the router interface facing the client subnet, configured with the unicast IP address of the remote DHCP server.
Configuring a DHCP relay agent (or ip helper-address) on the interface connected to the client subnet intercept broadcasts on UDP port 67 and forwards them as unicast traffic directly to the DHCP server's IP address on the remote subnet.

Step-by-Step Solution

1
Identify the cause of the DHCP failure
Routers drop broadcast traffic by default, preventing DHCP Discover broadcast messages originating on VLAN 10 from reaching the server on VLAN 20.
DHCP discovery relies on Layer 2 and Layer 3 broadcasts (255.255.255.255) which cannot cross network boundaries without assistance.
2
Determine the necessary mechanism to bridge subnets for DHCP traffic
A DHCP Relay Agent (configured via 'ip helper-address' in Cisco IOS) must be enabled on the client-facing router interface.
The relay agent intercepts client DHCP broadcasts on the local interface, encapsulates them into unicast IP packets directed to the designated DHCP server IP, and relays the unicast replies back to the client.

Key Concept

DHCP Relay Services and IP Helper Address
Question 669Question →

A technician is configuring IPv4 address settings for workstations on the 192.168.10.0/26192.168.10.0/26 subnet. Which of the following IP addresses are valid usable host addresses that can be assigned to devices on this subnet? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: 192.168.10.15192.168.10.15; 192.168.10.50192.168.10.50

Answer

The valid usable host IP addresses are 192.168.10.15192.168.10.15 and 192.168.10.50192.168.10.50.
For the subnet 192.168.10.0/26192.168.10.0/26, the block size is 6464 addresses (262^6). The network address is 192.168.10.0192.168.10.0, and the broadcast address is 192.168.10.63192.168.10.63. Consequently, the assignable usable host address range spans from 192.168.10.1192.168.10.1 to 192.168.10.62192.168.10.62. The addresses 192.168.10.15192.168.10.15 and 192.168.10.50192.168.10.50 fall directly within this range.

Step-by-Step Solution

1
Determine the total block size for a /26/26 subnet prefix.
A /26/26 subnet mask leaves 32−26=632 - 26 = 6 host bits, yielding a total block size of 26=642^6 = 64 addresses.
Identifying block size helps establish subnet boundaries.
2
Identify the network address, broadcast address, and usable host range.
Network address: 192.168.10.0192.168.10.0. Broadcast address: 192.168.10.63192.168.10.63. Usable host range: 192.168.10.1192.168.10.1 through 192.168.10.62192.168.10.62.
Network and broadcast addresses must be subtracted from the total pool to find assignable addresses.
3
Evaluate each given IP address against the usable range.
192.168.10.15192.168.10.15 and 192.168.10.50192.168.10.50 reside between 192.168.10.1192.168.10.1 and 192.168.10.62192.168.10.62, making them valid host addresses.
Only IP addresses within the usable boundaries can be assigned to endpoint interfaces.

Key Concept

Usable IPv4 Host Address Range Calculation
Question 670Question →

A network administrator is inspecting the interface configurations on two interconnecting enterprise switches, SW-CORP-A and SW-CORP-B. The partial configuration snippet for interface GigabitEthernet1/0/12 on SW-CORP-A is shown below:

interface GigabitEthernet1/0/12
switchport mode trunk
switchport trunk native vlan 50
switchport trunk allowed vlan 10,20,30

The corresponding interconnecting interface GigabitEthernet1/0/12 on SW-CORP-B is configured as follows:

interface GigabitEthernet1/0/12
switchport mode trunk
switchport trunk native vlan 10
switchport trunk allowed vlan 10,20,50

If an untagged broadcast frame originating from an end-host assigned to VLAN 50 arrives at SW-CORP-A, what will happen to the frame when SW-CORP-A attempts to process it for transmission across interface GigabitEthernet1/0/12?

Show answer & explanation

Answer: The frame is dropped at the egress interface of SW-CORP-A because VLAN 50 is excluded from the trunk's allowed VLAN list.

Answer

The frame is dropped at the egress interface of SW-CORP-A because VLAN 50 is excluded from the trunk's allowed VLAN list.
On IEEE 802.1Q trunk interfaces, the allowed VLAN list ('switchport trunk allowed vlan') defines which VLANs are permitted to send and receive frames across the link. Omitting a VLAN from this list prevents all frame transmission for that VLAN over the trunk. Even though VLAN 50 is specified as the native VLAN on SW-CORP-A, its exclusion from the allowed VLAN list (which only permits VLANs 10, 20, and 30) causes SW-CORP-A to drop any egress frames for VLAN 50 at interface GigabitEthernet1/0/12.

Step-by-Step Solution

1
Examine the trunk interface configuration on SW-CORP-A for GigabitEthernet1/0/12.
The interface is set to trunk mode with 'switchport trunk native vlan 50' and 'switchport trunk allowed vlan 10,20,30'.
Determining both the native VLAN setting and the allowed VLAN list is required to evaluate frame forwarding logic.
2
Evaluate the interaction between the native VLAN configuration and the allowed VLAN list on SW-CORP-A.
VLAN 50 is configured as the native VLAN, but VLAN 50 is NOT included in the allowed VLAN list (10, 20, 30).
On 802.1Q trunks, the allowed VLAN list acts as an absolute filter for egress and ingress traffic.
3
Determine the frame handling behavior for ingress VLAN 50 traffic on SW-CORP-A exiting out GigabitEthernet1/0/12.
Because VLAN 50 is absent from the allowed VLAN list, SW-CORP-A drops all egress frames belonging to VLAN 50 at the interface.
Excluding a VLAN from the allowed list prevents any frames for that VLAN from traversing the trunk, overriding native VLAN tagging/untagging parameters.

Key Concept

802.1Q Trunk Allowed VLAN Filtering vs Native VLAN Configuration
Question 671Question →

Place the steps of a DHCP Relay Agent forwarding a client's IP address request across subnets into the correct chronological order, from initial client transmission to destination server delivery.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence starts with the client broadcasting a DHCPDISCOVER message, followed by the relay agent receiving the broadcast, inserting its gateway IP address (giaddr) and converting the frame to unicast, and finally transmitting the unicast packet to the remote DHCP server.
The sequence begins when an IP-less client broadcasts a DHCPDISCOVER message on its local VLAN. The default gateway's DHCP relay agent intercepts this broadcast, inserts its local interface address into the Gateway IP Address (giaddr) field, rewrites the packet as a unicast message, and forwards it to the target DHCP server.

Step-by-Step Solution

1
Identify the client's initial discovery attempt
The client broadcasts a DHCPDISCOVER packet on the local segment.
Clients without an IP address cannot communicate via unicast and must use local broadcasts.
2
Identify the relay agent's packet interception
The router interface acting as a DHCP relay agent receives the broadcast packet.
Routers block broadcasts by default unless configured to act as relay agents on that interface.
3
Identify the packet header modification and re-encapsulation
The relay agent populates the giaddr field with its own interface IP address and converts the frame from broadcast to unicast.
The giaddr field tells the central DHCP server which IP pool to allocate an address from.
4
Identify final delivery to the server
The unicast DHCPDISCOVER packet is routed across subnets to the DHCP server.
Unicast packets can be routed through Layer 3 infrastructure to remote subnets.

Key Concept

DHCP Relay Broadcast-to-Unicast Conversion and Gateway IP (giaddr) Tagging
Question 672Question →

A network administrator is provisioning an IPv4 network segment for a branch office using the subnet mask 255.255.255.224255.255.255.224 (/2727). What is the maximum number of usable host IPv4 addresses that can be assigned to devices on this subnet?

Show answer & explanation

Answer: 30

Answer

30 usable host IPv4 addresses
A subnet mask of 255.255.255.224255.255.255.224 corresponds to CIDR notation /2727. Subtracting 27 network bits from the 32 total bits in an IPv4 address leaves 5 host bits. The total number of IP addresses in this block is 25=322^5 = 32. Subtracting the network address and the broadcast address (32−232 - 2) results in 30 usable host IP addresses.

Step-by-Step Solution

1
Determine the number of host bits in the subnet.
5 host bits (32−27=532 - 27 = 5).
An IPv4 address contains 32 bits in total, and a /2727 prefix reserves 27 bits for the network portion.
2
Calculate the total number of available IP addresses.
25=322^5 = 32 total IP addresses.
The total address capacity is determined by 2n2^n, where nn is the number of host bits.
3
Subtract reserved addresses to find the usable host count.
32−2=3032 - 2 = 30 usable host addresses.
The first address (network ID) and the last address (broadcast address) cannot be assigned to individual hosts.

Key Concept

Calculating usable host IPv4 addresses using CIDR notation and subnet masks
Estimated Time:45s
Question 673Question →

A system administrator is configuring perimeter firewall rules for a newly provisioned edge appliance. The appliance requires remote encrypted web-based management and secure file upload capabilities via Secure Shell (SSH). Which TWO of the following port and transport protocol combinations must be opened inbound on the firewall to satisfy these operational requirements? (Select TWO)

Select all that apply

Show answer & explanation

Answer: TCP port 443 for HTTPS administrative access; TCP port 22 for SSH-based Secure File Transfer Protocol (SFTP)

Answer

The firewall rules must permit inbound traffic on TCP port 443 for HTTPS administrative access and TCP port 22 for SFTP secure file transfer.
TCP port 443 is the standard port reserved for HTTPS traffic, providing encrypted web management capabilities over connection-oriented TCP. TCP port 22 is the standard port for Secure Shell (SSH) services, which includes SFTP for secure file operations.

Step-by-Step Solution

1
Determine the default transport protocol and port for encrypted web access.
HTTPS uses TCP port 443.
Hypertext Transfer Protocol Secure (HTTPS) requires reliable connection-oriented delivery over TCP port 443.
2
Determine the default transport protocol and port for secure SSH file management.
SFTP uses TCP port 22.
SSH File Transfer Protocol (SFTP) runs as an subsystem of SSH, utilizing TCP port 22.

Key Concept

Default transport protocols and port assignments for encrypted web and remote management services
Question 674Question →

A network architect is implementing Variable Length Subnet Masking (VLSM) on the assigned address block 172.16.48.0/20172.16.48.0/20 to provision distinct functional zones in a datacenter. Zone A requires host capacity for at least 500 IP addresses, while Zone B requires host capacity for at least 250 IP addresses. Subnets are allocated sequentially starting from the lowest available network address of the block, with Zone A allocated first, followed immediately by Zone B. What is the broadcast address of the subnet allocated to Zone B?

Show answer & explanation

Answer: 172.16.50.255172.16.50.255

Answer

The broadcast address of the subnet allocated to Zone B is 172.16.50.255172.16.50.255.
To host at least 500 devices in Zone A, 9 host bits are needed (29−2=5102^9 - 2 = 510), resulting in a /23/23 subnet (172.16.48.0172.16.48.0 to 172.16.49.255172.16.49.255). The next available network address for Zone B is 172.16.50.0172.16.50.0. To host 250 devices in Zone B, 8 host bits are required (28−2=2542^8 - 2 = 254), resulting in a /24/24 subnet (172.16.50.0172.16.50.0 to 172.16.50.255172.16.50.255). The broadcast address of this subnet is the final address in its range, 172.16.50.255172.16.50.255.

Step-by-Step Solution

1
Determine prefix length and block size for Zone A (500 hosts required)
Host bits needed: 2h−2≥500  ⟹  h=92^h - 2 \ge 500 \implies h = 9 bits (29−2=5102^9 - 2 = 510 usable hosts). Subnet mask prefix is 32−9=/2332 - 9 = /23, which gives a block size of 29=5122^9 = 512 addresses.
Zone A requires at least 500 hosts, so a /24/24 (254 usable hosts) is insufficient and a /23/23 must be used.
2
Calculate the address boundary for Zone A
Zone A starts at 172.16.48.0/23172.16.48.0/23 and spans 172.16.48.0172.16.48.0 through 172.16.49.255172.16.49.255.
A /23/23 block starting at 172.16.48.0172.16.48.0 increments the third octet by 2 (48+2=5048 + 2 = 50).
3
Determine prefix length and block size for Zone B (250 hosts required)
Starting address for Zone B is 172.16.50.0172.16.50.0. Host bits needed: 2h−2≥250  ⟹  h=82^h - 2 \ge 250 \implies h = 8 bits (28−2=2542^8 - 2 = 254 usable hosts). Subnet mask prefix is 32−8=/2432 - 8 = /24, giving a block size of 28=2562^8 = 256 addresses.
250 hosts require 8 host bits, corresponding to a /24/24 subnet mask.
4
Determine the broadcast address of Zone B
The subnet range for Zone B is 172.16.50.0172.16.50.0 to 172.16.50.255172.16.50.255. The broadcast address is 172.16.50.255172.16.50.255.
The last address in an IP subnet block is reserved as the broadcast address.

Key Concept

Variable Length Subnet Masking (VLSM) calculations, subnet block size determination, and broadcast address identification.
Question 675Question →

Match each 802.1Q trunking configuration scenario on the left with its corresponding frame encapsulation and ingress/egress processing behavior on the right.

Click a left item, then click its matching right item

Items

An untagged ethernet frame arrives on an 802.1Q trunk interface configured with native VLAN 40.
An ethernet frame containing an 802.1Q tag with VLAN ID 0 (priority-tagged frame) arrives on a trunk port configured with native VLAN 40.
An ethernet frame tagged with VLAN ID 40 arrives on an 802.1Q trunk interface where VLAN 40 is explicitly configured as the native VLAN.
An ethernet frame tagged with VLAN ID 10 arrives on a standard single-VLAN access port configured with access VLAN 10.

Matches

Show answer & explanation

Answer

Untagged frames on a trunk port map to the native VLAN and egress untagged on matching native trunks. Priority-tagged frames (VLAN 0) preserve 802.1p CoS bits while placing traffic into the native VLAN. Explicitly tagged native VLAN frames cause ingress processing anomalies unless native tagging is globally forced. Standard access ports drop incoming tagged frames.
Each match correctly reflects IEEE 802.1Q standards and switchport mechanics: untagged frames map directly to the native VLAN; VID 0 frames deliver 802.1p CoS priority to the native VLAN; explicitly tagged native frames violate default untagged expectations; and standard access ports reject tagged frames at ingress.

Step-by-Step Solution

1
Analyze untagged frame processing on trunk interfaces.
Untagged ingress traffic is assigned to the interface's configured native VLAN (VLAN 40) and transmitted untagged across native trunks.
IEEE 802.1Q backwards compatibility requires untagged traffic to be mapped to the native VLAN.
2
Analyze VLAN ID 0 (802.1p priority tagging) behavior.
VLAN ID 0 carries CoS priority bits but no VLAN ID, directing traffic to the native VLAN.
The 802.1Q header format reserves VID 0 exclusively for layer 2 QoS prioritization.
3
Analyze tagged native VLAN frame handling.
Receiving a tagged frame matching the native VLAN ID creates a mismatch condition unless tag native behavior is globally configured.
Standard 802.1Q trunking expects native VLAN traffic to be unencapsulated.
4
Evaluate ingress frame rules on access switchports.
Access ports drop 802.1Q tagged frames upon receipt.
Access interfaces operate purely in a single untagged broadcast domain.

Key Concept

802.1Q Tagging, Native VLAN Behavior, and Access vs Trunk Port Ingress Rules
Question 676Question →

A network technician is configuring an IEEE 802.1Q trunk link between two switch ports. Which of the following statements accurately describe the behavior and configuration requirements of the native VLAN on an 802.1Q trunk? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: Frames associated with the native VLAN are transmitted across the trunk link without an 802.1Q header tag.; The native VLAN ID configuration must match on both connected switch interfaces to prevent frame misdirection.

Answer

The correct statements are that frames belonging to the native VLAN are sent untagged across the 802.1Q trunk, and both switches on the trunk link must be configured with matching native VLAN IDs.
IEEE 802.1Q trunk links carry native VLAN traffic in an untagged state by default. Because receiving switch ports place untagged frames into their locally configured native VLAN, both ends of the trunk link must agree on the same native VLAN number to preserve proper network isolation.

Step-by-Step Solution

1
Analyze standard IEEE 802.1Q tagging behavior for native vs. tagged VLANs.
IEEE 802.1Q inserts a 4-byte VLAN tag into Ethernet frames for all non-native VLANs, while native VLAN frames are transmitted untagged.
This untagged capability provides backward compatibility for legacy hubs or untagged device connections.
2
Evaluate the operational requirement for matching native VLAN settings across switches.
If Switch 1 sends untagged frames from native VLAN 10, Switch 2 will place incoming untagged frames into its configured native VLAN.
A mismatch causes untagged frames sent by one VLAN to leak into a different VLAN on the opposite side of the trunk.

Key Concept

802.1Q Native VLAN Tagging and Matching Requirements
Estimated Time:1m 0s
Question 677Question →

A network administrator is configuring static IP addresses for user workstations in a newly provisioned branch office segment using the subnet 172.20.48.64/26172.20.48.64/26. Which of the following IPv4 addresses are valid usable host addresses that can be assigned to workstations on this subnet? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: 172.20.48.75172.20.48.75; 172.20.48.110172.20.48.110

Answer

The valid usable host IP addresses are 172.20.48.75172.20.48.75 and 172.20.48.110172.20.48.110.
For the subnet 172.20.48.64/26172.20.48.64/26, the block size is 6464 IPs (262^6). The network ID is 172.20.48.64172.20.48.64 and the broadcast address is 172.20.48.127172.20.48.127. The valid usable host IP range is from 172.20.48.65172.20.48.65 through 172.20.48.126172.20.48.126. The IP addresses 172.20.48.75172.20.48.75 and 172.20.48.110172.20.48.110 both fall cleanly within this usable host range.

Step-by-Step Solution

1
Calculate the block size and subnet mask from the prefix notation.
A /26/26 prefix leaves 32−26=632 - 26 = 6 host bits. The total address block size is 26=642^6 = 64 addresses, corresponding to a subnet mask of 255.255.255.192255.255.255.192.
Determining the block size establishes the boundary markers for adjacent subnets.
2
Identify the network ID and broadcast address for the subnet.
The network address is 172.20.48.64172.20.48.64. The next subnet begins at 172.20.48.64+64=172.20.48.128172.20.48.64 + 64 = 172.20.48.128. Therefore, the broadcast address for the target subnet is 172.20.48.127172.20.48.127.
The network ID and broadcast address serve specific control functions and are reserved.
3
Determine the usable host IP address range.
The usable range extends from 172.20.48.65172.20.48.65 (first usable host) to 172.20.48.126172.20.48.126 (last usable host).
Subtracting 2 from the total block size (64−2=6264 - 2 = 62) yields the set of valid host addresses.
4
Compare the provided candidate IP addresses against the usable range.
172.20.48.75172.20.48.75 and 172.20.48.110172.20.48.110 lie within the range [172.20.48.65,172.20.48.126][172.20.48.65, 172.20.48.126]. 172.20.48.64172.20.48.64 is the network ID, 172.20.48.127172.20.48.127 is the broadcast address, and 172.20.48.128172.20.48.128 is in the next subnet.
Only IP addresses strictly between the network ID and broadcast address can be statically configured on hosts.

Key Concept

IPv4 CIDR Subnetting and Usable Host Range Calculation

Alternative Method

Calculate total host space as 2^(32-26) = 64 addresses. Identify the subnet range from .64 to .127. Subtract the first (.64) and last (.127) numbers to confirm valid hosts must be between .65 and .126.
Estimated Time:1m 30s
Question 678Question →

A network administrator is configuring switch ports connected to desktop environments where IP phones and workstations share a single physical Ethernet drop. The switchports are configured to handle untagged data traffic on VLAN 10 and tagged voice traffic on VLAN 20. Which of the following statements correctly describe Layer 2 switch forwarding and VLAN broadcast domain behavior in this setup? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: Untagged Ethernet frames entering the port are processed internally within VLAN 10.; Broadcast traffic originating on VLAN 10 is constrained to ports belonging to VLAN 10 and will not cross into VLAN 20 at Layer 2.

Answer

Untagged Ethernet frames entering the port are processed internally within VLAN 10, and broadcast traffic originating on VLAN 10 is constrained to ports belonging to VLAN 10 and will not cross into VLAN 20 at Layer 2.
VLANs operate by logically partitioning a single physical switch into distinct Layer 2 broadcast domains. Untagged frames received on a port configured for a specific data VLAN are assigned to that VLAN's broadcast domain. Consequently, broadcast frames generated on VLAN 10 are forwarded only to ports assigned to VLAN 10, maintaining full isolation from VLAN 20.

Step-by-Step Solution

1
Analyze how untagged frames are handled on switch ports configured for voice and data VLANs.
Untagged frames are mapped to the access/data VLAN ID (VLAN 10) for internal Layer 2 forwarding.
Switches assign incoming untagged Ethernet frames to the port's configured access/native VLAN.
2
Evaluate the broadcast domain boundary enforcement of VLANs.
Broadcast frames (destination MAC FF:FF:FF:FF:FF:FF) on VLAN 10 are replicated only to ports assigned to VLAN 10.
VLANs segment physical switch infrastructure into independent Layer 2 broadcast domains.
3
Verify inter-VLAN routing requirements.
Traffic moving between VLAN 10 and VLAN 20 requires a Layer 3 routing mechanism.
Layer 2 switches cannot forward frames directly between different VLANs without IP routing.

Key Concept

VLAN Broadcast Domain Segmentation and Layer 2 Frame Processing
Question 679Question →

A network administrator is troubleshooting an IEEE 802.1Q trunk link configured between two enterprise switches, `SW-BUILDING-A` and `SW-BUILDING-B`, on interface `GigabitEthernet1/0/48`.

The administrator inspects the running configuration on both switch interfaces:

`SW-BUILDING-A# show running-config interface GigabitEthernet1/0/48`
text
interface GigabitEthernet1/0/48
switchport mode trunk
switchport trunk native vlan 10
switchport trunk allowed vlan 10,20,30

`SW-BUILDING-B# show running-config interface GigabitEthernet1/0/48`
text
interface GigabitEthernet1/0/48
switchport mode trunk
switchport trunk native vlan 20
switchport trunk allowed vlan 10,20,30

Based on these interface configurations, which of the following operational outcomes or symptoms will occur across this trunk link? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: Untagged traffic originating from VLAN 10 on SW-BUILDING-A will be placed into VLAN 20 upon ingress at SW-BUILDING-B.; System log messages indicating a native VLAN mismatch will be periodically generated by discovery protocols such as CDP or LLDP.

Answer

Untagged traffic originating from VLAN 10 on SW-BUILDING-A will be placed into VLAN 20 upon ingress at SW-BUILDING-B, and system log messages indicating a native VLAN mismatch will be periodically generated by protocols like CDP or LLDP.
In IEEE 802.1Q trunking, traffic belonging to the native VLAN is forwarded untagged. When SW-BUILDING-A transmits untagged frames for its native VLAN 10, SW-BUILDING-B receives them untagged and assigns them to its configured native VLAN 20. This causes cross-VLAN traffic leakage and breaks single-VLAN Layer 2 continuity. Furthermore, management protocols like CDP or LLDP periodically exchange interface parameters and issue native VLAN mismatch syslog notifications upon detecting mismatched native VLAN IDs.

Step-by-Step Solution

1
Analyze the 802.1Q native VLAN behavior on the sending switch (SW-BUILDING-A).
SW-BUILDING-A has `switchport trunk native vlan 10`. Frames originating on VLAN 10 are sent out GigabitEthernet1/0/48 untagged (without an 802.1Q header).
By standard 802.1Q specification, traffic matching the local native VLAN ID on a trunk interface is transmitted unencapsulated.
2
Analyze frame ingress processing on the receiving switch (SW-BUILDING-B).
SW-BUILDING-B has `switchport trunk native vlan 20`. Any untagged frame arriving on GigabitEthernet1/0/48 is implicitly assigned to VLAN 20.
Receiving switches associate untagged trunk ingress traffic with their own locally configured native VLAN.
3
Evaluate protocol monitoring behaviors across mismatched native VLAN trunks.
Link-layer discovery protocols (CDP/LLDP) detect the mismatch in PDU exchange fields and log persistent console/syslog errors.
Network operating systems use management protocols to validate trunk parameters and alert administrators to configuration errors.

Key Concept

IEEE 802.1Q Native VLAN Mismatch and Frame Tagging Dynamics
Question 680Question →

Match each core IP addressing service or DNS record type to its primary operational function within enterprise network infrastructure.

Click a left item, then click its matching right item

Items

MX Record
Dynamic DNS (DDNS)
IP Address Management (IPAM)
Network Time Protocol (NTP)

Matches

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Answer

MX Record maps to directing incoming domain email traffic; Dynamic DNS maps to automatically updating DNS zone records upon IP assignment changes; IPAM maps to centralized tracking and management of IP address spaces; NTP maps to synchronizing system clocks across network devices.
Each addressing service is mapped strictly according to standard protocol definitions: MX records specify mail routing destinations; Dynamic DNS (DDNS) automates DNS record modification when DHCP leases change; IPAM centrally coordinates IP address allocation and scope management; NTP maintains unified time synchronization across all networked devices.

Step-by-Step Solution

1
Identify the purpose of domain email resolution records.
MX records specify designated mail transfer hostnames for a domain.
Domain name systems rely on specialized MX records rather than standard A records to process inbound mail traffic.
2
Analyze how variable IP addressing integrates with name resolution.
Dynamic DNS (DDNS) automatically registers changes between DHCP lease grants and DNS resource records.
Hostnames remain resolvable even when workstations receive changing IPv4 or IPv6 dynamic leases.
3
Determine the administrative system used for network addressing governance.
IPAM provides centralized management of subnets, scopes, and address utilization metrics.
Enterprise networks require IPAM to prevent IP address overlapping and streamline DHCP/DNS infrastructure configuration.
4
Evaluate time distribution protocol functionality across network infrastructure.
NTP synchronizes client and infrastructure system time over UDP port 123.
Consistent system time is required for accurate log correlation, Kerberos authentication, and certificate validation.

Key Concept

Core IP Addressing Services and DNS Record Functions
Estimated Time:1m 30s
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