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Network Implementation

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Question 61Question →

Match each VLAN configuration term on the left with its corresponding operational behavior on the right.

Click a left item, then click its matching right item

Items

Native VLAN
Access Port
802.1Q Tagging
Trunk Port

Matches

Show answer & explanation

Answer

Native VLAN matches with carrying untagged frame traffic across an 802.1Q trunk link without modifying the Ethernet frame header. Access Port matches with connecting to an end-host device and carrying traffic for only one assigned VLAN using standard untagged frames. 802.1Q Tagging matches with inserting a 4-byte header into the Ethernet frame to identify the source VLAN across switch-to-switch links. Trunk Port matches with carrying traffic for multiple VLANs simultaneously over a single physical link between network switches.
Each concept aligns directly with its fundamental operation: Native VLAN handles untagged trunk traffic, Access Ports connect end-user devices on a single VLAN, 802.1Q Tagging inserts a 4-byte VLAN identifier into frames, and Trunk Ports carry traffic for multiple VLANs across a single physical link.

Step-by-Step Solution

1
Identify the purpose of Native VLAN on 802.1Q trunks.
Determine that Native VLAN handles untagged frame traffic across trunk connections.
By default in 802.1Q trunking, frames belonging to the designated native VLAN are transmitted without an 802.1Q tag.
2
Distinguish between Access Ports and Trunk Ports.
Match Access Port to single-VLAN host connectivity and Trunk Port to multi-VLAN inter-switch links.
Access ports service host devices on a single broadcast domain, while trunk ports multiplex multiple VLANs across network switches.
3
Identify the mechanism of 802.1Q frame encapsulation.
Match 802.1Q Tagging with inserting a 4-byte header into Ethernet frames.
The 802.1Q standard identifies frame VLAN membership by inserting a 4-byte field containing the VLAN ID into frame headers.

Key Concept

VLAN Configuration and 802.1Q Trunking Fundamentals
Question 62Question →

A network architect is assigned the IPv4 address block 172.24.160.0/19172.24.160.0/19 to provision subnets for a new regional data center. To optimize routing and broadcast domain sizes, the architect must partition the entire block into equal-sized subnets such that each subnet can accommodate a minimum of 120120 usable host IP addresses. Using the longest possible CIDR prefix mask that satisfies this host requirement, what is the total maximum number of usable host IP addresses available across all subnets created?

Show answer & explanation

Answer: 8064

Answer

8064 usable host IP addresses
To support at least 120 usable host IP addresses per subnet while utilizing the longest possible CIDR prefix, a /25 subnet mask (7 host bits) is required, providing 27−2=1262^7 - 2 = 126 usable host addresses per subnet. Partitioning the parent 172.24.160.0/19172.24.160.0/19 block into /25 subnets yields 225−19=26=642^{25-19} = 2^6 = 64 subnets. Multiplying 64 subnets by 126 usable host IP addresses per subnet results in 8064 total usable host IP addresses.

Step-by-Step Solution

1
Determine the host bit requirement (hh) for at least 120 usable hosts per subnet.
7 host bits are required because 27−2=126≥1202^7 - 2 = 126 \ge 120 usable host addresses.
6 host bits (26−2=622^6 - 2 = 62) are insufficient to meet the 120 host requirement.
2
Calculate the subnet prefix length (CIDR notation).
Prefix length is /25/25 (32−7=2532 - 7 = 25).
Subtracting 7 host bits from the 32 total IPv4 bits leaves a 25-bit network prefix.
3
Calculate total /25/25 subnets created from the parent /19/19 block.
64 subnets (225−19=26=642^{25 - 19} = 2^6 = 64).
The difference between the new prefix (/25) and parent prefix (/19) is 6 subnet bits.
4
Calculate total usable host IP addresses across all 64 subnets.
8064 usable host IP addresses (64×126=806464 \times 126 = 8064).
Each of the 64 subnets reserves 2 IP addresses (network ID and broadcast address) out of 128 total addresses.

Key Concept

Partitioning IPv4 CIDR blocks using VLSM/subnetting and calculating total aggregate usable host space.
Question 63Question →

A network administrator is setting up an 802.1Q trunk link connecting two switches, SW-Dist01 and SW-Access01, on interface GigabitEthernet1/0/24.

On SW-Dist01, the interface is configured as follows:

switchport mode trunk
switchport trunk native vlan 50
switchport trunk allowed vlan 10,20,50

On SW-Access01, the interface is configured as follows:

switchport mode trunk
switchport trunk native vlan 1
switchport trunk allowed vlan 10,20,50

Based on these configurations, which of the following operational outcomes will occur on this link? (Select TWO)

Select all that apply

Show answer & explanation

Answer: Untagged frames transmitted from VLAN 50 on SW-Dist01 will be received into VLAN 1 on SW-Access01, creating cross-VLAN traffic leakage.; Cisco Discovery Protocol (CDP) or log monitoring services will generate native VLAN mismatch warning messages on both switches.

Answer

Untagged frames transmitted from VLAN 50 on SW-Dist01 will be received into VLAN 1 on SW-Access01 (causing cross-VLAN traffic leakage), and system management protocols (like CDP) will generate native VLAN mismatch warnings on both switches.
On an 802.1Q trunk, traffic matching the local native VLAN is forwarded untagged. When SW-Dist01 sends untagged VLAN 50 frames, SW-Access01 receives them without a tag and classifies them into its own native VLAN (VLAN 1), causing cross-VLAN traffic leakage. Furthermore, protocols such as CDP exchange VLAN configuration metadata and trigger explicit native VLAN mismatch logs when opposing ends do not match.

Step-by-Step Solution

1
Analyze frame egress behavior on SW-Dist01 for native VLAN 50
SW-Dist01 sends traffic originating from VLAN 50 across the 802.1Q trunk link untagged because native VLAN traffic is sent without an 802.1Q header tag by default.
802.1Q trunking rules dictate that traffic matching the port's configured native VLAN is sent unencapsulated.
2
Analyze frame ingress behavior on SW-Access01 for untagged frames
SW-Access01 receives the untagged frames and places them directly into its own native VLAN, which is VLAN 1.
An 802.1Q switch port assigns all incoming untagged frames to whichever VLAN ID is configured as its native VLAN.
3
Evaluate protocol monitoring and log detection mechanisms
Neighbor discovery protocols like CDP detect the discrepancy (VLAN 50 vs. VLAN 1) and periodically generate native VLAN mismatch alerts in the console/syslog logs.
CDP advertises local native VLAN configurations in management frames to detect misconfigurations.

Key Concept

802.1Q Native VLAN Operation and Mismatch Behavior
Question 64Question →

A network technician is configuring an IP addressing plan for an IoT security segment that requires static IP assignments for 6262 security cameras and 22 default gateway interfaces. Which CIDR prefix length represents the most space-efficient IPv4 subnet mask that satisfies this requirement?

Show answer & explanation

Answer: /25

Answer

/25
The total number of IP addresses required is 62+2=6462 + 2 = 64. Using the usable host formula 2n−22^n - 2, a subnet with n=6n = 6 host bits provides 26−2=622^6 - 2 = 62 usable addresses, which is not enough. Increasing to n=7n = 7 host bits yields 27−2=1262^7 - 2 = 126 usable host addresses. The resulting prefix length is 32−7=2532 - 7 = 25, making the /25 prefix length the most space-efficient valid choice.

Step-by-Step Solution

1
Calculate total usable host addresses required
62 cameras + 2 gateway interfaces = 64 usable host addresses
Both endpoint devices and gateway interfaces require individual usable IP addresses within the subnet.
2
Apply the usable host formula (2n−2≥642^n - 2 \ge 64)
For n=6n = 6 host bits, 26−2=622^6 - 2 = 62 usable hosts (insufficient). For n=7n = 7 host bits, 27−2=1262^7 - 2 = 126 usable hosts (sufficient).
Two IP addresses in every IPv4 subnet are reserved for the network ID and the broadcast address, so 2 must be subtracted from total addresses.
3
Determine the CIDR prefix length
32−7=2532 - 7 = 25, corresponding to /25 prefix length
Subtracting the required host bits from 32 total IPv4 bits gives the prefix length.

Key Concept

IPv4 Subnetting and Usable Host Calculation
Question 65Question →

A network administrator needs to implement an interior gateway protocol (IGP) that uses a link-state routing algorithm to maintain a complete map of the network topology. Which of the following protocols should the administrator deploy?

Show answer & explanation

Answer: Open Shortest Path First (OSPF)

Answer

Open Shortest Path First (OSPF)
Open Shortest Path First (OSPF) is an interior gateway link-state routing protocol. Each router running OSPF maintains a Link-State Database (LSDB) containing a complete topology map of the network area to compute the shortest path to every destination subnet.

Step-by-Step Solution

1
Identify the protocol class specified in the requirement
The scenario requires an interior gateway protocol operating on a link-state algorithm to maintain a complete topology map.
Link-state protocols construct a complete view of the network topology to compute shortest paths.
2
Evaluate the candidate options against link-state protocol characteristics
OSPF uses Dijkstra's Shortest Path First algorithm and Link-State Advertisements to map the entire network area.
OSPF is the standard link-state dynamic routing protocol among the choices provided.

Key Concept

Link-State Routing Protocols vs. Distance-Vector Protocols
Question 66Question →

A network engineer is configuring an 802.1Q trunk link between two switches, SW-CORE-01 and SW-ACCESS-02, to carry traffic for VLAN 10 (Data), VLAN 20 (Voice), and VLAN 99 (Native/Management). Interface GigabitEthernet0/1 on SW-CORE-01 is configured with `switchport trunk native vlan 99`. However, interface GigabitEthernet0/1 on SW-ACCESS-02 was accidentally left with default settings (`switchport trunk native vlan 1`). If a host on SW-ACCESS-02 transmits unencapsulated (untagged) Ethernet frames into GigabitEthernet0/1, how will SW-CORE-01 handle these incoming frames, and what security/operational issue will occur?

Show answer & explanation

Answer: SW-CORE-01 will associate the untagged frames with VLAN 99, causing traffic sent on VLAN 1 from SW-ACCESS-02 to leak into VLAN 99 on SW-CORE-01.

Answer

SW-CORE-01 will associate the untagged frames with VLAN 99, leading to cross-VLAN traffic leakage between VLAN 1 (on SW-ACCESS-02) and VLAN 99 (on SW-CORE-01).
IEEE 802.1Q trunking sends frames belonging to the native VLAN untagged over the link. When a receiving switch gets an untagged frame on a trunk interface, it maps that frame to its local native VLAN configuration. Because SW-CORE-01 has its native VLAN set to VLAN 99, any untagged frame sent by SW-ACCESS-02 (originating from its native VLAN 1) will be placed directly into VLAN 99 on SW-CORE-01, causing cross-VLAN traffic leakage and potential Spanning Tree Protocol (STP) inconsistencies.

Step-by-Step Solution

1
Analyze the trunk configuration and 802.1Q native VLAN operation
IEEE 802.1Q trunks transmit frames belonging to the configured native VLAN without adding a 4-byte 802.1Q tag.
By default, native VLAN traffic is sent untagged across the link.
2
Evaluate ingress frame processing on SW-CORE-01
When SW-CORE-01 receives an untagged frame on GigabitEthernet0/1, it assigns the frame to its own locally configured native VLAN, which is VLAN 99.
Trunk interfaces determine the VLAN membership of untagged incoming frames solely based on their local native VLAN assignment.
3
Identify the cross-VLAN traffic leakage anomaly
Frames originated on VLAN 1 on SW-ACCESS-02 are egressed untagged and subsequently placed into VLAN 99 upon ingress at SW-CORE-01.
A native VLAN mismatch bridges two separate broadcast domains at Layer 2 without a router, breaking VLAN segmentation.

Key Concept

802.1Q Native VLAN Ingress Processing and Mismatch Behavior
Question 67Question →

A network technician is configuring an IEEE 802.1Q trunk link between two switches. By default, how does an 802.1Q trunk port process outgoing frames belonging to the native VLAN?

Show answer & explanation

Answer: It forwards the frames across the trunk link without inserting an 802.1Q header tag.

Answer

By default, an IEEE 802.1Q trunk port forwards outgoing frames belonging to the native VLAN across the trunk link without inserting an 802.1Q header tag.
Under the IEEE 802.1Q standard, frames belonging to the native VLAN (VLAN 1 by default) are transmitted across trunk links without adding an 802.1Q tag header. When the receiving switch receives an untagged frame on a trunk interface, it automatically assigns that frame to its configured native VLAN.

Step-by-Step Solution

1
Identify the standard tagging mechanism used by IEEE 802.1Q on trunk links.
IEEE 802.1Q inserts a 4-byte tag into Ethernet frames to identify VLAN membership for tagged VLANs.
Tagging allows multiple VLANs to share a single physical trunk link while maintaining logical separation.
2
Determine the special exception for native VLAN traffic on an 802.1Q trunk.
Frames associated with the native VLAN are forwarded across the trunk without receiving an 802.1Q tag header.
Backward compatibility with legacy non-trunking devices requires one VLAN's traffic to remain untagged.

Key Concept

IEEE 802.1Q Native VLAN Tagging Behavior
Question 68Question →

An enterprise network utilizes a centralized DHCP server on a management subnet to serve clients across multiple remote VLANs. A host on VLAN 30 initiates dynamic address assignment via a Layer 3 relay switch configured with Option 82 agent information insertion. In what chronological sequence do the packet processing and forwarding steps occur during the initial address discovery phase?

Drag items to arrange them in the correct order

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Answer

The correct chronological sequence begins with the local client broadcasting a DHCPDISCOVER message on VLAN 30. Next, the Layer 3 relay agent intercepts the broadcast, appends Option 82 payload data, populates its own interface IP into the giaddr field, and forwards the packet via unicast UDP port 67 to the central server. The DHCP server inspects the giaddr value to allocate an available address from the VLAN 30 scope and sends a unicast DHCPOFFER back to the relay agent. Finally, the relay agent removes Option 82 header tags and forwards the DHCPOFFER to the client on VLAN 30.
The DHCP relay process begins with a local Layer 2 broadcast (DHCPDISCOVER). The relay agent converts this into a unicast frame augmented with Option 82 and giaddr IP information directed to the central DHCP server's UDP port 67. The central server uses the giaddr field to match the correct subnet pool and returns a unicast DHCPOFFER to the relay agent. The relay agent removes Option 82 tags and delivers the offer frame back to the local client.

Step-by-Step Solution

1
Client broadcast initiation
DHCPDISCOVER frame is transmitted onto the local Layer 2 broadcast domain.
Unconfigured hosts do not possess an IP address or knowledge of the server location, requiring a broadcast transmission.
2
Relay agent interception and encapsulation
The broadcast is converted to a unicast packet targeting the server on UDP port 67 with giaddr and Option 82 appended.
Routers drop Layer 2 broadcasts by default; the helper agent acts as a proxy to cross Layer 3 boundaries.
3
Server processing and unicast reply
DHCPOFFER packet is crafted and sent back to the giaddr IP of the relay agent.
The server uses giaddr to map the request to the matching IP pool subnet and relies on unicast routing back to the relay.
4
Relay agent decapsulation and local delivery
Option 82 data is stripped and the DHCPOFFER is delivered to the client on VLAN 30.
The client network card expects standard DHCP response formatting on its local segment.

Key Concept

DHCP Relay Agent and Option 82 Forwarding Sequence
Question 69Question →

A network administrator is reviewing the switchport configuration on switch `SW-ACCESS-01` after users in the Accounting department (VLAN 20) report a loss of connectivity to servers on the same VLAN across the trunk link to `SW-CORE-01`.

The administrator inspects the output of `show interfaces trunk` on `SW-ACCESS-01` for the inter-switch link interface `GigabitEthernet 0/1`:

text
Port Mode Encapsulation Status Native vlan
Gi0/1 on 802.1q trunking 1

Port Vlans allowed on trunk
Gi0/1 1-10, 30-100

Which command executed on interface `GigabitEthernet 0/1` of `SW-ACCESS-01` will resolve the connectivity issue for VLAN 20 traffic?

Show answer & explanation

Answer: switchport trunk allowed vlan add 20

Answer

Execute the command `switchport trunk allowed vlan add 20` on interface GigabitEthernet 0/1 to permit VLAN 20 traffic across the trunk link.
The output indicates that VLAN 20 is currently pruned (excluded) from the allowed VLAN list (`1-10, 30-100`) on interface GigabitEthernet 0/1. Issuing the command to add VLAN 20 to the allowed list permits frames tagged with VLAN 20 to be transmitted across the 802.1Q trunk link.

Step-by-Step Solution

1
Analyze the trunk configuration status output
The `Vlans allowed on trunk` list shows `1-10, 30-100`, which explicitly excludes VLAN 20.
When a VLAN is omitted from the allowed list on an 802.1Q trunk, frames tagged for that VLAN are dropped at ingress/egress.
2
Determine the corrective remediation command
Using `switchport trunk allowed vlan add 20` appends VLAN 20 to the existing allowed range without overwriting existing permitted VLANs.
This modifies the interface configuration to permit tagged frames for VLAN 20 across the trunk connection.

Key Concept

802.1Q Trunk Allowed VLAN List Configuration
Question 70Question →

A network administrator is designing an IP addressing scheme for a newly provisioned server VLAN. The VLAN must accommodate static IPv4 address assignments for 30 application servers, 1 default gateway router interface, and 1 monitoring appliance (a total of 32 usable host IP addresses). Which of the following represents the smallest CIDR prefix that satisfies this requirement?

Show answer & explanation

Answer: /26

Answer

The correct subnet mask is /26 because it provides 62 usable host IP addresses, which is the smallest CIDR prefix capable of accommodating all 32 required host devices.
To host 32 devices, a subnet must supply at least 32 usable IP addresses. Because 2 addresses are reserved for network and broadcast identification (2h−22^h - 2), 5 host bits (/27) only provide 30 usable addresses. Therefore, 6 host bits (/26) are required, offering 62 usable host IP addresses, making it the smallest prefix that meets the 32-device requirement.

Step-by-Step Solution

1
Calculate the total number of usable host IP addresses required.
30 application servers + 1 default gateway + 1 monitoring appliance = 32 usable host IP addresses.
Every network-attached interface requires a unique usable IPv4 host address.
2
Apply the usable host formula 2h−2≥required hosts2^h - 2 \ge \text{required hosts} to find the necessary host bits (hh).
Testing h=5h = 5: 25−2=302^5 - 2 = 30 usable hosts (insufficient). Testing h=6h = 6: 26−2=622^6 - 2 = 62 usable hosts (sufficient).
Two IP addresses in every subnet are reserved for the network ID (all host bits 0) and the broadcast address (all host bits 1).
3
Calculate the CIDR prefix from the required host bits.
32 total IPv4 bits - 6 host bits = /26 subnet prefix.
The CIDR prefix notation represents the number of network bits (32−h32 - h).

Key Concept

IPv4 Subnet Mask Calculation and Reserved Addresses
Question 71Question →

An enterprise network architect is assigned the IPv4 block 10.180.64.0/1910.180.64.0/19. Using Variable Length Subnet Masking (VLSM), the architect sequentially allocates subnets starting from the lowest available IP address to satisfy the following host requirements:

- Subnet A (Data Center): 1,2001,200 usable hosts
- Subnet B (Corporate HQ): 500500 usable hosts
- Subnet C (Voice Gateway): 250250 usable hosts
- Subnet D (Branch Office): 120120 usable hosts

Each subnet is provisioned using the smallest viable CIDR block. After allocating these four subnets back-to-back without leaving gaps between them, what is the maximum number of usable host IP addresses that can be supported by the single largest contiguous subnet that can be formed within the remaining unallocated space of the 10.180.64.0/1910.180.64.0/19 prefix?

Show answer & explanation

Answer: 4094

Answer

The maximum number of usable host IP addresses in the single largest contiguous subnet remaining is 4094.
The sequentially allocated subnets occupy IP addresses up to 10.180.75.127. The unallocated space spans 10.180.75.128 to 10.180.95.255. Because IPv4 subnets must align on binary power-of-two boundaries matching their size, the largest valid contiguous subnet that can be formed in this space is 10.180.80.0/20 (size 4,096 IPs). Subtracting 2 for the network and broadcast addresses yields 4,094 usable hosts.

Step-by-Step Solution

1
Determine the minimum prefix size for each required subnet
Subnet A requires /21 (2048 IPs), Subnet B requires /23 (512 IPs), Subnet C requires /24 (256 IPs), Subnet D requires /25 (128 IPs)
Host requirement formula is 2^n - 2. Subnet A: 2^11 - 2 = 2046 >= 1200; Subnet B: 2^9 - 2 = 510 >= 500; Subnet C: 2^8 - 2 = 254 >= 250; Subnet D: 2^7 - 2 = 126 >= 120.
2
Map sequential subnet allocations across address space
Allocated address range spans 10.180.64.0 through 10.180.75.127
Subnet A: 10.180.64.0/21 (10.180.64.0 - 10.180.71.255). Subnet B: 10.180.72.0/23 (10.180.72.0 - 10.180.73.255). Subnet C: 10.180.74.0/24 (10.180.74.0 - 10.180.74.255). Subnet D: 10.180.75.0/25 (10.180.75.0 - 10.180.75.127).
3
Identify remaining contiguous IP space and evaluate boundary alignments
Unallocated space consists of 10.180.75.128/25, 10.180.76.0/22, and 10.180.80.0/20
Address 10.180.80.0 is divisible by 16 in the 3rd octet, aligning perfectly on a /20 boundary (4096 IPs) covering 10.180.80.0 through 10.180.95.255.
4
Calculate usable hosts for the largest valid contiguous subnet block (/20)
4,094 usable host IP addresses
Formula: 2^(32 - 20) - 2 = 2^12 - 2 = 4,096 - 2 = 4,094.

Key Concept

Variable Length Subnet Masking (VLSM) and Bit Boundary Alignment
Estimated Time:3m 0s
Question 72Question →

A network technician is configuring a branch office router connected to headquarters via two WAN links. The primary connection uses a static route targeting the headquarters network 10.2.0.0/1610.2.0.0/16, configured with an administrative distance of 120120. A secondary backup connection runs Open Shortest Path First (OSPF) to reach the same network prefix. During testing, traffic unexpectedly routes through the secondary OSPF link even while the primary link remains operational. Which configuration change will ensure that traffic prioritizes the primary static route over the backup OSPF route?

Show answer & explanation

Answer: Reduce the administrative distance of the static route to a value lower than 110110.

Answer

Reduce the administrative distance of the static route to a value lower than 110.
When a router receives multiple routes to the exact same destination prefix from different sources, it evaluates Administrative Distance (AD). Standard OSPF has an AD of 110. Because the static route was manually set to an AD of 120, the router preferred OSPF. Lowering the static route's AD below 110 (for example, to 1) restores its priority over the OSPF route.

Step-by-Step Solution

1
Identify the routing source selection process.
The router compares Administrative Distance (AD) first when choosing between routes to the exact same prefix learned from different routing protocols or static configurations.
Lower AD values indicate higher preference and trustworthiness.
2
Compare the AD values of the two configured paths.
The static route currently has an AD of 120, while the standard OSPF route has a default AD of 110.
Because 110 is lower than 120, the router selects the OSPF path.
3
Determine the necessary configuration change.
Lowering the static route's AD below 110 (such as setting it to the default static AD of 1) forces the router to install the static route into the routing table.
This guarantees the primary static connection is preferred over the backup dynamic route.

Key Concept

Administrative Distance vs Protocol Metrics in Route Selection
Estimated Time:1m 30s
Question 73Question →

A network engineer is manually assigning static IPv4 addresses to industrial control gateways within an automated manufacturing segment using the address block 172.20.64.0/22172.20.64.0/22. Which of the following IPv4 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: 172.20.64.1172.20.64.1; 172.20.67.254172.20.67.254

Answer

The addresses 172.20.64.1172.20.64.1 and 172.20.67.254172.20.67.254 are valid, usable host IP addresses within the 172.20.64.0/22172.20.64.0/22 subnet.
The subnet 172.20.64.0/22172.20.64.0/22 covers the IP range 172.20.64.0172.20.64.0 through 172.20.67.255172.20.67.255. Subtracting the network address (172.20.64.0172.20.64.0) and the broadcast address (172.20.67.255172.20.67.255) yields a usable host IP address range of 172.20.64.1172.20.64.1 to 172.20.67.254172.20.67.254. Both 172.20.64.1172.20.64.1 and 172.20.67.254172.20.67.254 fall inside this range.

Step-by-Step Solution

1
Determine the subnet mask and block size for a /22/22 prefix length.
A /22/22 prefix corresponds to a subnet mask of 255.255.252.0255.255.252.0. The block size in the third octet is 256−252=4256 - 252 = 4.
Calculating the block size identifies the boundary increments between adjacent networks.
2
Identify the network ID and broadcast address for the 172.20.64.0/22172.20.64.0/22 network.
The network ID is 172.20.64.0172.20.64.0. The next subnet begins at 172.20.68.0172.20.68.0, making the broadcast address 172.20.67.255172.20.67.255.
Network addresses and broadcast addresses are reserved and cannot be assigned as usable host IPs.
3
Determine the valid host address range.
The usable host range spans from 172.20.64.1172.20.64.1 (first host) to 172.20.67.254172.20.67.254 (last host).
Usable host addresses consist of all IP addresses strictly between the network ID and the broadcast address.
4
Evaluate the choices against the usable host range.
172.20.64.1172.20.64.1 and 172.20.67.254172.20.67.254 lie within the usable range, whereas 172.20.64.0172.20.64.0 is the network ID, 172.20.67.255172.20.67.255 is the broadcast address, and 172.20.68.1172.20.68.1 belongs to another network.
Only IP addresses inside the calculated host range are valid host static assignments.

Key Concept

Usable IPv4 Host Range Calculation for Subnets Larger than /24/24
Question 74Question →

An enterprise campus utilizes a Layer 3 core switch handling routing for multiple virtual LANs. Host devices on VLAN 50 (10.50.0.0/2410.50.0.0/24) fail to acquire dynamic IP address assignments upon boot, resulting in self-assigned APIPA addresses (169.254.x.x169.254.x.x). Packet analysis on the core switch reveals that incoming broadcast DHCPDISCOVERDHCPDISCOVER messages arrive at the VLAN 50 interface, but the switch fails to encapsulate and forward these requests as unicast traffic to the remote DHCP server located at 172.16.100.25172.16.100.25. Which configuration issue on the core switch is preventing clients on VLAN 50 from receiving DHCP leases?

Show answer & explanation

Answer: The ip helper-address statement is omitted from the Switch Virtual Interface (SVI) for VLAN 50.

Answer

The ip helper-address statement is omitted from the Switch Virtual Interface (SVI) for VLAN 50.
For a Layer 3 switch or router to act as a DHCP relay agent, the relay directive (such as `ip helper-address`) must be applied to the specific Layer 3 ingress interface (SVI or physical interface) that receives the client's broadcast traffic. When clients on VLAN 50 broadcast a DHCPDISCOVER message, the core switch checks SVI 50 for a helper address. If missing, the broadcast is discarded at the Layer 3 boundary. When properly configured on SVI 50, the switch converts the broadcast into a unicast UDP packet (destined for port 67 on the DHCP server) and populates the `giaddr` field with SVI 50's IP address.

Step-by-Step Solution

1
Analyze client frame traffic characteristics
Clients on VLAN 50 broadcast DHCPDISCOVER frames to destination 255.255.255.255, bounded by the Layer 2 broadcast domain.
Routers and Layer 3 switches drop broadcast packets by default unless explicitly configured to intercept and relay them.
2
Identify the ingress interface for DHCP relay placement
The core switch must intercept broadcasts at the gateway interface serving the client broadcast domain (VLAN 50 SVI).
The relay agent records the interface IP address in the giaddr (gateway IP address) field of the DHCP packet header to inform the DHCP server which address scope to allocate from.
3
Verify missing interface directive
Without the `ip helper-address 172.16.100.25` directive on SVI 50, the switch silently discards client DHCP broadcasts.
The missing command directly explains why no outbound unicast packets were generated toward the DHCP server.

Key Concept

DHCP Relay Agent Placement and Gateway SVI Configuration
Question 75Question →

A network administrator is designing an IPv4 subnetting scheme for a branch office using the 192.168.10.0/24192.168.10.0/24 network address block. Each departmental VLAN requires support for at least 2828 usable host IP addresses. Which subnet mask will satisfy this host requirement while maximizing the total number of subnets that can be created?

Show answer & explanation

Answer: 255.255.255.224255.255.255.224

Answer

The subnet mask 255.255.255.224255.255.255.224 satisfies the requirement.
To support at least 2828 hosts, 55 bits are required for the host portion because 25−2=302^5 - 2 = 30 usable IP addresses. A /27/27 prefix (255.255.255.224255.255.255.224) leaves 55 host bits, yielding 88 subnets from a /24/24 block, which maximizes the subnet count while satisfying the host constraint.

Step-by-Step Solution

1
Calculate the minimum number of host bits (hh) required using the formula 2h−2≥hosts2^h - 2 \ge \text{hosts}.
25−2=30≥282^5 - 2 = 30 \ge 28, so h=5h = 5 host bits are needed.
Two IP addresses per subnet are reserved for the network ID and the broadcast address.
2
Determine the network prefix length from the total number of IPv4 bits (3232).
Prefix length =32−5=/27= 32 - 5 = /27.
Subtracting host bits from 3232 leaves the number of network bits.
3
Convert the CIDR prefix /27/27 into dotted-decimal subnet mask format.
/27=255.255.255.224/27 = 255.255.255.224.
The fourth octet contains three binary 11 s (111000002=22411100000_2 = 224).

Key Concept

Calculating required host bits and dotted-decimal subnet mask to maximize subnets while satisfying host capacity constraints
Question 76Question →

During the operation of IEEE 802.1D Spanning Tree Protocol (STP), a switch port transitions through multiple operational states. Which of the following activities occur specifically while a port is in the Learning state? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: It populates the switch MAC address table using source MAC addresses from incoming data frames.; It transmits and processes Bridge Protocol Data Units (BPDUs) received from neighboring switches.

Answer

In the IEEE 802.1D Learning state, the switch port processes BPDUs and populates its MAC address table, but it does not yet forward user data frames.
During the standard IEEE 802.1D Learning state, a switch port actively processes and sends BPDUs while learning source MAC addresses from received frames to populate the MAC address table. However, to prevent switching loops during convergence, it does not forward user payload frames.

Step-by-Step Solution

1
Identify the primary functions of the 802.1D Learning port state.
The Learning state is designed to prepare the switch for forwarding traffic by learning MAC addresses without actually forwarding frames.
Populating the MAC table prior to forwarding prevents excessive unicast flooding upon entering the Forwarding state.
2
Evaluate control plane vs. data plane behavior during the Learning state.
Control plane BPDUs are sent and received, MAC table entries are updated from incoming frames, but user data traffic is not forwarded.
Restricting data frame forwarding during convergence prevents temporary switching loops while topology details stabilize.

Key Concept

802.1D Spanning Tree Protocol Port State Operations
Question 77Question →

A network architect is subnetting the IP address block 10.112.0.0/1910.112.0.0/19 to provision dedicated VLANs for expanding branch offices. Each VLAN requires support for at least 127127 usable host IP addresses while conserving as much address space as possible. Subnets are allocated sequentially starting from the beginning of the block. What is the broadcast address of the 5th5^{\text{th}} subnet allocated?

Show answer & explanation

Answer: 10.112.4.25510.112.4.255

Answer

The broadcast address of the 5th allocated subnet is 10.112.4.25510.112.4.255.
To support 127127 usable hosts, the subnet must provide at least 127127 host IP addresses. Using the formula 2h−22^h - 2, h=7h=7 host bits yields 128−2=126128 - 2 = 126 usable hosts, which does not meet the requirement. Therefore, h=8h=8 host bits must be used (28−2=2542^8 - 2 = 254 usable hosts), corresponding to a /24/24 prefix with a block size of 256256 IP addresses per subnet. Counting sequentially from 10.112.0.010.112.0.0, the 5th subnet is 10.112.4.0/2410.112.4.0/24, which has a broadcast address of 10.112.4.25510.112.4.255.

Step-by-Step Solution

1
Determine the required number of host bits (hh).
h=8h = 8 host bits.
The usable host formula is 2h−2≥required hosts2^h - 2 \ge \text{required hosts}. For 127127 hosts, h=7h=7 gives 27−2=1262^7 - 2 = 126 usable hosts, which is insufficient. Selecting h=8h=8 gives 28−2=2542^8 - 2 = 254 usable hosts.
2
Calculate the subnet prefix and block size.
Prefix is /24/24 (32−8=2432 - 8 = 24), and block size is 28=2562^8 = 256 addresses per subnet.
Each /24/24 subnet increments the third octet by 11.
3
List the sequential /24/24 subnets starting from 10.112.0.0/1910.112.0.0/19.
Subnet 1: 10.112.0.0/2410.112.0.0/24
Subnet 2: 10.112.1.0/2410.112.1.0/24
Subnet 3: 10.112.2.0/2410.112.2.0/24
Subnet 4: 10.112.3.0/2410.112.3.0/24
Subnet 5: 10.112.4.0/2410.112.4.0/24
Subnets are assigned continuously starting from the network ID 10.112.0.010.112.0.0.
4
Identify the broadcast address for the 5th5^{\text{th}} subnet (10.112.4.0/2410.112.4.0/24).
Broadcast address is 10.112.4.25510.112.4.255.
The broadcast address is the last IP address in the subnet block immediately preceding the next subnet (10.112.5.010.112.5.0).

Key Concept

IPv4 Subnet Mask Calculation and Boundary Identification
Question 78Question →

A workstation on a remote VLAN requires an IP address assignment from a central DHCP server located on a different subnet across a Layer 3 router. What is the correct sequence of packet delivery steps during the discovery and offering stage of this relayed DHCP request?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The sequence begins when the client broadcasts a DHCPDISCOVER packet on its local subnet. The DHCP relay agent receives this broadcast, inserts its own interface address as the gateway IP (giaddr), and forwards the packet as a unicast message to the central DHCP server. The DHCP server matches the giaddr to the appropriate IP scope and returns a unicast DHCPOFFER packet to the relay agent. Finally, the relay agent forwards the DHCPOFFER packet to the local subnet for the client host.
In a cross-subnet DHCP environment, the host initiates the process by broadcasting a DHCPDISCOVER packet locally. The relay agent on the router interface receives this broadcast, populates the giaddr header field with its interface IP, and forwards the packet as a unicast message to the central DHCP server. The server reads the giaddr, allocates an IP from the corresponding scope, and returns a unicast DHCPOFFER packet to the relay agent. The relay agent then forwards the DHCPOFFER message onto the host's subnet.

Step-by-Step Solution

1
Identify the initial client request broadcast.
The unconfigured host initiates communication by broadcasting a DHCPDISCOVER message within its local broadcast domain.
Clients lacking IP configurations rely on local broadcast packets to reach DHCP listening services.
2
Determine how the relay agent processes the broadcast request.
The configured router interface intercepts the broadcast, inserts its interface IP into the giaddr field, and sends a unicast DHCPDISCOVER packet to the remote server IP.
Routers block standard broadcast packets by default; the relay agent encapsulation enables cross-subnet transmission.
3
Trace the server response mechanism back to the relay agent.
The DHCP server identifies the subnet pool via the giaddr header and responds with a unicast DHCPOFFER packet addressed to the relay agent.
The server cannot reply directly to the unconfigured client's unrouted IP, so it targets the reachable relay agent interface.
4
Trace the final delivery to the client.
The relay agent forwards the DHCPOFFER packet onto the client's local subnet segment.
This completes the delivery of offered lease details to the requesting workstation host.

Key Concept

DHCP Relay Agent Packet Processing and Relay Sequence
Question 79Question →

A network security engineer is investigating a DNS operational failure between an internal primary name server and a DMZ-hosted secondary name server following a firewall policy hardening exercise. The secondary server fails to perform full zone transfers (AXFR), and external resolvers fail to receive DNSSEC-signed responses that exceed 512 bytes. The engineer executes the following diagnostic command from the secondary server:

text
$ dig @172.16.10.10 example.com AXFR +dnssec
;; Truncation detected; requesting TCP fallback...
;; Connection failed: connection timed out
;; Communications error to 172.16.10.10#53: timed out

Which protocol behavior explains these combined failures, and what resolution must be implemented on the perimeter firewall?

Show answer & explanation

Answer: Zone transfers (AXFR) and truncated DNS responses (TC flag set) require TCP port 53 transport; the firewall must be updated to allow inbound and outbound TCP port 53 traffic between the name servers.

Answer

Zone transfers (AXFR) and truncated DNS responses (TC flag set) require TCP port 53 transport; the firewall must be updated to allow inbound and outbound TCP port 53 traffic between the name servers.
DNS relies on both UDP and TCP on port 53. While standard lookup queries utilize UDP for speed and low overhead, zone transfers (AXFR) and responses that exceed maximum UDP payload sizes (such as large DNSSEC key responses) trigger the Truncation (TC) flag in the DNS header. When the TC flag is set, the client or secondary server MUST re-query over TCP port 53. If a firewall blocks TCP port 53, both zone transfers and large DNSSEC query resolutions will fail.

Step-by-Step Solution

1
Analyze the diagnostic command output.
The `dig` output shows `Truncation detected; requesting TCP fallback...` followed by a timeout on port 53.
When a DNS response exceeds the allowable UDP buffer size or when initiating zone transfers (AXFR/IXFR), DNS transitions from UDP to TCP port 53.
2
Identify transport layer requirements for DNS Infrastructure.
Both full zone transfers (AXFR) and truncated DNSSEC queries require functional TCP port 53 communication.
UDP port 53 is used for standard small queries, but TCP port 53 is mandatory for reliable bulk data delivery (zone transfers) and fallback for oversized responses.
3
Determine the firewall remediation.
Permit TCP port 53 traffic between the primary and secondary DNS servers on the perimeter firewall.
Allowing TCP port 53 enables secondary servers to complete AXFR zone synchronization and resolves DNSSEC payload truncation fallbacks.

Key Concept

DNS Transport Layer Selection (UDP 53 vs TCP 53) and AXFR Zone Transfer Operations
Estimated Time:3m 0s
Question 80Question →

A network administrator is inspecting a router's Network Address Translation (NAT) table to verify traffic flow. Which of the following terms specifically describes the private IPv4 address assigned to a host on the internal network before any address translation takes place?

Show answer & explanation

Answer: Inside Local

Answer

Inside Local is the standard network address translation term for the un-translated private IPv4 address assigned to a device on the internal network.
Inside Local is the technical term used in networking to identify the private IP address assigned to an end device on an internal network before Network Address Translation is performed.

Step-by-Step Solution

1
Identify the location of the host and the point of observation.
The host is on the internal network ('Inside'), and the address is observed before translation occurs ('Local').
Standard NAT naming conventions combine location descriptors (Inside/Outside) with visibility scope (Local/Global).
2
Select the NAT term matching an untranslated internal host address.
Combining 'Inside' and 'Local' yields 'Inside Local'.
'Inside Local' specifically denotes the original private IP address configured on an internal endpoint.

Key Concept

NAT Terminology and Address Types
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