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Network Implementation

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Question 181Question →

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 182Question →

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 183Question →

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 184Question →

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 185Question →

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 186Question →

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 187Question →

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 188Question →

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 189Question →

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 190Question →

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 191Question →

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 192Question →

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 193Question →

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 194Question →

A network architect is implementing a Variable Length Subnet Mask (VLSM) allocation starting at the beginning of the 10.150.32.0/2010.150.32.0/20 address block for an industrial facility. The design requires three consecutive subnets allocated sequentially in descending order of size:

1. Subnet A (Industrial Controllers): Minimum of 500500 usable host IP addresses
2. Subnet B (HMI Touch Panels): Minimum of 100100 usable host IP addresses
3. Subnet C (Telemetry Sensors): Minimum of 2525 usable host IP addresses

Which of the following statements correctly describe the resulting network subnetting plan? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: Subnet A is assigned the prefix 10.150.32.0/2310.150.32.0/23 and has a broadcast address of 10.150.33.25510.150.33.255.; Subnet C is assigned the prefix 10.150.34.128/2710.150.34.128/27 with a valid usable host IP range from 10.150.34.12910.150.34.129 through 10.150.34.15810.150.34.158.

Answer

The correct statements are that Subnet A is assigned the prefix 10.150.32.0/23 with a broadcast address of 10.150.33.255, and Subnet C is assigned the prefix 10.150.34.128/27 with a valid usable host IP range from 10.150.34.129 through 10.150.34.158.
For VLSM allocations, subnets must be ordered by size and allocated sequentially. Subnet A requires 500 hosts, necessitating 9 host bits (29−2=5102^9 - 2 = 510). A /23/23 subnet beginning at 10.150.32.010.150.32.0 covers 10.150.32.010.150.32.0 to 10.150.33.25510.150.33.255, making its broadcast address 10.150.33.25510.150.33.255. Subnet B uses 10.150.34.0/2510.150.34.0/25 (10.150.34.010.150.34.0 to 10.150.34.12710.150.34.127). Subnet C begins at 10.150.34.12810.150.34.128 and requires 5 host bits (25−2=302^5 - 2 = 30), forming the prefix 10.150.34.128/2710.150.34.128/27. Its usable hosts range from 10.150.34.12910.150.34.129 to 10.150.34.15810.150.34.158.

Step-by-Step Solution

1
Calculate host requirements and prefix length for Subnet A
500 hosts requires 9 host bits (29−2=510≥5002^9 - 2 = 510 \geq 500). Prefix = 32−9=/2332 - 9 = /23. Subnet A range: 10.150.32.0/2310.150.32.0/23 (10.150.32.010.150.32.0 to 10.150.33.25510.150.33.255). Broadcast address: 10.150.33.25510.150.33.255.
Subnets must be sized to accommodate usable host IP counts plus network and broadcast overhead.
2
Determine starting address and prefix length for Subnet B
Next available address is 10.150.34.010.150.34.0. 100 hosts requires 7 host bits (27−2=126≥1002^7 - 2 = 126 \geq 100). Prefix = 32−7=/2532 - 7 = /25. Subnet B range: 10.150.34.0/2510.150.34.0/25 (10.150.34.010.150.34.0 to 10.150.34.12710.150.34.127).
Subnet B is allocated sequentially immediately following Subnet A's address block.
3
Determine starting address, prefix length, and usable range for Subnet C
Next available address is 10.150.34.12810.150.34.128. 25 hosts requires 5 host bits (25−2=30≥252^5 - 2 = 30 \geq 25). Prefix = 32−5=/2732 - 5 = /27. Subnet C range: 10.150.34.128/2710.150.34.128/27. First host: 10.150.34.12910.150.34.129, Last host: 10.150.34.15810.150.34.158, Broadcast: 10.150.34.15910.150.34.159.
Sequential VLSM allocation places Subnet C right after Subnet B, using the minimum required block size.

Key Concept

Variable Length Subnet Masking (VLSM) allocation based on usable host address formulas (2h−22^h - 2).
Question 195Question →

A network technician connects four switches into a redundant topology. All four switches retain their default Spanning Tree Protocol (STP) bridge priority of 32768. The switches have the following MAC addresses:
- Switch A: 00:11:22:AA:BB:CC
- Switch B: 00:11:22:11:22:33
- Switch C: 00:11:22:FF:EE:DD
- Switch D: 00:11:22:44:55:66

Which switch will be elected as the STP root bridge?

Show answer & explanation

Answer: Switch B

Answer

Switch B will be elected as the STP root bridge because it possesses the lowest MAC address among all switches sharing the same default priority.
In Spanning Tree Protocol (STP), the root bridge is selected based on the lowest Bridge ID (BID), which consists of the Bridge Priority and the MAC address. Because all switches share the default priority of 32768, the switch with the lowest MAC address (Switch B, with 00:11:22:11:22:33) wins the root bridge election.

Step-by-Step Solution

1
Compare the STP bridge priority values of all switches.
All switches have an identical priority of 32768.
Since bridge priorities are equal, the tiebreaker process must evaluate the MAC address portion of the Bridge ID (BID).
2
Compare the MAC addresses of each switch to find the lowest numerical value.
Switch B (00:11:22:11:22:33) is lower than Switch D (00:11:22:44:55:66), Switch A (00:11:22:AA:BB:CC), and Switch C (00:11:22:FF:EE:DD).
STP elects the switch with the lowest Bridge ID (Priority + MAC address) as the Root Bridge.

Key Concept

STP Root Bridge Election Process
Question 196Question →

A network technician is configuring an IPv4 interface on a server assigned a network prefix of /28/28. What is the corresponding subnet mask in dotted-decimal notation?

Show answer & explanation

Answer: 255.255.255.240

Answer

The subnet mask in dotted-decimal notation for a /28/28 prefix is 255.255.255.240255.255.255.240.
A /28/28 CIDR prefix specifies that 2828 bits are allocated for the network portion of the address. The first three octets account for 2424 bits (255.255.255255.255.255). The remaining 44 network bits in the fourth octet (11110000211110000_2) sum to 128+64+32+16=240128 + 64 + 32 + 16 = 240, giving the dotted-decimal subnet mask 255.255.255.240255.255.255.240.

Step-by-Step Solution

1
Determine the distribution of network bits across the four 8-bit octets.
A /28/28 prefix allocates 2828 network bits out of 3232 total bits, leaving 44 bits for host addressing (32−28=432 - 28 = 4).
An IPv4 address consists of four 8-bit octets (3232 bits total).
2
Write the subnet mask in binary form.
The binary representation is 11111111.11111111.11111111.1111000011111111.11111111.11111111.11110000.
The first three octets are completely filled (8×3=248 \times 3 = 24 bits), and the fourth octet contains 44 network bits.
3
Convert each binary octet to decimal notation.
The fourth octet binary value 11110000211110000_2 equals 128+64+32+16=240128 + 64 + 32 + 16 = 240, resulting in 255.255.255.240255.255.255.240.
Converting binary bit positions (128,64,32,16128, 64, 32, 16) to decimal yields the final subnet mask.

Key Concept

CIDR prefix notation to dotted-decimal subnet mask conversion

Alternative Method

Calculate the host block size by using 2host bits=24=162^{\text{host bits}} = 2^4 = 16. Subtract the block size from 256256 in the interesting octet: 256−16=240256 - 16 = 240. Thus, the subnet mask is 255.255.255.240255.255.255.240.
Estimated Time:45s
Question 197Question →

An enterprise router configured with OSPF receives a Link-State Update (LSU) packet containing a newer Link-State Advertisement (LSA) for a remote network segment. In what chronological order does the router process this update from initial receipt to global RIB installation?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence for processing an incoming OSPF LSU is: 1) Validate the incoming LSU packet and confirm a higher LSA sequence number; 2) Update the local LSDB and flood the LSU out active OSPF interfaces; 3) Run Dijkstra's Shortest Path First (SPF) algorithm; 4) Calculate new cumulative path costs and update the internal OSPF routing table; 5) Offer the candidate route to the global Routing Information Base (RIB) using Administrative Distance.
Upon receiving an LSU, the router verifies that the LSA sequence number is newer than its existing LSDB record. It then updates the LSDB and immediately floods the update out other interfaces to propagate the change across the OSPF area. Next, the router executes the SPF calculation to build an updated shortest-path tree, calculates cumulative metric path costs for OSPF routing, and finally submits the route to the global RIB where Administrative Distance comparison determines final forwarding entry installation.

Step-by-Step Solution

1
Inspect packet headers and sequence numbers upon LSU arrival.
The router confirms the LSA is valid and newer than its stored copy.
Prevents database corruption and processing of redundant topology updates.
2
Store the new LSA in the LSDB and flood it to neighboring routers.
Neighboring OSPF routers receive the update immediately.
Flooding precedes path computation to minimize convergence latency across the network.
3
Execute the Dijkstra SPF algorithm.
A new shortest path tree is constructed.
SPF evaluates interface costs to determine loop-free optimal paths.
4
Derive path metrics and populate internal OSPF routing entries.
OSPF identifies its best intra-protocol paths.
Protocol metrics must be finalized before submitting candidate routes to the global table.
5
Compare Administrative Distance for RIB candidate entry.
The route is written to the global RIB if OSPF has the lowest Administrative Distance for the prefix.
The router's main RIB resolves route precedence between multiple dynamic and static sources.

Key Concept

OSPF Link-State Processing and Convergence Sequence
Question 198Question →

Client workstations located on VLAN 10 are unable to automatically receive IP addressing information because the network's DHCP server resides on VLAN 20 across a Layer 3 device. Which service must be configured on the router interface servicing VLAN 10 to enable DHCP request delivery to the server?

Show answer & explanation

Answer: DHCP Relay Agent (IP helper address)

Answer

Configuring a DHCP Relay Agent (IP helper address) on the gateway router interface enables local broadcast DHCP requests to be forwarded as unicast packets to a DHCP server located on a separate subnet.
Configuring a DHCP Relay Agent (or IP helper address) on the local router interface allows the router to accept broadcast DHCP DISCOVER/REQUEST packets on the client subnet, encapsulate them into unicast packets, and forward them directly to the remote DHCP server's IP address across Layer 3 boundaries.

Step-by-Step Solution

1
Identify the cause of the DHCP request failure across subnets.
Recognize that Layer 2 DHCP broadcast packets (DISCOVER) cannot cross Layer 3 router boundaries by default.
Routers drop broadcast packets to maintain broadcast domain isolation between subnets.
2
Select the appropriate network service to bridge DHCP traffic between subnets.
Determine that a DHCP Relay Agent (configured via IP helper address) must be placed on the client's default gateway interface.
The relay agent listens for client broadcasts on UDP port 67 and encapsulates them into unicast UDP packets destined for the remote DHCP server's IP address.

Key Concept

DHCP Relay Agent and IP Helper Functionality
Question 199Question →

A network technician is troubleshooting an IEEE 802.1Q trunk link between two switches, `SW-DIST-01` and `SW-ACCESS-02`. The trunk is designed to transport traffic for VLAN 10 (Data), VLAN 20 (Voice), and VLAN 99 (Management), with VLAN 99 designated as the native VLAN.

On `SW-DIST-01`, interface GigabitEthernet1/0/10 is configured with the following commands:
text
interface GigabitEthernet1/0/10
switchport mode trunk
switchport trunk native vlan 99
switchport trunk allowed vlan 10,20,99

On `SW-ACCESS-02`, interface GigabitEthernet0/1 is configured as follows:
text
interface GigabitEthernet0/1
switchport mode trunk
switchport trunk allowed vlan 10,20

Management traffic for VLAN 99 fails to cross the trunk link, and switch log messages alert the administrator to a native VLAN mismatch. Which of the following best explains why VLAN 99 management traffic is failing across this link?

Show answer & explanation

Answer: SW-ACCESS-02 defaults to native VLAN 1 and drops egress VLAN 99 traffic because VLAN 99 is omitted from its allowed list, while incoming untagged native frames from SW-DIST-01 are incorrectly processed on VLAN 1.

Answer

The connectivity failure occurs because SW-ACCESS-02 defaults to native VLAN 1 and omits VLAN 99 from its allowed VLAN list. Consequently, untagged native traffic sent from SW-DIST-01 (VLAN 99) is ingested on SW-ACCESS-02 into VLAN 1, while return frames belonging to VLAN 99 are pruned at the interface egress filter.
On an 802.1Q trunk, frames associated with the designated native VLAN are transmitted untagged. When SW-DIST-01 (configured with native VLAN 99) sends untagged frames, SW-ACCESS-02 (configured with default native VLAN 1) accepts them into VLAN 1. Furthermore, because SW-ACCESS-02 explicitly restricts allowed trunk traffic to VLANs 10 and 20 (`switchport trunk allowed vlan 10,20`), any frame tagged for VLAN 99 or destined for VLAN 99 across the trunk is pruned and dropped.

Step-by-Step Solution

1
Analyze the default native VLAN behavior on switch interfaces.
Unless explicitly configured with `switchport trunk native vlan <id>`, Cisco/IEEE switchports default to VLAN 1 as the native VLAN.
SW-DIST-01 transmits VLAN 99 traffic across the trunk untagged because VLAN 99 is specified as its native VLAN. SW-ACCESS-02 receives these untagged frames and maps them to its default native VLAN 1.
2
Analyze the allowed VLAN list configuration on SW-ACCESS-02.
The command `switchport trunk allowed vlan 10,20` explicitly restricts allowed traffic to VLANs 10 and 20, pruning VLAN 99.
Any frame originating from or destined to VLAN 99 on SW-ACCESS-02 is dropped at the trunk interface because VLAN 99 is excluded from the trunk's active allowed list.
3
Synthesize the dual failure mechanism resulting in native VLAN mismatch log errors.
Untagged ingress frames land on VLAN 1 instead of VLAN 99, and egress VLAN 99 frames are pruned, producing native VLAN mismatch log warnings and complete management unreachability.
Resolving this issue requires adding `switchport trunk native vlan 99` and appending VLAN 99 to the allowed list on SW-ACCESS-02.

Key Concept

802.1Q Trunking, Native VLAN Ingress/Egress Processing, and Allowed VLAN List Filtering
Question 200Question →

A network administrator is provisioning IPv4 subnets from the block 172.16.96.0/21172.16.96.0/21 to support server clusters requiring at least 200200 usable host IP addresses per subnet. To maximize the number of subnets created while meeting this host requirement, which subnet mask must be configured, and what is the broadcast address of the third allocated subnet?

Show answer & explanation

Answer: Subnet mask 255.255.255.0255.255.255.0 and broadcast address 172.16.98.255172.16.98.255

Answer

Subnet mask 255.255.255.0255.255.255.0 and broadcast address 172.16.98.255172.16.98.255
To support 200200 usable hosts, 88 host bits are required because 28−2=2542^8 - 2 = 254 usable IP addresses (77 host bits only provide 126126). Subtracting 88 host bits from 3232 yields a /24/24 prefix (255.255.255.0255.255.255.0). Starting from 172.16.96.0/21172.16.96.0/21, the subnets sequence as: Subnet 1 (172.16.96.0/24172.16.96.0/24), Subnet 2 (172.16.97.0/24172.16.97.0/24), and Subnet 3 (172.16.98.0/24172.16.98.0/24). The broadcast address of the third subnet is the highest IP in its range, 172.16.98.255172.16.98.255.

Step-by-Step Solution

1
Determine the minimum host bits needed
8 host bits (28−2=254≥2002^8 - 2 = 254 \ge 200 usable hosts)
7 host bits only yield 126126 usable hosts (27−2=1262^7 - 2 = 126), which is insufficient for 200 hosts.
2
Calculate the CIDR prefix and dotted-decimal subnet mask
Prefix /24/24 (32−8=2432 - 8 = 24), subnet mask 255.255.255.0255.255.255.0
Subnetting a /21/21 block into /24/24 subnets maximizes the subnet count while guaranteeing at least 200200 usable hosts per segment.
3
Identify the network boundaries of the subnets
1st subnet: 172.16.96.0/24172.16.96.0/24, 2nd subnet: 172.16.97.0/24172.16.97.0/24, 3rd subnet: 172.16.98.0/24172.16.98.0/24
Each /24/24 subnet increments the third octet by 11.
4
Determine the broadcast address of the 3rd subnet
Broadcast address is 172.16.98.255172.16.98.255
The broadcast address is the last IP address within the 172.16.98.0/24172.16.98.0/24 network range.

Key Concept

Subnet Sizing and VLSM Host Masking
Estimated Time:2m 0s
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