All practice questions

Question 681Question →

A network architect is implementing a Variable Length Subnet Mask (VLSM) allocation starting at the beginning of the $10.150.32.0/20$ 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 $500$ usable host IP addresses
2. Subnet B (HMI Touch Panels): Minimum of $100$ usable host IP addresses
3. Subnet C (Telemetry Sensors): Minimum of $25$ usable host IP addresses

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

Select all that apply

Subnet A is assigned the prefix

10.150.32.0/23

and has a broadcast address of

10.150.33.255

.

Subnet A can be accommodated using a

/24

prefix because

2^8 = 256

total addresses provides sufficient capacity when combined across adjacent IP boundaries.

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

.

Subnet B begins at

10.150.33.0/25

immediately following the first

/24

boundary of Subnet A.

Subnet C utilizes a subnet mask of

255.255.255.240

(

/28

) because 25 hosts require a 4-bit host portion.

Show answer & explanation

Answer: Subnet A is assigned the prefix $10.150.32.0/23$ and has a broadcast address of $10.150.33.255$ .; Subnet C is assigned the prefix $10.150.34.128/27$ with a valid usable host IP range from $10.150.34.129$ through $10.150.34.158$ .

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 (

2^9 - 2 = 510

). A

/23

subnet beginning at

10.150.32.0

covers

10.150.32.0

to

10.150.33.255

, making its broadcast address

10.150.33.255

. Subnet B uses

10.150.34.0/25

(

10.150.34.0

to

10.150.34.127

). Subnet C begins at

10.150.34.128

and requires 5 host bits (

2^5 - 2 = 30

), forming the prefix

10.150.34.128/27

. Its usable hosts range from

10.150.34.129

to

10.150.34.158

.

Step-by-Step Solution

1

Calculate host requirements and prefix length for Subnet A

500 hosts requires 9 host bits (

2^9 - 2 = 510 \geq 500

). Prefix =

32 - 9 = /23

. Subnet A range:

10.150.32.0/23

(

10.150.32.0

to

10.150.33.255

). Broadcast address:

10.150.33.255

.

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

10.150.34.0

. 100 hosts requires 7 host bits (

2^7 - 2 = 126 \geq 100

). Prefix =

32 - 7 = /25

. Subnet B range:

10.150.34.0/25

(

10.150.34.0

to

10.150.34.127

).

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

10.150.34.128

. 25 hosts requires 5 host bits (

2^5 - 2 = 30 \geq 25

). Prefix =

32 - 5 = /27

. Subnet C range:

10.150.34.128/27

. First host:

10.150.34.129

, Last host:

10.150.34.158

, Broadcast:

10.150.34.159

.

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 (

2^h - 2

).

Question 682Question →

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?

Switch B

Switch C

Switch A

Switch D

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 683Question →

A network technician is configuring an IPv4 interface on a server assigned a network prefix of $/28$ . 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

prefix is

255.255.255.240

.

A

/28

CIDR prefix specifies that

28

bits are allocated for the network portion of the address. The first three octets account for

24

bits (

255.255.255

). The remaining

4

network bits in the fourth octet (

11110000_2

) sum to

128 + 64 + 32 + 16 = 240

, giving the dotted-decimal subnet mask

255.255.255.240

.

Step-by-Step Solution

1

Determine the distribution of network bits across the four 8-bit octets.

A

/28

prefix allocates

28

network bits out of

32

total bits, leaving

4

bits for host addressing (

32 - 28 = 4

).

An IPv4 address consists of four 8-bit octets (

32

bits total).

2

Write the subnet mask in binary form.

The binary representation is

11111111.11111111.11111111.11110000

.

The first three octets are completely filled (

8 \times 3 = 24

bits), and the fourth octet contains

4

network bits.

3

Convert each binary octet to decimal notation.

The fourth octet binary value

11110000_2

equals

128 + 64 + 32 + 16 = 240

, resulting in

255.255.255.240

.

Converting binary bit positions (

128, 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

2^{\text{host bits}} = 2^4 = 16

. Subtract the block size from

256

in the interesting octet:

256 - 16 = 240

. Thus, the subnet mask is

255.255.255.240

.

Estimated Time:45s

Question 684Question →

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 685Question →

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?

DHCP Relay Agent (IP helper address)

TCP port 67 port-forwarding rule

Client-side configuration to send requests over UDP port 68

VLAN trunking allowing client broadcast domains to span both subnets without routing

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 686Question →

A network administrator is creating an Access Control List (ACL) rule on a core router to permit secure, encrypted email retrieval for clients connecting to an internal mail server using IMAPS. Which transport protocol and destination port pair must be explicitly permitted in the ACL configuration?

TCP port 993

UDP port 993

TCP port 143

TCP port 995

Show answer & explanation

Answer: TCP port 993

Answer

TCP port 993 must be permitted in the ACL rule.

IMAPS (Internet Message Access Protocol Secure) uses TCP port 993 by default to establish an encrypted TLS/SSL session for retrieving email messages from a server.

Step-by-Step Solution

1

Identify the requested network application protocol and requirement

The scenario specifies IMAPS (Internet Message Access Protocol Secure) for encrypted email retrieval.

Determining the exact service and security level dictates the appropriate protocol standard.

2

Determine the transport layer protocol required for email retrieval

Email retrieval requires reliable, connection-oriented delivery, which uses TCP rather than UDP.

TCP guarantees ordered delivery and packet acknowledgment necessary for message transfer.

3

Identify the default port number for IMAPS

Standard unencrypted IMAP uses TCP port 143, while IMAPS (implicit TLS/SSL) uses TCP port 993.

Port 993 is assigned by IANA specifically for IMAPS traffic.

Key Concept

Common Ports and Protocols (IMAPS over TCP 993)

Estimated Time:1m 0s

Question 687Question →

A network administrator connects two Layer 2 switches via an 802.1Q trunk port. Switch A is configured with native VLAN $10$ , while Switch B is configured with native VLAN $20$ . If a host assigned to VLAN $10$ sends an untagged frame into Switch A that is forwarded over the trunk, how will Switch B handle the frame upon arrival?

Switch B will accept the untagged frame and forward it to hosts in VLAN 20.

Switch B will automatically drop the frame because devices in different VLANs require a Layer 3 router to exchange data.

Switch B will encapsulate the Layer 2 frame into a Layer 3 IP packet to route it to VLAN 10.

Switch B will change its Spanning Tree Protocol root bridge priority to match VLAN 10.

Show answer & explanation

Answer: Switch B will accept the untagged frame and forward it to hosts in VLAN 20.

Answer

Switch B will accept the untagged frame and process it within VLAN 20, leading to a native VLAN mismatch leak.

Under 802.1Q trunking, traffic associated with the native VLAN is sent across the link untagged. When Switch A transmits a VLAN 10 frame untagged, Switch B accepts the untagged frame and assigns it to its local native VLAN (VLAN 20), resulting in unintended traffic leakage between VLAN 10 and VLAN 20.

Step-by-Step Solution

1

Analyze trunk encapsulation behavior for the native VLAN on Switch A

Switch A receives the frame on VLAN 10. Since VLAN 10 is configured as the native VLAN on Switch A's trunk port, Switch A transmits the frame across the trunk without adding an 802.1Q VLAN header tag.

By definition, IEEE 802.1Q does not tag frames associated with the native VLAN.

2

Analyze frame reception behavior on Switch B

Switch B receives an untagged frame on its trunk port. Switch B assigns all incoming untagged frames to its own configured native VLAN, which is VLAN 20.

Switches map incoming untagged trunk traffic directly to their locally configured native VLAN.

3

Determine the resulting network behavior

Traffic originating on VLAN 10 is delivered into VLAN 20, creating a security and operational issue known as native VLAN mismatch frame leakage.

Mismatched native VLAN configurations break Layer 2 broadcast domain isolation across trunk links.

Key Concept

802.1Q Native VLAN Trunking and Mismatch Behavior

Estimated Time:1m 15s

Question 688Question →

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?

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.

IEEE 802.1Q trunk interfaces automatically shut down unless all switches dynamically negotiate matching native VLAN IDs using Dynamic Trunking Protocol (DTP).

Untagged frames sent across an 802.1Q trunk are automatically dropped by the receiving switch unless 802.1Q tag header insertion is forcibly enabled for every configured VLAN.

Devices assigned to VLAN 99 on different switches cannot communicate across a trunk unless a Layer 3 router performs inter-VLAN routing between the two switch interfaces.

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 689Question →

Match each network topology classification to its corresponding operational characteristic or architectural design pattern.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Star Topology matches with central device isolation; Token Ring Topology matches with token-passing sequential transmission; Point-to-Multipoint Topology matches with central base station to subscriber stations; Partial Mesh Topology matches with redundant paths balancing fault tolerance and cost.

Star Topology connects all hosts to a central switch. Token Ring uses deterministic token-passing for access control. Point-to-Multipoint connects one central transmitter to multiple receivers. Partial Mesh provides targeted link redundancy to balance cost and fault tolerance.

Step-by-Step Solution

1

Analyze the structural configuration of Star Topology

Identify that Star Topology relies on a central switch/hub connecting all hosts independently.

Centralized star layouts isolate point-to-point physical cable segments.

2

Evaluate media access methods for Token Ring Topology

Match Token Ring to deterministic token-passing sequence around a logical loop.

Token Ring avoids packet collisions by passing an electronic token sequentially.

3

Examine Point-to-Multipoint communication patterns

Match Point-to-Multipoint to central base station wireless transmission to multiple subscribers.

P2MP designs establish one central node distributing to multiple endpoints.

4

Differentiate Partial Mesh from other redundant topologies

Match Partial Mesh to selective redundant links balancing availability and cabling budget.

Partial mesh interconnects essential routers/switches without linking every node pair.

Key Concept

Classification and operational characteristics of physical and logical network topologies.

Question 690Question →

A network technician is verifying the IP configuration of a workstation with the IPv4 address $192.168.1.68/27$ . What is the broadcast address for the subnet to which this workstation belongs?

192.168.1.95

192.168.1.94

192.168.1.63

192.168.1.127

Show answer & explanation

Answer: $192.168.1.95$

Answer

The broadcast address for the subnet is

192.168.1.95

.

A /27 CIDR prefix corresponds to a subnet mask of 255.255.255.224, creating subnets in block sizes of 32 in the fourth octet. For host address 192.168.1.68, the network ID is 192.168.1.64 and the next network ID is 192.168.1.96. The broadcast address is the final IP before the next network boundary, which is 192.168.1.95.

Step-by-Step Solution

1

Determine the block size from the CIDR prefix.

A /27 prefix corresponds to subnet mask 255.255.255.224, giving a block size of

256 - 224 = 32

addresses in the fourth octet.

Calculating the block size identifies the step increment between network boundaries.

2

Identify the network ID for the host address 192.168.1.68.

Subnet boundaries increment by 32 (0, 32, 64, 96...). The host IP 192.168.1.68 falls into the

192.168.1.64

subnet block.

Finding the network ID establishes the lower boundary of the subnet.

3

Calculate the broadcast address.

The broadcast address is one less than the next network address (

192.168.1.96 - 1 = 192.168.1.95

).

The broadcast address is always the final address in a subnet block.

Key Concept

IPv4 Subnet Boundaries and Broadcast Address Calculation

Estimated Time:1m 0s

Question 691Question →

A network administrator is provisioning IPv4 subnets from the block $172.16.96.0/21$ to support server clusters requiring at least $200$ 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?

Subnet mask

255.255.255.0

and broadcast address

172.16.98.255

Subnet mask

255.255.255.128

and broadcast address

172.16.97.127

Subnet mask

255.255.255.0

and broadcast address

172.16.98.254

Subnet mask

255.255.255.0

and broadcast address

172.16.99.255

Show answer & explanation

Answer: Subnet mask $255.255.255.0$ and broadcast address $172.16.98.255$

Answer

Subnet mask

255.255.255.0

and broadcast address

172.16.98.255

To support

200

usable hosts,

8

host bits are required because

2^8 - 2 = 254

usable IP addresses (

7

host bits only provide

126

). Subtracting

8

host bits from

32

yields a

/24

prefix (

255.255.255.0

). Starting from

172.16.96.0/21

, the subnets sequence as: Subnet 1 (

172.16.96.0/24

), Subnet 2 (

172.16.97.0/24

), and Subnet 3 (

172.16.98.0/24

). The broadcast address of the third subnet is the highest IP in its range,

172.16.98.255

.

Step-by-Step Solution

1

Determine the minimum host bits needed

8 host bits (

2^8 - 2 = 254 \ge 200

usable hosts)

7 host bits only yield

126

usable hosts (

2^7 - 2 = 126

), which is insufficient for 200 hosts.

2

Calculate the CIDR prefix and dotted-decimal subnet mask

Prefix

/24

(

32 - 8 = 24

), subnet mask

255.255.255.0

Subnetting a

/21

block into

/24

subnets maximizes the subnet count while guaranteeing at least

200

usable hosts per segment.

3

Identify the network boundaries of the subnets

1st subnet:

172.16.96.0/24

, 2nd subnet:

172.16.97.0/24

, 3rd subnet:

172.16.98.0/24

Each

/24

subnet increments the third octet by

1

.

4

Determine the broadcast address of the 3rd subnet

Broadcast address is

172.16.98.255

The broadcast address is the last IP address within the

172.16.98.0/24

network range.

Key Concept

Subnet Sizing and VLSM Host Masking

Estimated Time:2m 0s

Question 692Question →

An enterprise network topology features four switches operating standard IEEE 802.1D Spanning Tree Protocol (STP) with the following bridge configurations:
- Switch-A: Priority 32768, MAC 00:00:0C:11:11:11
- Switch-B: Priority 28672, MAC 00:00:0C:33:33:33
- Switch-C: Priority 28672, MAC 00:00:0C:22:22:22
- Switch-D: Priority 32768, MAC 00:00:0C:44:44:44

The switches are connected in a physical ring topology: Switch-A connects to Switch-B, Switch-B connects to Switch-C, Switch-C connects to Switch-D, and Switch-D connects to Switch-A. All inter-switch links are Gigabit Ethernet (STP path cost = 4), except the link between Switch-B and Switch-C, which is 10-Gigabit Ethernet (STP path cost = 2). Which switch will be elected as the Root Bridge, and what role will Switch-D assign to its port on the link connecting directly to Switch-A?

Switch-C is elected as the Root Bridge, and Switch-D assigns the Designated Port role to its interface connected to Switch-A.

Switch-B is elected as the Root Bridge, and Switch-D assigns the Designated Port role to its interface connected to Switch-A.

Switch-C is elected as the Root Bridge, and Switch-D assigns the Alternate/Blocking Port role to its interface connected to Switch-A.

Switch-B is elected as the Root Bridge, and Switch-D assigns the Root Port role to its interface connected to Switch-A.

Show answer & explanation

Answer: Switch-C is elected as the Root Bridge, and Switch-D assigns the Designated Port role to its interface connected to Switch-A.

Answer

Switch-C is elected as the Root Bridge, and Switch-D assigns the Designated Port role to its interface connected to Switch-A.

Switch-C is elected as the Root Bridge because it ties for the lowest priority (28672) with Switch-B and has a lower MAC address (00:00:0C:22:22:22). For port roles, Switch-D's direct link to Switch-C yields a Root Path Cost of 4. Switch-A reaches Switch-C via Switch-B with a Root Path Cost of 6 (4 + 2). On the segment connecting Switch-A and Switch-D, Switch-D advertises a superior path cost to the Root (4 vs. 6), causing Switch-D to assume the Designated Port role on that link.

Step-by-Step Solution

1

Determine the Root Bridge election result across all four switches.

Switch-B and Switch-C share the lowest numerical bridge priority value of 28672. Comparing their MAC addresses, Switch-C (00:00:0C:22:22:22) is lower than Switch-B (00:00:0C:33:33:33). Therefore, Switch-C is elected Root Bridge.

STP elects the Root Bridge based first on the lowest Bridge Priority, and breaks ties using the lowest MAC address.

2

Calculate the Root Path Cost for Switch-D and identify its Root Port.

Switch-D's direct link to Switch-C has a path cost of 4 (Gigabit Ethernet). Switch-D's alternate path (Switch-D -> Switch-A -> Switch-B -> Switch-C) has a cost of 4 + 4 + 2 = 10. Switch-D designates its direct port to Switch-C as its Root Port with a Root Path Cost of 4.

Each non-root bridge selects the interface offering the lowest cumulative path cost back to the Root Bridge.

3

Calculate the Root Path Cost for Switch-A and evaluate the segment between Switch-A and Switch-D.

Switch-A's best path to Root is via Switch-B (cost 4 to Switch-B + cost 2 to Switch-C = 6). On the link between Switch-A and Switch-D, Switch-D advertises a Root Path Cost of 4, while Switch-A advertises a Root Path Cost of 6. Because 4 < 6, Switch-D's interface on this link becomes the Designated Port.

On any shared segment between two switches, the switch that advertises the lower Root Path Cost wins the Designated Port role for that segment.

Key Concept

Spanning Tree Protocol Root Bridge election and port role calculation based on lowest priority, MAC tie-breaking, and path cost evaluation.

Question 693Question →

A network administrator upgrades an enterprise core network link between two OSPF-enabled routers from $1\text{ Gbps}$ to $10\text{ Gbps}$ . However, after the upgrade, the routers continue to load-balance traffic equally between the new $10\text{ Gbps}$ primary link and a parallel $1\text{ Gbps}$ backup link. Upon inspecting the OSPF database, both interfaces are assigned an OSPF cost metric of $1$ . Which of the following best explains why OSPF calculates an identical path metric for both the $1\text{ Gbps}$ and $10\text{ Gbps}$ links?

The OSPF default reference bandwidth is

100\text{ Mbps}

, which calculates a cost fractional value below

1

for both links and rounds up to a minimum interface cost of

1

.

OSPF uses Administrative Distance rather than interface bandwidth to determine path metrics, assigning a fixed metric of

1

to all internal link-state routes.

The split horizon mechanism prevents OSPF routers from advertising metric costs lower than

1

across interfaces operating within the same backbone area.

An autonegotiation failure forced a duplex mismatch on the

10\text{ Gbps}

interface, causing OSPF to fall back to a default legacy cost of

1

.

Show answer & explanation

Answer: The OSPF default reference bandwidth is $100\text{ Mbps}$ , which calculates a cost fractional value below $1$ for both links and rounds up to a minimum interface cost of $1$ .

Answer

The OSPF default reference bandwidth is 100 Mbps, which calculates a cost fractional value below 1 for both links and rounds up to a minimum interface cost of 1.

OSPF calculates path cost using the formula Cost = Reference Bandwidth / Interface Bandwidth. Because the default reference bandwidth in standard OSPF deployments is 100 Mbps, any interface speed equal to or greater than 100 Mbps (such as 1 Gbps or 10 Gbps) calculates to a value less than or equal to 1. Since OSPF costs must be positive integers, both 1 Gbps and 10 Gbps links receive a cost of 1, treating them as equal-cost paths until the auto-cost reference-bandwidth command is configured.

Step-by-Step Solution

1

Identify the OSPF metric calculation formula

\text{OSPF Cost} = \frac{\text{Reference Bandwidth}}{\text{Interface Bandwidth}}

OSPF determines interface metrics based on relative link speed.

2

Calculate the default cost for a

1\text{ Gbps}

link

\frac{100\text{ Mbps}}{1000\text{ Mbps}} = 0.1 \rightarrow \text{rounded up to } 1

The default reference bandwidth in OSPF is

100\text{ Mbps}

, and minimum OSPF metric cost is

1

.

3

Calculate the default cost for a

10\text{ Gbps}

link

\frac{100\text{ Mbps}}{10000\text{ Mbps}} = 0.01 \rightarrow \text{rounded up to } 1

Since OSPF metrics must be positive integers, any calculation yielding less than 1 defaults to 1.

4

Determine the operational outcome

Both paths evaluate to cost

1

, triggering Equal-Cost Multi-Path (ECMP) routing.

To differentiate speeds greater than

100\text{ Mbps}

, the network engineer must manually adjust the `auto-cost reference-bandwidth` setting.

Key Concept

OSPF Reference Bandwidth and Cost Calculation

Question 694Question →

A network administrator is provisioning a small subnet for point-to-multipoint router interconnects using a subnet mask of $255.255.255.248$ (/ $29$ ). What is the maximum number of usable host IP addresses that can be assigned to devices within this subnet?

6

8

14

4

Show answer & explanation

Answer: 6

Answer

The maximum number of usable host IP addresses in a /29 subnet is 6.

A /29 subnet mask leaves 3 bits for host addressing (

32 - 29 = 3

). Calculating total IP addresses gives

2^3 = 8

. Subtracting the reserved network ID and broadcast ID (

8 - 2

) results in 6 usable host addresses.

Step-by-Step Solution

1

Determine the number of host bits in the /29 subnet mask

Host bits =

32 - 29 = 3

bits

An IPv4 address consists of 32 bits. Subtracting the CIDR prefix length yields the host portion.

2

Calculate the total number of IP addresses in the subnet

Total IP addresses =

2^3 = 8

The total number of addresses supported by

n

host bits is given by

2^n

.

3

Subtract network and broadcast addresses to find usable host IPs

Usable host IPs =

8 - 2 = 6

The first IP address represents the network ID and the last IP address represents the subnet broadcast ID; neither can be assigned to a host interface.

Key Concept

Usable Host Calculation for IPv4 Subnetting

Estimated Time:45s

Question 695Question →

Place the standard 802.1D Spanning Tree Protocol (STP) port state transitions in the correct chronological order from initial switch port activation to full operational traffic handling.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of standard 802.1D STP port states from activation to active traffic forwarding is Blocking State, Listening State, Learning State, and Forwarding State.

Standard 802.1D Spanning Tree Protocol transitions switch ports through four sequential states upon activation: Blocking (prevents loops, reads BPDUs), Listening (elects root bridge and determines port roles), Learning (populates the MAC address table without forwarding data), and Forwarding (fully processes user data frames).

Step-by-Step Solution

1

Determine initial state upon switch port initialization

The port starts in the Blocking state, dropping data frames to prevent topology loops while reading BPDU control frames.

STP must enforce loop prevention immediately before allowing active participation.

2

Identify the topology election phase

The port transitions to the Listening state to process BPDUs and decide root bridge and port role assignments.

Roles must be settled before the switch modifies its MAC address tables.

3

Identify the MAC address table population phase

The port enters the Learning state, recording source MAC addresses without forwarding user data frames.

Pre-populating the MAC table prevents unnecessary frame flooding when data transmission begins.

4

Identify the final operational state

The port moves to the Forwarding state to send and receive all network traffic.

Convergence is complete and the link is verified as loop-free.

Key Concept

Spanning Tree Protocol (802.1D) Port State Transitions

Question 696Question →

Match each enterprise network administrative service or protocol on the left with its corresponding default port number and primary transport layer protocol on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches are: TACACS+ corresponds to TCP Port 49; BGP corresponds to TCP Port 179; TFTP corresponds to UDP Port 69; and SIP corresponds to UDP Port 5060.

Each service maps to its precise default port and transport protocol: TACACS+ uses TCP port 49 for administrative access control; BGP uses TCP port 179 for inter-domain routing session maintenance; TFTP uses UDP port 69 for basic configuration transfers; and SIP uses UDP port 5060 for VoIP session management.

Step-by-Step Solution

1

Identify the primary transport layer protocol and default port assigned to TACACS+.

TACACS+ requires reliable transmission over TCP port 49 for network device management.

TACACS+ handles full-payload encryption and requires connection-oriented TCP to prevent lost accounting packets.

2

Determine the transport mechanism and port used by BGP for routing update exchanges.

BGP operates over TCP port 179.

BGP relies on TCP to guarantee ordered, reliable delivery of routing updates between peering routers.

3

Identify the lightweight file transfer protocol (TFTP) specifications.

TFTP uses UDP port 69.

TFTP omits complex authentication and session state overhead, making UDP port 69 ideal for device firmware loading.

4

Select the default session initiation protocol (SIP) port for voice/video signaling.

SIP uses UDP port 5060.

SIP uses UDP port 5060 for unencrypted session setup and tear-down of VoIP sessions.

Key Concept

Standard Service Port and Transport Protocol Mappings

Question 697Question →

A layer 3 switch on a multi-site network receives a frame carrying an IP packet addressed to host $10.120.4.138$ . The switch's Routing Information Base (RIB) contains four active candidate route entries matching the destination address:

Prefix	Protocol Source	Administrative Distance	Metric	Next-Hop Interface
$10.120.4.0/23$	Static	1	0	GigabitEthernet0/1
$10.120.4.128/25$	OSPF	110	45	GigabitEthernet0/2
$10.120.4.128/26$	RIPv2	120	2	GigabitEthernet0/3
$10.120.4.136/29$	EIGRP	90	1560	GigabitEthernet0/4

Which next-hop interface will the layer 3 switch select to forward this packet?

GigabitEthernet0/4, because the EIGRP entry provides the longest prefix match for the destination address.

GigabitEthernet0/1, because the static route entry possesses the lowest administrative distance value.

GigabitEthernet0/3, because the RIPv2 path has the lowest numerical metric value.

GigabitEthernet0/2, because OSPF link-state announcements automatically supersede distance-vector protocol entries.

Show answer & explanation

Answer: GigabitEthernet0/4, because the EIGRP entry provides the longest prefix match for the destination address.

Answer

GigabitEthernet0/4 is selected because the

10.120.4.136/29

route provides the longest subnet mask match for destination address

10.120.4.138

.

When a router or layer 3 switch forwards a packet, it compares the packet's destination IP address against all entries in its routing table. The primary criteria for path selection is Longest Prefix Match (LPM)—the route with the highest number of matching network bits (most specific subnet mask). Here,

10.120.4.136/29

covers

10.120.4.136

through

10.120.4.143

, which includes

10.120.4.138

. Because

/29

is more specific than

/26

,

/25

, or

/23

, the switch forwards the packet out GigabitEthernet0/4 immediately.

Step-by-Step Solution

1

Evaluate subnet coverage for destination address

10.120.4.138

against each candidate route prefix in the RIB.

All four routes cover the host address:

10.120.4.0/23

(

10.120.4.0

–

10.120.5.255

),

10.120.4.128/25

(

10.120.4.128

–

10.120.4.255

),

10.120.4.128/26

(

10.120.4.128

–

10.120.4.191

), and

10.120.4.136/29

(

10.120.4.136

–

10.120.4.143

).

Before comparing protocols, the router must identify all valid matching prefixes.

2

Determine prefix length specificity for all matching candidate routes.

The prefix masks are

/23

,

/25

,

/26

, and

/29

. The

/29

mask has 29 network bits, making it the most specific route.

The fundamental routing logic rule dictates that the longest prefix match (LPM) always wins regardless of administrative distance or protocol metric.

3

Select the egress interface tied to the longest prefix match.

The route for

10.120.4.136/29

points to next-hop interface GigabitEthernet0/4.

Because

/29

is the longest matching prefix, the router immediately selects GigabitEthernet0/4 without evaluating administrative distance or metric values.

Key Concept

Longest Prefix Match (LPM) Path Selection Hierarchy

Estimated Time:1m 30s

Question 698Question →

A network engineer configures an IP helper address on a Layer 3 switch's VLAN interface ( $192.168.10.1/24$ ) to relay client requests to a centralized DHCP server at $10.100.1.50$ . When a workstation on VLAN 10 broadcasts a DHCPDISCOVER message, which set of actions does the DHCP relay agent perform on the packet before sending it to the server?

It populates the giaddr field with

192.168.10.1

, sets the source IP to its outgoing interface address, and unicasts the packet to

10.100.1.50

via UDP port 67.

It encapsulates the DHCP broadcast inside a TCP port 67 session established with

10.100.1.50

to ensure reliable delivery across the routed network.

It changes the destination UDP port from 67 to 68 so that the central DHCP server receives the request on its listening service port.

It floods the unmodified Layer 2 broadcast frame across all active Layer 3 routed ports until it reaches the VLAN where the DHCP server resides.

Show answer & explanation

Answer: It populates the giaddr field with $192.168.10.1$ , sets the source IP to its outgoing interface address, and unicasts the packet to $10.100.1.50$ via UDP port 67.

Answer

The DHCP relay agent populates the giaddr (gateway IP address) field with the local interface address (

192.168.10.1

), converts the client's Layer 2 broadcast into a Layer 3 unicast packet, and forwards it directly to the DHCP server (

10.100.1.50

) using UDP destination port 67.

When a Layer 3 device functions as a DHCP relay agent (via `ip helper-address`), it intercepts client DHCPDISCOVER broadcast packets on the receiving interface. It modifies the packet by inserting its own local interface IP address into the `giaddr` (Gateway IP Address) header field and replaces the broadcast destination address with the specific unicast IP address of the DHCP server. It then forwards the packet across the routed infrastructure using standard UDP port 67.

Step-by-Step Solution

1

Analyze incoming broadcast request

The client sends a local broadcast DHCPDISCOVER frame to 255.255.255.255 on UDP port 67.

Clients lacking an IP address must broadcast locally to find a DHCP server.

2

Process and modify packet header at the relay agent

The Layer 3 switch relay agent intercepts the broadcast, inserts its own interface IP (

192.168.10.1

) into the `giaddr` field, sets its egress IP as the source IP, and sets the destination IP to

10.100.1.50

.

The `giaddr` field informs the remote DHCP server which subnet scope must be used to select an available IP address.

3

Unicast packet transport across routed network

The packet is transmitted via standard Layer 3 routing to destination IP

10.100.1.50

on UDP port 67.

Unicast IP packets can be forwarded across routers, whereas Layer 2 broadcasts are blocked at subnet boundaries.

Key Concept

DHCP Relay Agent and GIADDR Operation

Question 699Question →

A network architect is partitioning the IP address space within the block $10.88.24.0/22$ to provision a dedicated subnet for a server farm that requires a minimum of $110$ usable host IP addresses. To maximize IP space efficiency, the architect assigns the smallest standard CIDR prefix that satisfies this host requirement. Assuming the server farm is allocated the first available subnet block starting at $10.88.24.0$ , what is the broadcast address for this specific server subnet?

10.88.24.127

10.88.24.126

10.88.24.128

10.88.24.255

Show answer & explanation

Answer: $10.88.24.127$

Answer

The broadcast address for the allocated server subnet is

10.88.24.127

.

To accommodate 110 usable hosts, 7 host bits are required (

2^7 - 2 = 126

usable hosts). Subtracting 7 from 32 gives a

/25

prefix (block size of 128 addresses). The first

/25

subnet starting at

10.88.24.0

spans from

10.88.24.0

to

10.88.24.127

. Thus, the broadcast address is

10.88.24.127

.

Step-by-Step Solution

1

Determine the number of host bits required for 110 usable hosts

7 host bits are needed because

2^7 - 2 = 126 \ge 110

usable addresses (

2^6 - 2 = 62

is insufficient).

Calculating

2^h - 2

accounts for reserving the network ID and broadcast address.

2

Calculate the CIDR prefix length and block size

The prefix length is

32 - 7 = /25

(

255.255.255.128

). The total block size is

2^7 = 128

IP addresses.

Subtracting host bits from 32 total IPv4 address bits yields the subnet prefix length.

3

Determine the network range and broadcast address for the first subnet

The subnet range starts at network ID

10.88.24.0

and spans 128 addresses up to

10.88.24.127

. The last IP address in the block (

10.88.24.127

) is the broadcast address.

The broadcast address is always the final address in a subnet block before the next network segment begins.

Key Concept

Calculating custom CIDR prefixes and broadcast boundaries based on host count requirements

Estimated Time:2m 0s

Question 700Question →

A network engineer is configuring an IEEE 802.1Q trunk link between two switches, SW-DIST-01 and SW-ACCESS-02. Switch SW-DIST-01 has port GigabitEthernet1/0/1 configured with 'switchport trunk native vlan 99'. Switch SW-ACCESS-02 has port GigabitEthernet0/1 configured as a trunk using the default native VLAN (VLAN 1). Which of the following statements accurately describe the operational impact of this configuration or the steps required to resolve it? (Select TWO.)

Select all that apply

Untagged frames transmitted from VLAN 99 on SW-DIST-01 will be improperly received into VLAN 1 on SW-ACCESS-02.

Tagged VLAN traffic, such as VLAN 20 and VLAN 30, will be automatically blocked across the trunk until the native VLAN mismatch is resolved.

Executing 'switchport trunk native vlan 99' on interface GigabitEthernet0/1 of SW-ACCESS-02 will align both ends and eliminate cross-VLAN traffic leakage.

Changing the port mode on SW-DIST-01 to switchport mode access will maintain trunking capabilities while resolving the tag mismatch.

Show answer & explanation

Answer: Untagged frames transmitted from VLAN 99 on SW-DIST-01 will be improperly received into VLAN 1 on SW-ACCESS-02.; Executing 'switchport trunk native vlan 99' on interface GigabitEthernet0/1 of SW-ACCESS-02 will align both ends and eliminate cross-VLAN traffic leakage.

Answer

Untagged frames transmitted from VLAN 99 on SW-DIST-01 will be improperly received into VLAN 1 on SW-ACCESS-02, and executing 'switchport trunk native vlan 99' on interface GigabitEthernet0/1 of SW-ACCESS-02 will align both ends and eliminate cross-VLAN traffic leakage.

On an IEEE 802.1Q trunk, frames belonging to the native VLAN are transmitted without an 802.1Q tag header. When a native VLAN mismatch exists between two switches, untagged traffic sent by one switch is placed into whichever native VLAN is configured on the receiving switch port, causing cross-VLAN frame leakage. Aligning the native VLAN configurations on both trunk ports ensures consistent frame interpretation and resolves the mismatch.

Step-by-Step Solution

1

Analyze 802.1Q native VLAN frame encapsulation behavior across trunk endpoints.

802.1Q trunks leave native VLAN frames untagged. When SW-DIST-01 sends untagged VLAN 99 frames, SW-ACCESS-02 receives them on a port expecting VLAN 1 to be untagged, misplacing the frames into VLAN 1.

Trunk links do not insert an 802.1Q tag header for frames belonging to the configured native VLAN.

2

Determine the effect of a native VLAN mismatch on tagged VLANs traversing the trunk.

Frames with explicit 802.1Q tags (VLAN 20, VLAN 30) are correctly identified and processed by both switches based on their tag headers.

Native VLAN mismatch affects only untagged traffic and control plane protocols; tagged traffic is processed independently based on the 4-byte 802.1Q header.

3

Identify the proper remediation command.

Configuring matching native VLAN IDs (VLAN 99) on both trunk interfaces aligns ingress and egress frame processing.

Matching native VLAN configurations prevent VLAN leaking and clear STP/CDP native VLAN mismatch warnings.

Key Concept

802.1Q Trunking Native VLAN Mismatch Consequences and Configuration Alignment

Estimated Time:1m 30s

Answer

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