Network Fundamentals

398 questions

All types398 Multiple choice205 Multiple select92 Numeric18 Matching68 Ordering7 Short answer8

An enterprise network administrator is implementing a Variable Length Subnet Masking (VLSM) scheme using the assigned summary prefix $172.28.128.0/18$ . The design requires allocating three subnets sequentially from the beginning of this address space, prioritizing the largest requirements first:

- Subnet A (Data Center): requires support for at least $1,000$ usable host IP addresses.
- Subnet B (App Servers): requires support for at least $500$ usable host IP addresses.
- Subnet C (Management): requires support for at least $250$ usable host IP addresses.

After allocating these three subnets efficiently without leaving unallocated gaps between them, what is the valid usable host IPv4 address range for the next available subnet that can accommodate a dev/test segment requiring at least $120$ usable host IP addresses?

172.28.135.1

172.28.135.126

172.28.135.0

172.28.135.127

172.28.136.1

172.28.136.126

172.28.135.1

172.28.135.254

Show answer & explanation

Answer: $172.28.135.1$ to $172.28.135.126$

Answer

The valid usable host address range for the next available subnet is

172.28.135.1

172.28.135.126

Allocating subnets sequentially based on host requirements yields: Subnet A (

172.28.128.0/22

, ending at

172.28.131.255

), Subnet B (

172.28.132.0/23

, ending at

172.28.133.255

), and Subnet C (

172.28.134.0/24

, ending at

172.28.134.255

). The next block starts at

172.28.135.0

. For

120

usable hosts, a

/25

subnet (

126

usable hosts) is required, giving a usable IP host range of

172.28.135.1

172.28.135.126

Step-by-Step Solution

Calculate prefix length and boundary for Subnet A (

1,000

usable hosts)

2^{10} - 2 = 1,022 \ge 1,000 \implies /22

prefix (

1,024

total addresses). Allocated range:

172.28.128.0

172.28.131.255

A /22 subnet provides 1,022 usable hosts across 4 third-octet block values.

Calculate prefix length and boundary for Subnet B (

500

usable hosts)

2^9 - 2 = 510 \ge 500 \implies /23

prefix (

512

total addresses). Starting IP:

172.28.132.0

. Allocated range:

172.28.132.0

172.28.133.255

A /23 subnet provides 510 usable host addresses.

Calculate prefix length and boundary for Subnet C (

250

usable hosts)

2^8 - 2 = 254 \ge 250 \implies /24

prefix (

256

total addresses). Starting IP:

172.28.134.0

. Allocated range:

172.28.134.0

172.28.134.255

A /24 subnet provides 254 usable host addresses.

Determine the network address and usable host range for the dev/test subnet (

120

usable hosts)

Next available start address is

172.28.135.0

2^7 - 2 = 126 \ge 120 \implies /25

prefix. Subnet range:

172.28.135.0/25

. Usable host range:

172.28.135.1

172.28.135.126

A /25 prefix accommodates up to 126 usable hosts, excluding the network ID (.0) and broadcast address (.127).

Key Concept

Variable Length Subnet Masking (VLSM) host requirement sizing and sequential block allocation

Estimated Time:3m 0s

Question 242Question →

An enterprise network architect is migrating a primary data center from a traditional 3-Tier (Access-Aggregation-Core) hierarchy to a 2-Tier Spine-Leaf (Clos) architecture to better handle high-volume East-West microservices traffic. Which TWO statements correctly describe the design rules and operational characteristics of this Spine-Leaf topology? (Choose two.)

Select all that apply

Every leaf switch connects to every spine switch, but leaf switches do not connect directly to other leaf switches.

Equal-Cost Multi-Path (ECMP) routing is leveraged across Layer 3 links to utilize all available bandwidth concurrently between leaf and spine switches.

Spine switches operate as Layer 2 switches that rely on Spanning Tree Protocol (STP) to block redundant uplinks and isolate broadcast domains.

Enabling STP PortFast on interswitch trunk links between leaf and spine switches is recommended to eliminate convergence delays when adding new switches.

Show answer & explanation

Answer: Every leaf switch connects to every spine switch, but leaf switches do not connect directly to other leaf switches.; Equal-Cost Multi-Path (ECMP) routing is leveraged across Layer 3 links to utilize all available bandwidth concurrently between leaf and spine switches.

Answer

The correct statements are that every leaf switch connects to every spine switch without direct leaf-to-leaf connections, and that Equal-Cost Multi-Path (ECMP) routing is utilized to load-balance traffic across all available links.

In a 2-Tier Spine-Leaf design, the fabric forms a bipartite graph where every leaf switch connects to every spine switch, with no direct connections between switches of the same tier. Traffic between leaves passes through spine switches using Layer 3 routing and Equal-Cost Multi-Path (ECMP), enabling deterministic low-latency, active-active forwarding across all available links.

Step-by-Step Solution

Analyze physical connectivity rules in a 2-Tier Spine-Leaf (Clos) architecture.

Leaf switches connect to all spine switches in a full mesh (bipartite graph). Leaves do not connect to leaves, and spines do not connect to spines.

Ensures predictable latency where every access point is exactly two hops away from any other access point.

Evaluate control plane forwarding mechanisms between leaf and spine nodes.

Layer 3 routing protocols are implemented across interswitch links to enable Equal-Cost Multi-Path (ECMP).

ECMP enables active-active packet forwarding over all physical links, avoiding the link blocking penalties of traditional Layer 2 STP topologies.

Identify misconceptions regarding Layer 2 switching and PortFast configuration on interswitch links.

STP blocking and PortFast on trunk links are inappropriate for leaf-spine interconnects.

Layer 3 forwarding prevents STP loops organically, and applying PortFast on switch trunks risks bridging loops.

Key Concept

Spine-Leaf (Clos) Topology Connectivity Rules and ECMP Forwarding

Estimated Time:2m 0s

Question 243Question →

A network interface with a MAC address of $7069.5\text{A}11.2233$ is configured on a subnet with the assigned IPv6 prefix $2001:\text{db8}:1000:\text{abc}::/64$ . The interface automatically generates its IPv6 addresses using IEEE EUI-64 format. Which two statements regarding the resulting IPv6 configuration on this interface are correct?

Select all that apply

The calculated EUI-64 interface identifier for the host is 7269:5aff:fe11:2233.

The complete global unicast address formed by the interface is 2001:db8:1000:abc:7269:5aff:fe11:2233.

The interface identifier is constructed as 7069:5aff:fe11:2233 because the first byte of the MAC address remains unmodified during EUI-64 generation.

The /64 prefix allows

2^{64} - 2

usable host addresses because IPv6 reserves the first address for network identification and the last address for broadcast.

Show answer & explanation

Answer: The calculated EUI-64 interface identifier for the host is 7269:5aff:fe11:2233.; The complete global unicast address formed by the interface is 2001:db8:1000:abc:7269:5aff:fe11:2233.

Answer

The two correct statements are that the calculated EUI-64 interface identifier is 7269:5aff:fe11:2233 and the complete global unicast address formed by the interface is 2001:db8:1000:abc:7269:5aff:fe11:2233.

The EUI-64 process inserts 0xFFFE between the OUI and vendor portion of the MAC address and flips the 7th bit of the first byte (from 0x70 to 0x72), resulting in the interface ID 7269:5aff:fe11:2233. Appending this to prefix 2001:db8:1000:abc::/64 generates the global unicast address 2001:db8:1000:abc:7269:5aff:fe11:2233.

Step-by-Step Solution

Insert FFFE into the 48-bit MAC address

Splitting MAC 70:69:5A:11:22:33 into 70:69:5A and 11:22:33 and inserting FFFE produces 70:69:5A:FF:FE:11:22:33.

EUI-64 formatting expands 48-bit MAC addresses to 64 bits by inserting 16 bits (0xFFFE) in the middle.

Invert the Universal/Local (7th) bit of the first octet

First octet 0x70 in binary is 0111 0000. Inverting the 7th bit yields 0111 0010, which is 0x72 in hex. The interface ID becomes 7269:5aff:fe11:2233.

IEEE EUI-64 standards invert the 7th bit to reflect local scope management in IPv6 host addresses.

Combine prefix and EUI-64 interface ID

2001:db8:1000:abc::/64 + 7269:5aff:fe11:2233 = 2001:db8:1000:abc:7269:5aff:fe11:2233.

The subnet prefix forms the upper 64 bits and the EUI-64 ID forms the lower 64 bits of the global unicast address.

Key Concept

EUI-64 Interface Identifier Generation and IPv6 Addressing

Question 244Question →

A Cisco Catalyst switch receives an Ethernet frame on ingress port GigabitEthernet0/1 with Source MAC address 0010.ab12.3456 and Destination MAC address 0020.cd78.90ab. The switch's current MAC address table records 0010.ab12.3456 as learned on GigabitEthernet0/3, but has no entry for 0020.cd78.90ab. Which TWO actions will the switch perform upon processing this ingress frame? (Select two.)

Select all that apply

Update the MAC address table entry for 0010.ab12.3456 to associate it with port GigabitEthernet0/1.

Flood the frame out all operational ports within the same VLAN, excluding ingress port GigabitEthernet0/1.

Forward the frame directly out GigabitEthernet0/3 based on the previous MAC table entry for the host.

Drop the frame immediately because a MAC address security violation has occurred due to a station move.

Send an ICMP Redirect message back out GigabitEthernet0/1 to notify the sender of the topology update.

Show answer & explanation

Answer: Update the MAC address table entry for 0010.ab12.3456 to associate it with port GigabitEthernet0/1.; Flood the frame out all operational ports within the same VLAN, excluding ingress port GigabitEthernet0/1.

Answer

The switch updates the MAC address table entry for 0010.ab12.3456 to GigabitEthernet0/1 and floods the unknown unicast frame out all operational ports on the VLAN except GigabitEthernet0/1.

When an Ethernet frame arrives at a switch port, the switch first performs MAC address learning by checking the frame's source MAC address. If the source MAC address exists in the MAC table but is associated with a different interface, the switch updates the table entry with the new ingress port (station move update). Next, the switch looks up the frame's destination MAC address. Because the destination MAC address is not present in the MAC table, the switch treats the frame as unknown unicast and floods it out all active ports in the same VLAN except the receiving port.

Step-by-Step Solution

Examine the source MAC address handling (MAC Learning & Station Move).

The source MAC address 0010.ab12.3456 was previously listed on GigabitEthernet0/3. Since it arrived on GigabitEthernet0/1, the switch rewrites the table entry to associate 0010.ab12.3456 with GigabitEthernet0/1.

Layer 2 switches dynamically update dynamic MAC table entries whenever a known MAC address arrives on a different port.

Examine the destination MAC address handling (Forwarding Decision).

The destination MAC address 0020.cd78.90ab is missing from the MAC address table.

When a destination MAC address is not found in the Layer 2 forwarding table, the frame is classified as an unknown unicast frame.

Execute unknown unicast forwarding logic.

The frame is flooded to all active ports assigned to the same VLAN, except the receiving interface (GigabitEthernet0/1).

Unknown unicast traffic must be flooded across the broadcast domain so the intended destination host can receive it and respond.

Key Concept

Layer 2 MAC Table Station Learning and Unknown Unicast Flooding Mechanics

Question 245Question →

A network administrator assigns the IPv4 subnet mask $255.255.255.224$ to a VLAN interface on a Cisco switch. What is the maximum number of usable host IPv4 addresses that can be assigned to end devices in this subnet?

Show answer & explanation

Answer: 30

Answer

The maximum number of usable host IPv4 addresses is 30.

A subnet mask of

255.255.255.224

corresponds to a CIDR prefix of /27, leaving 5 bits for host addressing. Using the formula

2^n - 2

(where

n = 5

), the calculation yields

2^5 - 2 = 32 - 2 = 30

usable host IP addresses.

Step-by-Step Solution

Determine the number of host bits (

n

) available in the

255.255.255.224

subnet mask.

The mask has 27 network bits and 5 host bits (

32 - 27 = 5

The last octet value

224

in binary is

11100000_2

, meaning 3 bits belong to the network/subnet portion and 5 bits belong to the host portion.

Calculate the number of usable host IP addresses using the formula

2^n - 2

2^5 - 2 = 32 - 2 = 30

usable host addresses.

The formula

2^n - 2

accounts for the total IP addresses (

2^n

) minus the reserved network identifier (all host bits set to 0) and the broadcast address (all host bits set to 1).

Key Concept

Usable Host Address Calculation

Question 246Question →

A network engineer is designing a wireless LAN for a multi-story corporate building and must account for radio frequency (RF) propagation differences between the $2.4\text{ GHz}$ and $5\text{ GHz}$ bands. Which two statements accurately describe the physical characteristics and channel allocations of these frequency bands? (Select two.)

Select all that apply

The

2.4\text{ GHz}

frequency band experiences lower free-space path loss and attenuates less through physical walls, resulting in a larger coverage area per access point.

The

5\text{ GHz}

frequency band provides a significantly greater number of non-overlapping

20\text{ MHz}

channels than the

2.4\text{ GHz}

band, which helps minimize co-channel interference in high-density enterprise environments.

The

2.4\text{ GHz}

frequency band supports up to 24 non-overlapping

20\text{ MHz}

channels, making it preferable for high-density deployment of real-time voice applications.

Higher frequency signals in the

5\text{ GHz}

band penetrate dense building materials such as concrete and brick more effectively than

2.4\text{ GHz}

signals.

The

2.4\text{ GHz}

spectrum is immune to non-802.11 interference because it operates exclusively within licensed frequency ranges.

Show answer & explanation

Answer: The $2.4\text{ GHz}$ frequency band experiences lower free-space path loss and attenuates less through physical walls, resulting in a larger coverage area per access point.; The $5\text{ GHz}$ frequency band provides a significantly greater number of non-overlapping $20\text{ MHz}$ channels than the $2.4\text{ GHz}$ band, which helps minimize co-channel interference in high-density enterprise environments.

Answer

The correct statements are that the 2.4 GHz band experiences lower free-space path loss and penetrates physical walls better (providing a larger coverage area per AP), and the 5 GHz band provides significantly more non-overlapping 20 MHz channels to mitigate co-channel interference.

The 2.4 GHz band uses longer wavelengths, giving it superior range and obstacle penetration through walls compared to 5 GHz. Additionally, the 5 GHz band provides up to 24 non-overlapping 20 MHz channels in North America (UNII bands), which provides significantly higher capacity and reduces co-channel interference (CCI) compared to the limited 3 non-overlapping channels (1, 6, 11) available in the 2.4 GHz band.

Step-by-Step Solution

Analyze the RF propagation characteristics of 2.4 GHz versus 5 GHz frequency bands.

Lower frequencies (2.4 GHz) have longer wavelengths, yielding lower free-space loss and superior penetration through physical structures relative to higher frequencies (5 GHz).

RF signal attenuation increases as frequency increases.

Evaluate the channel availability for both frequency bands in enterprise WLAN design.

The 2.4 GHz spectrum has only 3 non-overlapping 20 MHz channels (1, 6, 11), whereas 5 GHz provides up to 24 non-overlapping channels across UNII-1, UNII-2 (DFS), and UNII-3 bands.

Greater channel availability allows high-density access point deployments without co-channel interference (CCI).

Key Concept

RF Propagation and Channel Allocation Differences (2.4 GHz vs 5 GHz)

Question 247Question →

A network engineer analyzes a packet capture of an active application stream during heavy network load. The capture reveals that the receiving host continuously updates the 16-bit Window field in its acknowledgment segments to smaller values, causing the sending host to reduce the amount of unacknowledged data sent before waiting for an ACK. Which statement correctly identifies this transport layer mechanism and its operational contrast with UDP?

TCP employs dynamic windowing for receiver-driven flow control to prevent buffer overflow, whereas UDP lacks flow control mechanisms and a window size header field entirely.

TCP uses the window field to perform intermediate router rate limiting, whereas UDP relies on its 8-byte header to sequence packets and request retransmissions.

TCP windowing triggers automatic segment retransmission upon packet corruption, whereas UDP uses a variable 20-byte header to negotiate transmission window sizes.

TCP reduces the window size to signal Path MTU Discovery (PMTUD) updates, whereas UDP provides sequence tracking to manage buffer exhaustion without acknowledgments.

Show answer & explanation

Answer: TCP employs dynamic windowing for receiver-driven flow control to prevent buffer overflow, whereas UDP lacks flow control mechanisms and a window size header field entirely.

Answer

TCP employs dynamic windowing for receiver-driven flow control to prevent buffer overflow, whereas UDP lacks flow control mechanisms and a window size header field entirely.

The correct answer accurately describes TCP flow control. The TCP header includes a 16-bit Window field that allows the receiving host to advertise its available buffer space. As the buffer fills, the receiver reduces the window size, instructing the sender to transmit less data before awaiting an ACK. In contrast, UDP has a simple, fixed 8-byte header (Source Port, Destination Port, Length, Checksum) and does not provide flow control, windowing, or rate management.

Step-by-Step Solution

Analyze the observed protocol behavior in the packet capture.

The receiver is modifying the 16-bit Window field in acknowledgment segments to restrict the sender's unacknowledged data output.

This behavior is known as dynamic windowing (sliding window flow control), which protects receiver buffers from being overwhelmed.

Compare TCP header capabilities with UDP header structure.

TCP headers (minimum 20 bytes) contain fields for Sequence Number, Acknowledgment Number, and Window Size. UDP headers (fixed 8 bytes) contain only Source Port, Destination Port, Length, and Checksum.

UDP is connectionless and stateless, prioritizing low overhead and speed over flow control, sequencing, or error recovery.

Evaluate the option choices based on transport layer mechanics.

The option identifying TCP receiver-driven flow control and UDP's complete lack of windowing and flow control is the only technically accurate comparison.

UDP leaves application-level buffer management entirely to upper-layer protocols if needed.

Key Concept

TCP Sliding Window Flow Control vs. UDP Stateless Header Overhead

Estimated Time:2m 0s

Question 248Question →

A network administrator is deploying a virtualized server environment using bare-metal Type 1 hypervisors connected to an enterprise campus network. Which two operational roles and functions are performed directly by the Type 1 hypervisors in this infrastructure? (Select two.)

Select all that apply

Abstracting underlying physical hardware resources to allocate vCPUs, system memory, and virtual network interfaces directly to hosted virtual machines

Facilitating Layer 2 frame forwarding between virtual machines residing on the same physical host via an embedded virtual switch

Operating as an application software layer on top of an existing general-purpose host operating system such as Windows Server or Linux

Terminating Layer 3 broadcast domains and making IP packet forwarding decisions between subnets across physical servers

Show answer & explanation

Answer: Abstracting underlying physical hardware resources to allocate vCPUs, system memory, and virtual network interfaces directly to hosted virtual machines; Facilitating Layer 2 frame forwarding between virtual machines residing on the same physical host via an embedded virtual switch

Answer

The two correct functions performed directly by a Type 1 hypervisor are abstracting physical hardware resources to allocate vCPUs, memory, and virtual network interfaces to guest virtual machines, and facilitating Layer 2 frame forwarding between virtual machines on the same physical host through an integrated virtual switch.

Type 1 bare-metal hypervisors run directly on server hardware, allowing efficient hardware abstraction and direct distribution of vCPU, memory, and virtual interfaces to guest VMs. In addition, hypervisors feature built-in Layer 2 virtual switches (vSwitches) to forward frames locally between virtual machines sharing the same physical host.

Step-by-Step Solution

Identify the architectural deployment model of a Type 1 hypervisor.

A Type 1 hypervisor runs directly on the bare-metal physical hardware without relying on a underlying general-purpose host operating system.

This bare-metal positioning gives the hypervisor direct control over hardware virtualization.

Analyze the resource management and network forwarding capabilities built into the hypervisor.

The hypervisor partitions physical CPU, RAM, and NICs into virtual components for VMs and includes a virtual switch (vSwitch) to forward Layer 2 frames locally.

Local VM-to-VM traffic within the same host is bridged at Layer 2 by the internal vSwitch, while inter-subnet Layer 3 routing is handed off to network infrastructure devices.

Key Concept

Type 1 hypervisor bare-metal architecture and virtual switching roles

Question 249Question →

During the standard EUI-64 process used to derive a 64-bit IPv6 interface identifier from a 48-bit MAC address, what 16-bit hexadecimal value is inserted into the middle of the MAC address?

Show answer & explanation

Answer: FFFE; fffe; 0xFFFE; 0xfffe

Answer

FFFE

To create a 64-bit interface identifier using the IEEE EUI-64 format, a host takes its 48-bit MAC address, splits it into two 24-bit halves (OUI and NIC identifier), and inserts the 16-bit hexadecimal value FFFE (0xFFFE) in the middle. Additionally, the 7th bit (Universal/Local bit) of the first byte is inverted.

Step-by-Step Solution

Split the 48-bit MAC address into two equal 24-bit halves

The first 24 bits represent the Organizationally Unique Identifier (OUI) and the last 24 bits represent the Extension Identifier.

EUI-64 modification requires inserting a specific 16-bit sequence between these two halves.

Insert the 16-bit value FFFE into the midpoint between the two 24-bit halves

The 48-bit address expands to a 64-bit value containing FFFE in the center.

Inserting 0xFFFE creates the mandatory 64-bit structure required for an EUI-64 interface identifier.

Key Concept

EUI-64 Interface Identifier Generation

Question 250Question →

A switch receives an Ethernet frame on port FastEthernet0/1. The source MAC address is 0000.1111.1111, and the unicast destination MAC address is 0000.2222.2222. The switch checks its MAC address table and finds no entry for 0000.2222.2222. What action does the switch take to handle this frame?

Floods the frame out all operational interfaces except the ingress port FastEthernet0/1.

Drops the frame immediately because the destination address is missing from the MAC address table.

Floods the frame out all active interfaces, including the ingress port FastEthernet0/1.

Forwards the frame directly to the default gateway interface for Layer 3 resolution.

Show answer & explanation

Answer: Floods the frame out all operational interfaces except the ingress port FastEthernet0/1.

Answer

The switch floods the frame out all operational interfaces within the VLAN except the ingress port FastEthernet0/1.

When a Layer 2 switch receives a unicast frame whose destination MAC address is not currently in its MAC address table, it performs unknown unicast flooding. This forwards the frame out every active port belonging to the same VLAN, excluding the port on which the frame was received.

Step-by-Step Solution

Examine the source MAC address processing.

The switch records MAC address 0000.1111.1111 in its MAC address table mapped to port FastEthernet0/1 if it is not already present.

Switches learn MAC addresses dynamically from incoming frames' source MAC addresses.

Examine the destination MAC address lookup.

The destination MAC address 0000.2222.2222 is determined to be an unknown unicast address because it is not present in the MAC address table.

Forwarding decisions are made based on looking up the destination MAC address in the switch MAC address table.

Determine the forwarding behavior for unknown unicast traffic.

The switch floods the frame out all active switch ports in the VLAN, except for the incoming port FastEthernet0/1.

Flooding unknown unicast traffic ensures the destination host receives the packet regardless of which port it is connected to.

Key Concept

Unknown Unicast Flooding and MAC Address Table Lookup Operations

Question 251Question →

When a Layer 2 Ethernet switch receives an ingress frame, it processes the header information in a specific sequential order. Place the steps taken by the switch into the correct chronological order from first to last.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order of switch operation is: 1) The frame arrives on an ingress port, 2) The switch learns or updates the source MAC address and port mapping in the MAC address table, 3) The switch performs a table lookup for the destination MAC address, and 4) The switch forwards or floods the frame out the appropriate port(s).

A Layer 2 Ethernet switch always follows a standard pipeline: frame reception on ingress, source MAC learning and table updating, destination MAC address lookup, and finally executing unicast forwarding or VLAN flooding.

Step-by-Step Solution

Receive ingress frame

The physical frame is received on an active Layer 2 switch port.

Initial reception triggers Layer 2 inspection.

Source MAC learning

The source MAC address is mapped to the ingress port in the MAC address table.

Switches always learn source MAC locations before executing forwarding logic.

Destination MAC lookup

The MAC address table is searched for the destination MAC address.

Determines whether the egress interface is known or unknown.

Frame forwarding decision

The frame is unicast forwarded to a single port or flooded across the VLAN (excluding ingress port).

Completes the Layer 2 delivery cycle.

Key Concept

Layer 2 Frame Ingress Processing and MAC Table Pipeline

Question 252Question →

An enterprise organization with 40 remote branch sites needs to connect all sites to a central data center. Direct communication between branch sites is rarely required, but predictable bandwidth to the central data center and minimal WAN circuit licensing costs are strict requirements. If a single branch WAN link fails, only that site should lose connectivity. Which WAN topology architecture best meets these operational and financial requirements?

Hub-and-spoke topology

Full-mesh topology

Single ring topology

Point-to-point daisy-chain topology

Show answer & explanation

Answer: Hub-and-spoke topology

Answer

The hub-and-spoke topology best satisfies the requirements by concentrating connections at the central hub, minimizing total WAN circuit costs, and isolating individual branch link failures.

The hub-and-spoke topology connects each remote branch directly to a central hub (the data center). This provides independent connectivity per branch, requires only one WAN link per branch site, and minimizes overall cost while aligning perfectly with centralized traffic patterns.

Step-by-Step Solution

Analyze WAN traffic flow requirements

Traffic flow is primarily client-to-server (branch-to-central data center), with minimal branch-to-branch communication required.

Understanding traffic flows dictates whether direct site-to-site WAN circuits are necessary.

Evaluate cost and link scaling across candidate topologies

A hub-and-spoke design scales linearly with

N-1

links (39 links for 40 sites), whereas a full-mesh design requires

\frac{40 \times 39}{2} = 780

links.

Minimizing WAN circuit licensing and circuit operational expenditure requires choosing a cost-efficient topology.

Verify fault domain isolation capabilities

In a hub-and-spoke architecture, each spoke operates on its own WAN connection to the hub, so a failure on one spoke circuit affects only that specific branch.

Ring and linear/daisy-chain topologies risk impacting multiple downstream sites upon a single link outage.

Key Concept

WAN Topology Architectures (Hub-and-Spoke vs Mesh vs Ring)

Estimated Time:2m 0s

Question 253Question →

An enterprise network architect is designing an IP addressing plan within the RFC 1918 private address space using the allocated $172.16.0.0/12$ block. The design requires server LAN subnets that can each support at least $1,000$ usable host IP addresses, as well as point-to-point infrastructure subnets requiring $/30$ prefixes. Which two statements regarding this subnetting design and IPv4 address allocation are correct?

Select all that apply

/22

subnet prefix provides

1,022

usable host addresses, with

172.24.4.0/22

spanning from network ID

172.24.4.0

to broadcast address

172.24.7.255

/22

subnet prefix provides

1,024

usable host addresses because

2^{10} = 1024

host addresses are available per subnet.

The subnet

172.31.255.252/30

represents the final valid

/30

point-to-point subnet available within the assigned RFC 1918

172.16.0.0/12

private block.

The subnet

172.32.0.0/30

is a valid point-to-point subnet within the allocated RFC 1918 Class B private block.

Show answer & explanation

Answer: A $/22$ subnet prefix provides $1,022$ usable host addresses, with $172.24.4.0/22$ spanning from network ID $172.24.4.0$ to broadcast address $172.24.7.255$ .; The subnet $172.31.255.252/30$ represents the final valid $/30$ point-to-point subnet available within the assigned RFC 1918 $172.16.0.0/12$ private block.

Answer

The two correct statements are the assertion that a /22 prefix provides 1,022 usable host addresses with 172.24.4.0/22 spanning up to broadcast address 172.24.7.255, and the assertion that 172.31.255.252/30 is the final valid /30 subnet within the RFC 1918 172.16.0.0/12 private block.

A /22 subnet provides 10 host bits, which allows for 1,022 usable hosts (2^10 - 2) and spans 172.24.4.0 through 172.24.7.255. Furthermore, 172.31.255.252/30 is the final /30 subnet within the 172.16.0.0/12 RFC 1918 private block, which terminates at 172.31.255.255.

Step-by-Step Solution

Calculate host capacity and range for a /22 subnet prefix

Host bits = 32 - 22 = 10 bits. Total IP addresses = 2^10 = 1,024. Usable host addresses = 2^10 - 2 = 1,022. The 3rd octet mask value is 252 (255.255.252.0), creating a 3rd octet block increment of 256 - 252 = 4. Therefore, subnet 172.24.4.0/22 spans from network ID 172.24.4.0 to broadcast address 172.24.7.255.

Determining the required prefix length for 1,000 hosts requires at least 10 host bits (1,022 usable), and validating the subnet boundaries ensures proper addressing without overlap.

Evaluate the RFC 1918 Class B private range boundaries

RFC 1918 defines Class B private space as 172.16.0.0/12, which ranges from 172.16.0.0 to 172.31.255.255.

Verifying private IPv4 address boundaries prevents using public addresses (such as 172.32.0.0/30) internally.

Verify the last /30 subnet within the RFC 1918 Class B block

A /30 subnet uses a mask of 255.255.255.252 with a block size of 4 in the fourth octet. The last block in the 172.31.255.0/24 subnet starts at 172.31.255.252 and ends at 172.31.255.255 (broadcast).

Confirming the upper boundary of 172.16.0.0/12 validates that 172.31.255.252/30 is the last valid /30 private subnet.

Key Concept

Subnetting host capacity formulas (2^n - 2) and RFC 1918 private IPv4 address range boundaries (172.16.0.0/12).

Question 254Question →

An enterprise organization is upgrading its campus network infrastructure to enable high-speed inter-VLAN routing between local client subnets directly within the wiring closet. The required device must perform wire-speed hardware packet forwarding between local VLANs using Switched Virtual Interfaces (SVIs) while continuing to provide Layer 2 access port connectivity for endpoints. Which network component should be deployed to meet these requirements?

Layer 3 multilayer switch

Unmanaged Layer 2 switch

Type 2 hypervisor

Layer 2 discovery agent running LLDP

Show answer & explanation

Answer: Layer 3 multilayer switch

Answer

Layer 3 multilayer switch

A Layer 3 multilayer switch performs both Layer 2 frame switching and Layer 3 IP routing. By configuring Switched Virtual Interfaces (SVIs), it routes traffic between VLANs at hardware wire-speed directly in the access layer, eliminating the need to pass traffic to an external router.

Step-by-Step Solution

Analyze the operational requirements specified in the scenario.

The target device must handle local Layer 2 endpoint connectivity while simultaneously routing IP packets between different local VLANs using SVIs.

Inter-VLAN routing requires a device operating at Layer 3 of the OSI model.

Evaluate candidate network components based on their functional roles.

A Layer 3 switch (multilayer switch) integrates Application-Specific Integrated Circuits (ASICs) for hardware-based packet routing between VLANs alongside standard Layer 2 bridging.

This offloads inter-VLAN routing from core edge routers and fulfills both Layer 2 access and Layer 3 routing requirements.

Key Concept

Multilayer (Layer 3) Switch Operational Roles and Inter-VLAN Forwarding

Estimated Time:1m 0s

Question 255Question →

A network administrator observes severe latency and frequent frame retries on an enterprise floor using a 2.4 GHz wireless deployment. Spectrum analysis reveals high co-channel interference (CCI) among neighboring access points. Which operational characteristic of the 2.4 GHz band primarily causes this channel reuse constraint when compared to the 5 GHz band?

The 2.4 GHz spectrum offers only three non-overlapping 20 MHz channels in standard deployments, restricting spatial frequency reuse.

2.4 GHz radio signals experience higher free-space path loss and attenuation through solid walls than 5 GHz radio signals.

Access points operating in the 2.4 GHz band automatically disable DTLS encryption on CAPWAP control tunnels when client density rises.

The 2.4 GHz spectrum relies on Dynamic Frequency Selection (DFS) to assign adjacent access points to identical broadcast channels.

Show answer & explanation

Answer: The 2.4 GHz spectrum offers only three non-overlapping 20 MHz channels in standard deployments, restricting spatial frequency reuse.

Answer

The 2.4 GHz spectrum offers only three non-overlapping 20 MHz channels in standard deployments, restricting spatial frequency reuse.

In 802.11 wireless networks using the 2.4 GHz ISM band, 20 MHz channels are spaced only 5 MHz apart. Consequently, only channels 1, 6, and 11 do not overlap with each other. In dense multi-AP deployments, having only three usable non-overlapping channels causes neighboring access points to operate on the same channel, generating high co-channel interference (CCI). In contrast, the 5 GHz band provides a much larger selection of non-overlapping channels, enabling far more flexible channel reuse.

Step-by-Step Solution

Analyze the channel spacing and bandwidth in the 2.4 GHz Wi-Fi spectrum.

The 2.4 GHz ISM band allocates channels spaced 5 MHz apart, while standard 802.11 channels require 20 MHz of channel width.

Understanding channel width requirements identifies how channels overlap.

Determine the maximum number of non-overlapping channels in 2.4 GHz.

Only 3 channels (channels 1, 6, and 11) can operate simultaneously in the same physical space without overlapping frequencies.

This strict limit creates spatial planning constraints for network designers.

Compare with 5 GHz channel availability to assess co-channel interference causes.

Because 5 GHz supports over 20 non-overlapping 20 MHz channels, 2.4 GHz environments suffer far more co-channel interference (CCI) in dense deployments.

The constraint of having only 3 non-overlapping channels directly leads to high CCI in multi-AP designs.

Key Concept

2.4 GHz vs 5 GHz Non-Overlapping Channels and Co-Channel Interference

Question 256Question →

An administrator is connecting switch SwitchA to switch SwitchB via interface GigabitEthernet1/0/15. Auto-MDIX has been manually disabled on interface GigabitEthernet1/0/15 using the `no mdix auto` command, and a standard Category 6 straight-through Ethernet cable is used between the two switches. The output of the `show interfaces GigabitEthernet1/0/15` command displays the following:

text
GigabitEthernet1/0/15 is down, line protocol is down
Hardware is Gigabit Ethernet, address is 0007.7d00.4a15
MTU 1500 bytes, BW 1000000 Kbit/sec, DLY 10 usec,
reliability 255/255, txload 1/255, rxload 1/255

Which condition is the primary cause of the interface remaining in a down/down state?

An incorrect cable pinout is used for connecting two like network switches when auto-MDIX is disabled.

A native VLAN mismatch exists between the 802.1Q trunking configurations on SwitchA and SwitchB.

The neighbor discovery protocol TLV exchanges are failing to complete Layer 2 protocol negotiation.

Incompatible EtherChannel negotiation modes prevent the line protocol from reaching an operational state.

Show answer & explanation

Answer: An incorrect cable pinout is used for connecting two like network switches when auto-MDIX is disabled.

Answer

The primary cause is an incorrect cable pinout used for connecting two like network switches when auto-MDIX is disabled.

Connecting two switches (like-to-like devices) requires a crossover cable so that the transmit pins on one end connect to the receive pins on the other end. While Auto-MDIX dynamically detects and corrects cable pinout mismatches, disabling Auto-MDIX with the `no mdix auto` command forces the switch interface to rely strictly on correct physical cable wiring. Using a straight-through cable between two switches with Auto-MDIX disabled prevents physical link detection, causing the interface status and line protocol to remain in a down/down state.

Step-by-Step Solution

Analyze the CLI output for interface status and line protocol state.

The output shows 'GigabitEthernet1/0/15 is down, line protocol is down', indicating a physical Layer 1 connectivity failure.

When both the interface and line protocol are down, the device cannot establish a physical electrical link with the remote peer.

Evaluate physical cable pinout requirements for like devices.

Switch-to-switch connections (like devices) transmit on pins 1,2 and receive on pins 3,6 (for 10/100/1000BASE-T). Connecting two like devices with a straight-through cable connects transmit pins to transmit pins and receive pins to receive pins.

Without Auto-MDIX automatically swapping transmit and receive pin pairs, a crossover cable is strictly required to establish physical link alignment.

Determine the impact of disabling Auto-MDIX.

Disabling Auto-MDIX (`no mdix auto`) forces the interface to rely entirely on correct physical cable pinouts, keeping the interface in a down/down state when connected via a straight-through cable.

Correcting the cable type to a crossover cable or re-enabling Auto-MDIX resolves the physical layer signaling failure.

Key Concept

Auto-MDIX and Cable Pinout Compatibility in Layer 1 Troubleshooting

Question 257Question →

An network administrator is setting up a traditional Cisco three-tier campus network. Which network tier is primarily responsible for directly attaching workstations, IP phones, and wireless access points while providing boundary security features such as port security?

Access layer

Distribution layer

Core layer

Spine layer

Show answer & explanation

Answer: Access layer

Answer

The Access layer is responsible for direct endpoint connectivity and edge security control.

In a classic three-tier hierarchical campus network design, the Access layer switch provides direct physical and logical connectivity for end-user devices (workstations, printers, IP phones) and implements local port-level security features like Port Security and DHCP Snooping.

Step-by-Step Solution

Identify the primary functions required in the prompt

Directly connecting end-user endpoints (workstations, IP phones, APs) and applying edge security features such as port security.

Different network layers in Cisco hierarchical designs perform distinct operational tasks.

Map these functions to the correct Cisco three-tier layer

The Access layer provides end-node attachment and initial entry controls.

The Core layer focuses purely on high-speed switching, while the Distribution layer manages routing boundaries and aggregation.

Key Concept

Cisco 3-Tier Campus Architecture - Access Layer Role

Estimated Time:45s

Question 258Question →

Match each IPv6 address category on the left with its designated reserved prefix block on the right according to RFC standards.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Global Unicast corresponds to 2000::/3, Unique Local corresponds to fc00::/7, Link-Local corresponds to fe80::/10, and Multicast corresponds to ff00::/8.

Each IPv6 category matches its standards-based prefix block: Global Unicast matches 2000::/3, Unique Local matches fc00::/7, Link-Local matches fe80::/10, and Multicast matches ff00::/8.

Step-by-Step Solution

Identify the Global Unicast prefix allocation.

Global Unicast uses binary prefix 001, which maps to

2000::/3

RFC 4291 reserves

2000::/3

for publicly routable global unicast addresses.

Identify the Unique Local address range.

Unique Local uses binary prefix 1111 110, which maps to

fc00::/7

RFC 4193 reserves

fc00::/7

for private enterprise network addressing.

Identify the Link-Local address prefix.

Link-Local uses binary prefix 1111 1110 10, which maps to

fe80::/10

RFC 4291 defines

fe80::/10

for communications confined to a local physical or virtual link.

Identify the Multicast address range.

Multicast uses binary prefix 1111 1111, which maps to

ff00::/8

RFC 4291 assigns all IPv6 addresses starting with FF to multicast functionality.

Key Concept

IPv6 Address Scopes and Prefix Allocations

Question 259Question →

A network administrator is assigning IPv4 addresses to host interfaces within the $10.50.12.0/26$ subnet. Which of the following IPv4 addresses are valid usable host addresses on this subnet?

Select all that apply

10.50.12.1

10.50.12.50

10.50.12.0

10.50.12.63

Show answer & explanation

Answer: $10.50.12.1$ ; $10.50.12.50$

Answer

The addresses

10.50.12.1

and

10.50.12.50

are valid usable host addresses on the

10.50.12.0/26

subnet.

For a

/26

prefix, 6 host bits yield 64 total IP addresses. With a network address of

10.50.12.0

, the subnet broadcast address is

10.50.12.63

. The valid usable host range spans from

10.50.12.1

10.50.12.62

. Therefore,

10.50.12.1

and

10.50.12.50

are valid host addresses.

Step-by-Step Solution

Determine the subnet size from the CIDR prefix.

/26

prefix leaves

32 - 26 = 6

host bits. The total block size is

2^6 = 64

addresses.

Host bits determine the size of the IP block for a given prefix.

Identify the network address and broadcast address for the subnet.

The network address is

10.50.12.0

. The broadcast address is

10.50.12.63

(

0 + 64 - 1

The first address in a block is reserved for the network ID, and the last address is reserved for the broadcast ID.

Calculate the usable host address range.

The usable host address range is

10.50.12.1

through

10.50.12.62

Usable host addresses exclude both the network ID and the broadcast ID (

2^n - 2

Evaluate the given choices against the usable host range.

10.50.12.1

and

10.50.12.50

fall within the

[10.50.12.1, 10.50.12.62]

range, while

10.50.12.0

is the network ID and

10.50.12.63

is the broadcast ID.

Only addresses inside the calculated usable host range can be assigned to host interfaces.

Key Concept

Subnet Host Range Calculation

Estimated Time:45s

Question 260Question →

A network administrator is troubleshooting an addressing issue on a branch office switch. A host computer on VLAN 10 has been manually configured with the IP address $172.25.107.191$ and subnet mask $255.255.255.192$ . The switch SVI interface serving as the default gateway for VLAN 10 is assigned $172.25.107.129/26$ . The host is unable to ping its default gateway or communicate beyond its local segment. Which statement correctly identifies the cause of this connectivity failure?

The assigned address

172.25.107.191

is the broadcast address for the subnet and cannot be utilized as a valid host IP address.

The host address

172.25.107.191

falls outside the RFC 1918 private IPv4 range, causing the private default gateway to drop non-routable public packets.

The default gateway

172.25.107.129/26

provides a maximum of 62 usable addresses, meaning

172.25.107.191

is the first host address of the next adjacent subnet.

The host IP address

172.25.107.191

and gateway address

172.25.107.129

belong to different subnets because a /26 prefix divides the third octet into 64-host boundaries.

Show answer & explanation

Answer: The assigned address $172.25.107.191$ is the broadcast address for the subnet and cannot be utilized as a valid host IP address.

Answer

The host address 172.25.107.191 is the broadcast address for the 172.25.107.128/26 subnet and is invalid for host assignment.

The correct answer accurately identifies that 172.25.107.191 is the directed broadcast address for the 172.25.107.128/26 network segment. For a /26 prefix, the block size is 64, giving a network ID of 172.25.107.128 and a broadcast ID of 172.25.107.191. Because broadcast addresses cannot be configured as valid unicast endpoints, the host network stack fails to process traffic correctly.

Step-by-Step Solution

Determine the subnet mask prefix length and block size.

A subnet mask of

255.255.255.192

corresponds to CIDR notation

/26

. The block size in the fourth octet is calculated as

256 - 192 = 64

Calculating the block size identifies subnet boundaries.

Calculate the subnet boundaries for the gateway IP

172.25.107.129

Subnet multiples of 64 in the 4th octet are 0, 64, 128, and 192. The gateway IP

172.25.107.129

falls into the network

172.25.107.128/26

Identifying the network ID establishes the IP range for VLAN 10.

Calculate the network address, usable host range, and broadcast address.

Network Address:

172.25.107.128

. First Usable Host:

172.25.107.129

. Last Usable Host:

172.25.107.190

. Broadcast Address:

172.25.107.191

Formula

2^n - 2

yields 62 usable host IPs (

2^6 - 2 = 62

Evaluate the validity of the host's manually assigned IP address

172.25.107.191

The address

172.25.107.191

matches the subnet broadcast address exactly.

Broadcast addresses are reserved for network-wide broadcasts and cannot be bound to network adapters as valid unicast host IPs.

Key Concept

Subnet Broadcast Address Identification and Usable Host Range Calculation

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