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Network Fundamentals

398 questions

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Question 81Question →

A network engineer is reviewing a branch deployment where a bare-metal Type 1 hypervisor host running multiple virtual machines (VMs) across VLAN 10 and VLAN 20 is connected via a single interface to a physical Layer 2 switch. Users report that VMs on VLAN 10 cannot communicate with VMs on VLAN 20, and broadcast storms generated by a misconfigured endpoint on VLAN 10 are flooding all physical ports on the switch. Which statement correctly identifies the operational roles of the network components involved and the required architectural change?

Show answer & explanation

Answer: The Layer 2 switch limits collision domains per port but forwards broadcast frames across the entire VLAN; inter-VLAN traffic requires a Layer 3 device or router-on-a-stick, while the Type 1 hypervisor runs directly on hardware to manage virtual switch port groups.

Answer

The Layer 2 switch limits collision domains per port but forwards broadcast frames across the entire VLAN; inter-VLAN traffic requires a Layer 3 device or router-on-a-stick, while the Type 1 hypervisor runs directly on hardware to manage virtual switch port groups.
Layer 2 switches isolate collision domains on a per-port basis, but broadcast traffic is flooded across all ports belonging to the same VLAN. Furthermore, traffic traversing different VLANs (VLAN 10 to VLAN 20) requires a Layer 3 routing function (such as a Layer 3 switch or router-on-a-stick). A Type 1 hypervisor operates bare-metal directly on physical hardware to manage virtual switches and guest VMs.

Step-by-Step Solution

1
Analyze the role of a Layer 2 switch regarding domain boundaries.
A Layer 2 switch creates a separate collision domain for each connected port, but all ports assigned to a single VLAN belong to the same broadcast domain. Broadcast traffic will flood all ports in that VLAN.
Layer 2 switches forward frames based on MAC addresses and broadcast frames (FF:FF:FF:FF:FF:FF) out all unassigned/active ports in the VLAN.
2
Analyze the requirement for inter-VLAN communication between VLAN 10 and VLAN 20.
Communication across distinct VLANs (IP subnets) requires Layer 3 packet forwarding (routing). A standard Layer 2 switch cannot perform IP routing without a Layer 3 device or Layer 3 switch capabilities.
VLANs define logical Layer 2 boundaries; traversing between different VLANs requires rewriting Layer 2 headers via a default gateway router or Layer 3 switch.
3
Verify the role and definition of a Type 1 hypervisor.
A Type 1 hypervisor (bare-metal) installs directly on the physical host hardware and manages virtual switches (vSwitches) and VM access directly without relying on an intermediate desktop OS.
Understanding Type 1 hypervisor deployment clarifies that virtual networking components (port groups/VLAN tags) interface directly with the physical NICs.

Key Concept

Operational Roles of Layer 2 Switches, Routers, and Bare-Metal Type 1 Hypervisors
Question 82Question →

An organization is deploying a database cluster on a bare-metal Type 1 hypervisor architecture. A network engineer needs to determine how frame forwarding occurs between two virtual machines (VM-A and VM-B) located on the same physical host and assigned to the same VLAN. Which mechanism correctly describes how traffic is handled between these two virtual machines?

Show answer & explanation

Answer: The internal virtual switch (vSwitch) inside the hypervisor performs Layer 2 forwarding between vNICs without sending frames onto the physical network interface card (pNIC).

Answer

The internal virtual switch (vSwitch) inside the hypervisor performs Layer 2 forwarding between vNICs without sending frames onto the physical network interface card (pNIC).
In a Type 1 (bare-metal) hypervisor, a virtual switch (vSwitch) runs directly inside the hypervisor software layer. It maintains virtual ports connected to each virtual machine's vNIC. When two VMs residing on the same host and VLAN communicate, the vSwitch performs Layer 2 MAC address lookup and switches frames entirely in RAM, bypassing the physical NIC and external physical switches.

Step-by-Step Solution

1
Analyze the hypervisor architecture and VM placement
Identify that the environment uses a Type 1 bare-metal hypervisor hosting both virtual machines on the same physical host and VLAN.
Understanding hypervisor architecture determines how traffic flows between virtualized network components.
2
Evaluate the operational role of the virtual switch (vSwitch)
Recognize that the internal vSwitch maintains a MAC address table connecting the vNICs of VM-A and VM-B.
Because both VMs are attached to the same vSwitch on the same VLAN, the hypervisor can switch traffic internally in system memory.
3
Determine physical network interface involvement
Confirm that traffic remains within the hypervisor software and does not traverse the physical NIC (pNIC).
Physical network interfaces and physical switches are only engaged when sending frames to external hosts or across different physical hypervisors.

Key Concept

Virtual Switch (vSwitch) operation and intra-host Layer 2 frame forwarding in Type 1 hypervisor architectures
Estimated Time:1m 30s
Question 83Question →

A network administrator is troubleshooting poor file transfer performance on a switch interface connected to an enterprise application server. The administrator executes the `show interfaces gigabitethernet 0/1` command on the switch and receives the following CLI output:

GigabitEthernet0/1 is up, line protocol is up
Hardware is GigabitEthernet, address is 0007.ec29.2101
Full-duplex, 100Mb/s, media type is 100BaseTX
12458 input errors, 12458 CRC, 0 frame, 0 overrun, 0 ignored
0 output errors, 45211 collisions, 1204 late collision, 0 deferred

Based on this output, what is the most likely cause of the interface errors?

Show answer & explanation

Answer: A duplex mismatch between the switch port and the connected server interface

Answer

A duplex mismatch between the switch port and the connected server interface
A duplex mismatch occurs when one end of an Ethernet segment is configured for full-duplex while the opposite end operates in half-duplex mode. Because full-duplex mode disables CSMA/CD collision detection, the full-duplex side transmits data whenever ready. The half-duplex side, which is actively listening for carrier sense, experiences collisions and late collisions during its transmissions. The full-duplex side receives colliding signals mid-transmission, registering them as input CRC errors.

Step-by-Step Solution

1
Analyze the interface status and line protocol state.
The link status is 'up/up', proving Layer 1 signaling and Layer 2 framing are established.
Eliminates total physical disconnection or wrong pinout issues that prevent link activation.
2
Examine the duplex settings and error counters in the CLI output.
The interface reports 'Full-duplex' along with non-zero 'collisions', 'late collision', and 'CRC' errors.
In true full-duplex operation, collisions should never occur because transmit and receive pathways are independent.
3
Correlate late collisions and CRC errors to physical layer operations.
Late collisions occur when the remote half-duplex side detects a collision after the slot time because the local full-duplex side transmits without checking CSMA/CD carrier sense.
This symptom pattern unequivocally points to a duplex mismatch between the two connected devices.

Key Concept

Duplex Mismatch Symptoms and Error Counters
Estimated Time:1m 30s
Question 84Question →

In a centralized Cisco wireless network deployment using Split-MAC architecture, operational tasks are divided between Lightweight Access Points (LAPs) and the Wireless LAN Controller (WLC). Which two functions are performed by the Wireless LAN Controller rather than the Lightweight Access Points? (Select two.)

Select all that apply

Show answer & explanation

Answer: Executing Radio Resource Management (RRM) to dynamically adjust RF channels and power levels across access points; Aggregating wireless client authentication traffic and managing integration with central RADIUS servers

Answer

The Wireless LAN Controller handles Radio Resource Management (RRM) for dynamic channel/power optimization and processes centralized client authentication requests.
In a centralized Split-MAC architecture, the Wireless LAN Controller performs high-level management and control-plane tasks. This includes Radio Resource Management (RRM) to continuously calculate optimal channel assignments and transmit power for all connected access points, as well as handling centralized client authentication and external RADIUS server communications.

Step-by-Step Solution

1
Identify control-plane and management functions in a centralized Split-MAC wireless architecture.
Radio Resource Management (RRM) and centralized AAA/RADIUS security integration belong to the WLC.
Lightweight Access Points handle real-time frame transmission, beacons, and signal measurements, while system-wide management is offloaded to the central WLC.
2
Evaluate the incorrect choices to identify misconceptions regarding component roles.
A WLC does not replace Layer 3 routers for broadcast domain isolation, nor does it function as a Type 2 hypervisor.
Layer 3 devices break broadcast domains, and physical wireless frame transmission is handled by AP radio hardware.

Key Concept

Wireless LAN Controller (WLC) vs Lightweight Access Point (LAP) Functional Roles
Question 85Question →

A Cisco Catalyst switch operating with default MAC address table aging timer settings (300 seconds) has an existing dynamic entry for Host A (MAC: 0010.a412.1111) on interface GigabitEthernet0/1 with a current age of 295 seconds. At time t=0t = 0 seconds, Host B (MAC: 0010.a412.2222) connected to GigabitEthernet0/2 sends a unicast frame destined to Host A. Host A sends no traffic during this period. At time t=10t = 10 seconds, Host B sends a second unicast frame destined to Host A. How does the switch process the frame sent at t=10t = 10 seconds?

Show answer & explanation

Answer: The switch floods the frame out all operational ports within the VLAN except GigabitEthernet0/2 because Host A's entry aged out 5 seconds prior.

Answer

The switch floods the frame out all operational ports within the VLAN except GigabitEthernet0/2 because Host A's entry aged out 5 seconds prior.
Switches update and refresh MAC address table entries strictly based on the Source MAC address of ingress frames. When Host B sends a frame destined for Host A at t = 0 seconds, Host A's entry timer is NOT reset because Host A did not transmit any frame. Consequently, Host A's dynamic entry reaches the 300-second threshold at t = 5 seconds and is removed from the MAC table. When Host B sends another frame at t = 10 seconds, the switch finds no matching destination entry and floods the unknown unicast frame out all active ports in the VLAN except the receiving port (GigabitEthernet0/2).

Step-by-Step Solution

1
Analyze MAC address table aging mechanics for the frame transmitted at t = 0 seconds.
Host B (0010.a412.2222) is learned on GigabitEthernet0/2. The frame is forwarded out GigabitEthernet0/1 using Host A's existing MAC table entry.
MAC address aging timers are refreshed ONLY when a frame is received containing that MAC address as the Source MAC. Destined (destination MAC) lookups do not reset the aging timer.
2
Calculate the expiration time of Host A's MAC address entry.
Host A's entry had an age of 295 seconds at t = 0. With a 300-second default timer and no incoming frames originating from Host A, the entry expires at t = 5 seconds (295 + 5 = 300 seconds).
Host A remained silent, so no frames with Source MAC 0010.a412.1111 entered the switch to reset the timer.
3
Evaluate the switch forwarding logic for the frame sent at t = 10 seconds.
At t = 10 seconds, the destination MAC 0010.a412.1111 is no longer in the MAC address table. The switch performs unknown unicast flooding.
When a destination unicast MAC address is not present in the L2 MAC table, Layer 2 switches flood the frame out all active interfaces in the same VLAN except the ingress port.

Key Concept

L2 MAC Table Aging and Unknown Unicast Flooding Logic
Question 86Question →

An enterprise network architect is designing a two-tier Spine-Leaf (Clos) topology for a data center hosting high-performance containerized workloads. The design deploys 4 Spine switches and multiple Leaf switches. Each Leaf switch is configured with 48 x 25GbE downlink ports for server connections and 4 x 100GbE uplink ports, with exactly one uplink connecting to each of the 4 Spine switches. Layer 3 routing with Equal-Cost Multi-Pathing (ECMP) is configured across all spine-leaf interconnections. Which statement correctly evaluates the architectural traits and bandwidth performance of this Spine-Leaf deployment?

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Answer: The Leaf switch layer exhibits a 3:1 oversubscription ratio for East-West traffic while delivering deterministic, equal-latency forwarding across the spine fabric between any two Leaf switches.

Answer

The Leaf switch layer exhibits a 3:1 oversubscription ratio for East-West traffic while delivering deterministic, equal-latency forwarding across the spine fabric between any two Leaf switches.
The total host downlink bandwidth per Leaf switch is 48×25 Gbps=1200 Gbps48 \times 25\text{ Gbps} = 1200\text{ Gbps}, while the total fabric uplink bandwidth is 4×100 Gbps=400 Gbps4 \times 100\text{ Gbps} = 400\text{ Gbps}. Dividing downlink by uplink capacity yields an oversubscription ratio of 1200:4001200:400, which simplifies to 3:13:1. Additionally, because every Leaf switch connects directly to all 4 Spine switches in a two-tier Clos architecture, any inter-leaf traffic takes exactly 2 switch hops (Leaf →\rightarrow Spine →\rightarrow Leaf), guaranteeing consistent, deterministic latency across equal-cost paths.

Step-by-Step Solution

1
Calculate total host-facing downlink capacity per Leaf switch.
48 ports × 25 Gbps = 1200 Gbps total downlink bandwidth.
Host access capacity is determined by multiplying the number of downlink access ports by their individual link speed.
2
Calculate total fabric-facing uplink capacity per Leaf switch.
4 ports × 100 Gbps = 400 Gbps total uplink bandwidth.
Fabric uplink capacity is determined by multiplying the number of uplink connections to the spine by their link speed.
3
Determine the oversubscription ratio of the Leaf layer.
1200 Gbps : 400 Gbps = 3:1 oversubscription ratio.
Oversubscription ratio is calculated as Total Ingress (Downlink) Bandwidth divided by Total Egress (Uplink) Bandwidth.
4
Analyze path latency and forwarding behavior of the Spine-Leaf (Clos) fabric.
Every leaf-to-leaf path is exactly 2 fabric hops (Leaf A → Spine → Leaf B) using Layer 3 ECMP across all available Spine switches.
Spine-Leaf architectures guarantee consistent East-West latency because all endpoints connected to different leaf switches are separated by the exact same number of switch hops.

Key Concept

Spine-Leaf Architecture Oversubscription and ECMP Operation
Question 87Question →

A network administrator is evaluating transport layer options for application traffic. Which two characteristics accurately describe User Datagram Protocol (UDP)? (Select two.)

Select all that apply

Show answer & explanation

Answer: It operates as a connectionless protocol without establishing a session prior to data transmission; It maintains a fixed 8-byte header to keep processing overhead low

Answer

User Datagram Protocol (UDP) is connectionless and features an 8-byte header overhead.
The correct responses highlight UDP's core design as a lightweight, connectionless protocol. UDP sends datagrams without a three-way handshake and uses an 8-byte header containing source port, destination port, length, and checksum fields.

Step-by-Step Solution

1
Analyze transport layer requirements and connection states.
Identify that UDP is a stateless, connectionless protocol that transmits data without initiating a preliminary handshake.
UDP avoids the setup latency required by connection-oriented protocols like TCP.
2
Compare protocol header sizes and operational mechanisms.
Confirm that the UDP header requires only 8 bytes of overhead, compared to TCP's 20-byte baseline header.
A small header minimizes network overhead and speeds up datagram parsing for real-time applications.

Key Concept

Characteristics of User Datagram Protocol (UDP)
Question 88Question →

An engineer configures a Cisco router interface GigabitEthernet0/0 with a MAC address of 0011.2589.ABCD using the command `ipv6 address 2001:DB8:ACAD:1::/64 eui-64`. Which IPv6 address will be automatically generated on this interface via the EUI-64 process?

Show answer & explanation

Answer: 2001:DB8:ACAD:1:211:25FF:FE89:ABCD

Answer

2001:DB8:ACAD:1:211:25FF:FE89:ABCD
The correct address is constructed by splitting MAC address 0011.2589.ABCD into 0011.25 and 89.ABCD, inserting FFFE into the center to get 0011.25FF.FE89.ABCD, and inverting the 7th bit of the first byte from 0 to 1 (changing 0x00 to 0x02). Appending this interface identifier to prefix 2001:DB8:ACAD:1::/64 yields 2001:DB8:ACAD:1:211:25FF:FE89:ABCD after suppressing leading zeros.

Step-by-Step Solution

1
Split the 48-bit MAC address into two 24-bit halves.
Left half: 00-11-25, Right half: 89-AB-CD.
The EUI-64 process expands a 48-bit MAC address into a 64-bit Interface Identifier by splitting it at the midpoint.
2
Insert the 16-bit hexadecimal constant FFFE between the two halves.
00-11-25-FF-FE-89-AB-CD.
FFFE is inserted into the center of a 48-bit MAC address to construct a 64-bit IEEE EUI-64 interface identifier.
3
Invert the 7th bit (Universal/Local bit) of the first byte.
First byte 0x00 (binary 0000 0000) becomes 0x02 (binary 0000 0010).
Inverting the 7th bit changes a universally administered MAC address bit into a 1 to indicate global uniqueness within IPv6 interface identifiers.
4
Combine the /64 prefix with the generated 64-bit interface identifier and apply standard IPv6 zero suppression.
2001:DB8:ACAD:1:0211:25FF:FE89:ABCD becomes 2001:DB8:ACAD:1:211:25FF:FE89:ABCD.
Leading zeros in any 16-bit hex field must be omitted per standard IPv6 address formatting rules.

Key Concept

IPv6 EUI-64 Interface Identifier Generation
Estimated Time:1m 30s
Question 89Question →

A network security analyst is auditing border firewall traffic logs to ensure no unroutable internal addresses are leaked to the public Internet. Corporate security policy requires Network Address Translation (NAT) for all internal devices accessing public web services.

Which TWO IPv4 addresses identified in the firewall log belong to the RFC 1918 private address space and require source NAT before being routed across the public Internet? (Select two.)

Select all that apply

Show answer & explanation

Answer: 10.254.1.5; 172.30.100.45

Answer

The addresses 10.254.1.5 and 172.30.100.45 are RFC 1918 private IPv4 addresses that require Network Address Translation to be routed over the public Internet.
RFC 1918 defines three private IPv4 address blocks that are not routable across the public Internet: 10.0.0.0/8, 172.16.0.0/12 (172.16.0.0 through 172.31.255.255), and 192.168.0.0/16. The address 10.254.1.5 falls inside the 10.0.0.0/8 Class A block, and 172.30.100.45 falls inside the 172.16.0.0/12 Class B block. Both require source NAT to communicate over public networks.

Step-by-Step Solution

1
Identify the standard RFC 1918 private IPv4 address blocks.
Class A: 10.0.0.0/8 (10.0.0.0 – 10.255.255.255)
Class B: 172.16.0.0/12 (172.16.0.0 – 172.31.255.255)
Class C: 192.168.0.0/16 (192.168.0.0 – 192.168.255.255)
RFC 1918 designates these exact prefixes for private enterprise networks and dictates that they are non-routable on the public Internet.
2
Evaluate each address option against the private IPv4 boundary definitions.
10.254.1.5 resides within 10.0.0.0/8.
172.30.100.45 resides within 172.16.0.0/12.
172.32.50.10 is outside 172.16.0.0/12.
192.169.10.1 is outside 192.168.0.0/16.
Addresses starting with 172.32 and 192.169 belong to public IPv4 address space assigned to internet registries.

Key Concept

RFC 1918 Private IPv4 Addressing Ranges
Estimated Time:1m 15s
Question 90Question →

A network administrator needs to select a transport layer protocol for a real-time voice application where low latency is critical and missing packets should not be retransmitted. Which transport layer protocol satisfies these requirements?

Show answer & explanation

Answer: User Datagram Protocol (UDP)

Answer

User Datagram Protocol (UDP) is correct because it operates as a connectionless transport protocol with low header overhead and no retransmission mechanism, preserving low latency for real-time voice data.
User Datagram Protocol (UDP) operates without establishing a session connection or tracking packet delivery. Because it omits sequencing and retransmissions, it minimizes latency, making it the preferred transport layer protocol for time-sensitive traffic like voice and video.

Step-by-Step Solution

1
Identify application requirements
Real-time voice requires minimal latency and minimal overhead, and cannot afford the latency introduced by packet retransmissions.
Voice traffic relies on timely delivery; delayed retransmitted audio packets arrive too late to be usable.
2
Evaluate transport layer protocols
UDP provides lightweight, connectionless transmission without ordering or retransmission guarantees, whereas TCP adds latency via flow control and retransmissions.
UDP headers are only 8 bytes compared to TCP's 20-byte baseline header, eliminating setup handshakes and recovery delays.

Key Concept

Connectionless vs Connection-Oriented Transport Protocols
Question 91Question →

An engineer enters the following configuration command on a Cisco router interface whose hardware MAC address is `0A00.2712.3456`:

`R1(config-if)# ipv6 address 2001:db8:acad:b000::/64 eui-64`

Which fully compressed IPv6 global unicast address will be dynamically assigned to this interface?

Show answer & explanation

Answer: 2001:db8:acad:b000:800:27ff:fe12:3456

Answer

The correct dynamically generated address is 2001:db8:acad:b000:800:27ff:fe12:3456.
To derive the EUI-64 interface identifier from MAC address 0A00.2712.3456, insert FFFE between the 24-bit OUI and 24-bit NIC identifier to get 0A00:27FF:FE12:3456. Invert the 7th bit of the first hex byte (0A = 0000 1010 -> 08 = 0000 1000), resulting in 0800:27FF:FE12:3456. Combining this with prefix 2001:db8:acad:b000::/64 and suppressing the leading zero in 0800 yields 2001:db8:acad:b000:800:27ff:fe12:3456.

Step-by-Step Solution

1
Split MAC address and insert FFFE into the middle
MAC 0A00.2712.3456 becomes 0A00:27FF:FE12:3456.
EUI-64 generation requires expanding a 48-bit MAC address into a 64-bit interface ID by inserting FFFE (0xFFFE) between the OUI and device identifier.
2
Invert the 7th bit (Universal/Local bit) of the first byte
First byte 0A (binary 0000 1010) becomes 08 (binary 0000 1000).
The IEEE standard specifies that the 7th bit of the first byte must be inverted to indicate a globally unique, locally configured IPv6 interface identifier.
3
Combine the subnet prefix and interface ID, applying IPv6 compression rules
2001:db8:acad:b000:0800:27ff:fe12:3456 compresses to 2001:db8:acad:b000:800:27ff:fe12:3456.
Leading zeros within a hextet must be suppressed (0800 becomes 800), while non-zero fields like b000 retain their representation.

Key Concept

EUI-64 Interface Identifier Generation
Estimated Time:2m 0s
Question 92Question →

A network administrator is evaluating the division of responsibilities in a Cisco Split-MAC wireless architecture. Which two functions are performed directly by the Lightweight Access Point (LAP) in real time? (Select two.)

Select all that apply

Show answer & explanation

Answer: Transmitting 802.11 beacon and probe response frames; Executing real-time 802.11 frame encryption and decryption

Answer

The two functions performed directly by the Lightweight Access Point are transmitting 802.11 beacon and probe response frames, and executing real-time 802.11 frame encryption and decryption.
In Cisco Split-MAC architecture, time-sensitive real-time tasks are performed directly by the Lightweight Access Point (LAP). These include transmitting 802.11 beacons, responding to probe requests, and performing hardware encryption/decryption of RF frames. Management, authentication, and global radio management are handled centrally by the WLC.

Step-by-Step Solution

1
Analyze Split-MAC architecture roles
Split-MAC divides 802.11 protocol handling between real-time tasks (LAP) and management/control tasks (WLC).
Real-time functions require immediate RF handling and low latency.
2
Identify LAP real-time tasks
Beacon generation, probe responses, packet buffering, and MAC layer encryption/decryption are handled by the LAP.
These operations must occur instantaneously at the physical RF boundary.
3
Identify WLC centralized management tasks
802.1X authentication, RRM channel/power calculations, and frame payload translation occur at the controller.
Centralizing management provides coordinated control and security enforcement across the wireless network.

Key Concept

Cisco Split-MAC Architecture Function Division
Question 93Question →

A network administrator needs to segment an enterprise network to prevent broadcast traffic sent by one host from reaching devices in other departments. Which network component natively creates separate broadcast domains at Layer 3?

Show answer & explanation

Answer: Router

Answer

A router natively breaks up broadcast domains because it operates at Layer 3 and blocks Layer 2 broadcast traffic from passing between its interfaces.
Routers operate at Layer 3 (Network Layer) and break up broadcast domains by default because they do not forward Layer 2 broadcast frames between interfaces.

Step-by-Step Solution

1
Analyze the operational requirements specified in the scenario.
The requirement is to isolate broadcast traffic so broadcast frames are not flooded to all hosts.
Broadcast frames flood across all ports within a single broadcast domain.
2
Evaluate the function of Layer 2 vs. Layer 3 network components.
Layer 2 switches separate collision domains but forward broadcasts to all ports in the same VLAN. Routers work at Layer 3 and drop Layer 2 broadcast packets by default.
Routers terminate Layer 2 broadcast domains on each routed interface.

Key Concept

Network Components Roles and Functions
Question 94Question →

Match each Ethernet physical interface standard or transceiver type to its corresponding media specification, wavelength, and maximum distance limitation.

Click a left item, then click its matching right item

Items

10GBASE-SR SFP+ Transceiver
1000BASE-LX Transceiver (operating over MMF)
10GBASE-LR SFP+ Transceiver
100BASE-TX Interface

Matches

Show answer & explanation

Answer

10GBASE-SR SFP+ matches Multimode Fiber using 850 nm wavelength up to 300 m on OM3; 1000BASE-LX over MMF matches Multimode Fiber using 1310 nm wavelength requiring a mode-conditioning patch cord for 550 m reach; 10GBASE-LR SFP+ matches Single-Mode Fiber using 1310 nm wavelength up to 10 km; 100BASE-TX matches Category 5e/6 UTP Copper cable up to 100 m.
Each physical transceiver and interface type has specific operating wavelengths, core media requirements, and distance bounds: 10GBASE-SR uses 850 nm lasers on multimode fiber up to 300 meters (OM3); 1000BASE-LX can run over multimode fiber at 1310 nm up to 550 meters when using a mode-conditioning patch cord; 10GBASE-LR operates over 1310 nm single-mode fiber up to 10 kilometers; and 100BASE-TX uses twisted-pair copper with RJ-45 connectors up to 100 meters.

Step-by-Step Solution

1
Identify copper media standards among the options
Classify 100BASE-TX as unshielded twisted-pair (UTP) copper media operating up to 100 meters with RJ-45 terminations.
The suffix -TX in IEEE 802.3 standards denotes twisted-pair copper specifications.
2
Differentiate 10 Gigabit optical transceiver specifications
Pair 10GBASE-SR with 850 nm multimode fiber (300m reach on OM3) and 10GBASE-LR with 1310 nm single-mode fiber (10 km reach).
Short Range (-SR) optics rely on lower-cost 850 nm VCSEL lasers for multimode cores, whereas Long Range (-LR) optics require 1310 nm Fabry-Perot/DFB lasers for single-mode fiber.
3
Analyze legacy Gigabit Ethernet deployment nuances over MMF
Associate 1000BASE-LX operating on legacy MMF with the mode-conditioning patch cord requirement to reach 550 meters.
Coupling a single-mode laser directly into the center of a multimode fiber core causes multiple light modes to propagate asynchronously (differential mode delay), which is corrected by offset laser launch via a mode-conditioning patch cord.

Key Concept

Physical Interface, Fiber Optics, and Cabling Specifications
Estimated Time:2m 0s
Question 95Question →

A network engineer is designing a collapsed core enterprise campus topology connecting multiple access switches to a redundant pair of core switches. The design requires all redundant uplink interfaces from each access switch to forward traffic simultaneously for bandwidth optimization and load balancing, while preventing Layer 2 loops without forcing Spanning Tree Protocol (STP) to place any physical uplink into a blocking state. Which design architecture or technology satisfies these operational requirements?

Show answer & explanation

Answer: Implementation of multichassis link aggregation (such as Cisco StackWise or Virtual Switching System) across the collapsed core switches

Answer

Implementation of multichassis link aggregation (such as Cisco StackWise or Virtual Switching System) across the collapsed core switches
In a 2-Tier Collapsed Core architecture, dual core switches operating independently would trigger Spanning Tree Protocol (STP) to block one of the redundant uplinks from each access switch to prevent Layer 2 loops. By deploying multichassis technologies such as Cisco StackWise, Virtual Switching System (VSS), or StackWise Virtual, the two core switches merge into a single logical control plane. This enables access switches to configure Multi-chassis EtherChannel (MEC), allowing all redundant physical uplinks to actively forward traffic simultaneously with balanced load distribution.

Step-by-Step Solution

1
Identify topology requirements
The topology is a 2-Tier Collapsed Core model requiring redundant uplinks from access switches to dual core switches with active-active forwarding (no STP blocked ports).
Standard STP blocks redundant paths to prevent loops, reducing aggregate bandwidth if links are unaggregated.
2
Evaluate switch virtualization and aggregation mechanisms
Multichassis technology presents dual core switches as a single logical STP bridge.
Because the access switch sees only one logical upstream device, cross-chassis EtherChannel (MEC/LACP) bundles all physical links into a single logical channel.
3
Confirm operational behavior
Traffic is load balanced across all links in the EtherChannel bundle while STP views the bundle as a single loop-free connection.
This achieves full bandwidth efficiency and instantaneous hardware-based failover.

Key Concept

Collapsed Core Topology & Multichassis EtherChannel (MEC / VSS / StackWise)
Estimated Time:2m 0s
Question 96Question →

Match each network topology architecture on the left with its primary design characteristic on the right.

Click a left item, then click its matching right item

Items

Collapsed Core Architecture
Three-Tier Campus Architecture
Spine-Leaf Data Center Architecture
SOHO Network Architecture

Matches

Show answer & explanation

Answer

Collapsed Core Architecture pairs with combining core and distribution functions into a single tier. Three-Tier Campus Architecture pairs with using distinct Access, Distribution, and Core layers. Spine-Leaf Data Center Architecture pairs with providing predictable, low-latency east-west traffic flow. SOHO Network Architecture pairs with integrating routing, switching, wireless, and security into a single appliance.
Each topology architecture serves a distinct deployment scale and traffic profile. Collapsed Core simplifies management by consolidating distribution and core functions. Three-Tier provides clear boundary separation across large campuses. Spine-Leaf ensures low, uniform latency across data center workloads. SOHO designs consolidate multi-layer functions into a single physical appliance for small sites.

Step-by-Step Solution

1
Identify the structural definition of Collapsed Core architecture.
Collapsed Core merges the distribution and core layers into a single switch tier.
Smaller campus sites do not require a separate dedicated core layer, making a two-tier collapsed design more cost-effective.
2
Identify the structural definition of Three-Tier Campus architecture.
Three-Tier relies on dedicated Access, Distribution, and Core layers.
Large multi-building campus networks require modular tier separation for traffic control, policy enforcement, and scalability.
3
Identify the structural definition of Spine-Leaf Data Center architecture.
Spine-Leaf provides deterministic east-west latency with all leaf switches directly connected to all spine switches.
Modern data center applications generate significant server-to-server traffic, requiring consistent hop counts and ECMP routing.
4
Identify the structural definition of Small Office/Home Office (SOHO) network architecture.
SOHO architectures use a single multi-function wireless router appliance.
SOHO environments have very low user counts and simple connectivity demands handled by integrated consumer or small-business devices.

Key Concept

Enterprise Network Topology Architectures
Question 97Question →

A network engineer is designing a Variable Length Subnet Mask (VLSM) IP addressing scheme for a newly provisioned branch office using the summary block 192.168.10.0/24192.168.10.0/24. Subnets must be allocated sequentially without wasting address space, starting from the lowest available IP address in descending order of host requirements:

- Subnet A (Staff): Requires 5555 usable host IPv4 addresses
- Subnet B (Sales): Requires 2828 usable host IPv4 addresses
- Subnet C (VoIP): Requires 1212 usable host IPv4 addresses
- Subnet D (WAN Link): Requires 22 usable host IPv4 addresses

What is the broadcast address of Subnet C?

Show answer & explanation

Answer: 192.168.10.111

Answer

The broadcast address of Subnet C is 192.168.10.111.
Subnet C requires 12 host addresses. Satisfying 2n−2≥122^n - 2 \ge 12 requires n=4n=4 host bits (1414 usable IPs), which yields a /28/28 prefix with a block size of 16. Allocating Subnet A (5555 hosts →/26\rightarrow /26, range .0–.63) and Subnet B (2828 hosts →/27\rightarrow /27, range .64–.95) places Subnet C at network address 192.168.10.96. Adding the block size of 16 gives the subnet boundary 192.168.10.96 to 192.168.10.111, making 192.168.10.111 the broadcast address.

Step-by-Step Solution

1
Determine the prefix and range for Subnet A (55 hosts)
Subnet A requires a /26 prefix (block size 64). Subnet A spans 192.168.10.0 to 192.168.10.63.
Formula 2n−2≥552^n - 2 \ge 55 yields n=6n=6 host bits (6262 usable IPs). The prefix is /26/26 (32−632-6).
2
Determine the prefix and range for Subnet B (28 hosts)
Subnet B starts at 192.168.10.64 with a /27 prefix (block size 32). Subnet B spans 192.168.10.64 to 192.168.10.95.
Formula 2n−2≥282^n - 2 \ge 28 yields n=5n=5 host bits (3030 usable IPs). The prefix is /27/27 (32−532-5).
3
Calculate the network address, block size, and broadcast address for Subnet C (12 hosts)
Subnet C starts at 192.168.10.96 with a /28 prefix (block size 16). The network range is 192.168.10.96/28, giving a broadcast address of 192.168.10.111.
Formula 2n−2≥122^n - 2 \ge 12 yields n=4n=4 host bits (1414 usable IPs). The prefix is /28/28 (32−432-4). Adding the block size of 16 to the network start 192.168.10.96 yields a range of 192.168.10.96 to 192.168.10.111.

Key Concept

VLSM block allocation order and subnet broadcast address determination
Estimated Time:2m 0s
Question 98Question →

A network administrator is designing a network for a medium-sized office building. To reduce hardware costs and simplify maintenance while maintaining connectivity between access layer devices and external networks, the administrator decides to combine the responsibilities of the core layer and the distribution layer into a single set of switches. Which network topology architecture is being implemented in this enterprise design?

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Answer: Collapsed core architecture

Answer

The collapsed core architecture is being implemented because it merges the distribution layer and core layer roles into a single pair of switches.
In a collapsed core network design (also known as a two-tier campus design), the distribution layer and core layer functionality are merged into a single tier of redundant switches. This provides routing, security policy enforcement, and high-speed campus backbone connectivity on the same hardware, reducing overall cost and complexity for smaller site footprints.

Step-by-Step Solution

1
Analyze the requirements described in the scenario
The design merges core and distribution layer duties into one switch tier to save cost in a medium-sized campus.
Identifying key structural consolidations helps differentiate enterprise network models.
2
Match the requirement to the standard Cisco topology definitions
A 2-tier design where core and distribution layers are combined is specifically known as a collapsed core topology.
Collapsed core models reduce tier count while retaining access layer connections.

Key Concept

Collapsed Core Topology
Question 99Question →

A network administrator is documenting the RFC 1918 private IPv4 address space allocation for an enterprise network redesign. Match each RFC 1918 address block on the left with its corresponding CIDR prefix length and block composition description on the right.

Click a left item, then click its matching right item

Items

10.0.0.0 – 10.255.255.255
172.16.0.0 – 172.31.255.255
192.168.0.0 – 192.168.255.255

Matches

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Answer

10.0.0.0 – 10.255.255.255 matches '/8 prefix length comprising 1 Class A network block'; 172.16.0.0 – 172.31.255.255 matches '/12 prefix length comprising 16 contiguous Class B network blocks'; 192.168.0.0 – 192.168.255.255 matches '/16 prefix length comprising 256 contiguous Class C network blocks'.
Each RFC 1918 private IPv4 block aligns with a specific CIDR prefix and quantity of traditional classful networks: 10.0.0.0/8 represents one Class A block; 172.16.0.0/12 represents 16 contiguous Class B blocks (172.16.0.0 to 172.31.255.255); and 192.168.0.0/16 represents 256 contiguous Class C blocks (192.168.0.0 to 192.168.255.255).

Step-by-Step Solution

1
Identify the RFC 1918 Class A private address range
10.0.0.0 to 10.255.255.255 corresponds to 10.0.0.0/8 (1 Class A network).
RFC 1918 reserves the entire 10.0.0.0/8 network block for private IP addressing.
2
Identify the RFC 1918 Class B private address range
172.16.0.0 to 172.31.255.255 corresponds to 172.16.0.0/12, which spans 16 contiguous Class B networks (172.16.0.0/16 through 172.31.0.0/16).
A /12 prefix length allows 4 bits of subnetwork variation in the second octet (16 to 31), creating 16 Class B ranges.
3
Identify the RFC 1918 Class C private address range
192.168.0.0 to 192.168.255.255 corresponds to 192.168.0.0/16, which spans 256 contiguous Class C networks (192.168.0.0/24 through 192.168.255.0/24).
A /16 prefix length allows 8 bits of subnetwork variation in the third octet (0 to 255), creating 256 Class C ranges.

Key Concept

RFC 1918 Private IPv4 Address Space Specifications
Question 100Question →

An enterprise wireless administrator is selecting appropriate antenna types for various deployment scenarios across a corporate campus. Match each wireless antenna type on the left with its primary operational characteristic and recommended deployment scenario on the right.

Click a left item, then click its matching right item

Items

Omnidirectional dipole antenna
Patch directional antenna
Yagi directional antenna
Parabolic dish antenna

Matches

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Answer

Omnidirectional dipole antenna matches 360-degree horizontal coverage for open office environments; Patch directional antenna matches broad directional hemispherical pattern for wall mounting along corridors; Yagi directional antenna matches moderate gain focused beamwidth for short-to-medium outdoor bridges; Parabolic dish antenna matches extremely high gain tight beamwidth for long-distance point-to-point links.
Each antenna type is accurately paired with its corresponding radiation pattern, beamwidth, gain characteristic, and recommended deployment application.

Step-by-Step Solution

1
Analyze indoor open-space coverage requirements versus corridor coverage requirements.
Omnidirectional dipole antennas radiate 360 degrees horizontally for open spaces, whereas patch antennas direct RF energy forward in a hemispherical pattern ideal for hallways.
Matching antenna radiation pattern to physical space geometry prevents unnecessary RF bleed and signal degradation.
2
Evaluate outdoor point-to-point wireless bridging requirements based on distance and gain requirements.
Yagi antennas provide moderate gain for short-to-medium building links, while parabolic dish antennas provide high gain and narrow beamwidth for long-distance links.
Long-distance outdoor links require maximum directional gain to overcome free space path loss.

Key Concept

Wireless Antenna Types and RF Radiation Characteristics
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