Examkin

Network Fundamentals

398 questions

All types398Multiple choice205Multiple select92Numeric18Matching68Ordering7Short answer8
Question 201Question →

Match each RFC 1918 private IPv4 address prefix block to its corresponding scope and address range definition.

Click a left item, then click its matching right item

Items

10.0.0.0/8
172.16.0.0/12
192.168.0.0/16

Matches

Show answer & explanation

Answer

10.0.0.0/8 matches a single Class A network block (10.0.0.0 - 10.255.255.255); 172.16.0.0/12 matches 16 contiguous Class B network blocks (172.16.0.0 - 172.31.255.255); 192.168.0.0/16 matches 256 contiguous Class C network blocks (192.168.0.0 - 192.168.255.255).
RFC 1918 explicitly defines three private address blocks: one Class A block (10.0.0.0/8), 16 contiguous Class B blocks (172.16.0.0/12, spanning 172.16.0.0 to 172.31.255.255), and 256 contiguous Class C blocks (192.168.0.0/16, spanning 192.168.0.0 to 192.168.255.255).

Step-by-Step Solution

1
Identify the RFC 1918 range and structure for Class A private addressing.
10.0.0.0/8 covers addresses from 10.0.0.0 through 10.255.255.255, representing a single Class A block.
RFC 1918 allocates one /8 block for Class A private networks.
2
Identify the RFC 1918 range and structure for Class B private addressing.
172.16.0.0/12 covers addresses from 172.16.0.0 through 172.31.255.255, forming 16 contiguous Class B (/16) blocks.
RFC 1918 allocates 16 contiguous /16 blocks starting at 172.16.0.0 up to 172.31.255.255.
3
Identify the RFC 1918 range and structure for Class C private addressing.
192.168.0.0/16 covers addresses from 192.168.0.0 through 192.168.255.255, forming 256 contiguous Class C (/24) blocks.
RFC 1918 allocates 256 contiguous /24 blocks starting at 192.168.0.0 up to 192.168.255.255.

Key Concept

RFC 1918 Private IPv4 Address Space Allocation
Question 202Question →

Match each network architecture layer or topology design model on the left to its correct functional responsibility or operational trait on the right.

Click a left item, then click its matching right item

Items

Spine Layer (Data Center Spine-Leaf)
Leaf Layer (Data Center Spine-Leaf)
Distribution Layer (Traditional 3-Tier Campus)
Collapsed Core Layer (2-Tier Campus)

Matches

Show answer & explanation

Answer

Spine Layer matches high-speed L3 ECMP backbone forwarding; Leaf Layer matches endpoint attachment and VTEP encapsulation; Distribution Layer matches inter-VLAN routing, ACL filtering, and policy enforcement; Collapsed Core matches combining Core transport and Distribution policy into a single switch tier.
The pairings correctly align each tier with its defining design attributes: Spine switches provide non-blocking ECMP L3 transport; Leaf switches act as endpoint connection points and VTEPs; Distribution switches manage L2/L3 policy, ACLs, and routing boundaries; Collapsed Core switches combine Core and Distribution capabilities into one physical tier.

Step-by-Step Solution

1
Analyze Spine Layer responsibilities in Data Center Spine-Leaf (Clos) design.
Spine switches strictly interconnect Leaf switches, run L3 routing protocols, and provide equal-cost multipath transport without inter-spine links.
Spine nodes form the high-speed backbone core of the Clos fabric.
2
Analyze Leaf Layer responsibilities in Spine-Leaf design.
Leaf switches connect to servers, storage, and firewalls, serving as the edge access and VTEP overlay encapsulation boundary.
Endpoints attach exclusively to Leaf nodes in modern fabric architectures.
3
Analyze Distribution Layer responsibilities in traditional 3-Tier Campus design.
Distribution switches aggregate Access switches, perform inter-VLAN routing, and enforce security ACLs and QoS policies.
The Distribution tier isolates the Access tier from the Core tier and manages network policies.
4
Analyze Collapsed Core (2-Tier) campus design characteristics.
Collapsed Core integrates Distribution policy routing and Core high-speed switching into one combined physical tier.
It simplifies deployment for smaller networks that do not require separate core and distribution switch chassis.

Key Concept

Enterprise Campus (3-Tier & Collapsed Core) vs. Data Center (Spine-Leaf) Architectural Layers
Question 203Question →

A network technician connects SwitchA and SwitchB using interface GigabitEthernet1/0/1 on both devices. Auto-MDIX has been manually disabled on both switch ports. The technician executes `show interfaces gigabitethernet 1/0/1` on SwitchA and observes the following CLI output:

SwitchA# show interfaces gigabitethernet 1/0/1
GigabitEthernet1/0/1 is down, line protocol is down
Hardware is Gigabit Ethernet, address is 0007.7d0e.a101 (bia 0007.7d0e.a101)
MTU 1500 bytes, BW 1000000 Kbit/sec, DLY 10 usec
Encapsulation ARPA, loopback not set
Keepalive set (10 sec)
Auto-duplex, Auto-speed

Which underlying physical layer issue is causing the interface status and line protocol to both remain in the down state?

Show answer & explanation

Answer: A straight-through Ethernet cable is connected between similar switch interfaces while Auto-MDIX is disabled.

Answer

A straight-through Ethernet cable is connected between similar switch interfaces while Auto-MDIX is disabled.
When connecting two switches directly (like devices operating MDI-X to MDI-X), a crossover cable is required to connect transmit pins to receive pins. When Auto-MDIX is disabled, using a straight-through cable prevents the interfaces from detecting physical signals, keeping the status and line protocol down.

Step-by-Step Solution

1
Analyze the CLI output line status and line protocol.
Both interface status and line protocol report 'down', indicating a Layer 1 physical signal loss or cable issue.
When line status is down, the interface fails to detect electrical or optical carrier signals from the remote end.
2
Evaluate the cabling requirements for like-device connections without Auto-MDIX.
Switch-to-switch connections use MDI-X pinouts on both ends; without Auto-MDIX, a crossover cable is required.
A straight-through cable connects TX pins on SwitchA to TX pins on SwitchB and RX pins to RX pins, preventing physical carrier detection.

Key Concept

Physical interface troubleshooting and Auto-MDIX dependency on crossover vs straight-through cabling.
Estimated Time:1m 30s
Question 204Question →

An enterprise network operations team is replacing a legacy 3-Tier campus architecture with a 2-Tier Spine-Leaf (Clos) topology in their data center to optimize East-West traffic performance for containerized microservices. Which of the following architectural characteristics and operational design principles correctly describe this Spine-Leaf fabric when compared to the traditional 3-Tier model? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: Every Leaf switch connects to every Spine switch in a full mesh layout, guaranteeing a deterministic two-hop latency path for inter-leaf traffic.; Equal-Cost Multi-Pathing (ECMP) routing is implemented across Layer 3 uplinks to enable active-active load sharing without Spanning Tree Protocol port blocking.

Answer

The correct statements are that every Leaf switch connects directly to every Spine switch to provide a deterministic two-hop path for inter-leaf traffic, and that Layer 3 routing with Equal-Cost Multi-Pathing (ECMP) enables active-active load balancing across uplinks without Spanning Tree Protocol link blocking.
In a 2-Tier Spine-Leaf fabric, every Leaf switch connects directly to every Spine switch without horizontal intra-tier links. This structural layout provides a predictable two-hop transport path for East-West traffic across Leaf switches. Furthermore, deploying Layer 3 routing with Equal-Cost Multi-Pathing (ECMP) allows active-active forwarding across all Leaf-Spine links, eliminating Spanning Tree Protocol blocking.

Step-by-Step Solution

1
Analyze physical cabling topology constraints in a 2-Tier Spine-Leaf (Clos) architecture.
Verify that Spine-Leaf requires every Leaf node to connect to every Spine node, with zero direct links between nodes of the same tier.
This bipartite graph layout ensures a predictable, fixed path length of exactly two hops (Leaf to Spine to Leaf) for any East-West communication.
2
Evaluate control plane mechanisms and bandwidth utilization across the fabric.
Determine that Layer 3 routing combined with ECMP replaces Layer 2 Spanning Tree Protocol loop-prevention mechanisms.
ECMP utilizes all available uplinks simultaneously in an active-active forwarding state, avoiding the bandwidth degradation caused by STP port blocking.
3
Differentiate device roles and boundary placements between Leaf and Spine switches.
Confirm that end-system access ports and broadcast domain boundaries reside strictly on Leaf switches, while Spine switches act solely as high-speed transport infrastructure.
Isolating Spine switches from direct host connectivity and Layer 2 broadcast domains keeps the core stateless, fast, and scalable.

Key Concept

2-Tier Spine-Leaf (Clos) Architecture Characteristics vs 3-Tier Campus Design
Question 205Question →

A network engineer is auditing infrastructure hardware and operational roles across an enterprise campus network. Match each network component on the left with its primary operational function on the right.

Click a left item, then click its matching right item

Items

Core Switch
Access Switch
Lightweight Access Point (LAP)
Enterprise Router

Matches

Show answer & explanation

Answer

Core Switch matches high-speed backbone transport without heavy filtering; Access Switch matches connecting endpoints and enforcing port security; Lightweight Access Point matches real-time 802.11 processing with central controller management; Enterprise Router matches path determination between IP subnets and broadcast domain termination.
Each component is correctly matched to its standard Cisco three-tier hierarchical architecture role or split-MAC wireless model: Core switches provide low-latency backbone transport; Access switches connect endpoints; Lightweight APs process 802.11 frames while managed by a central controller; Routers make Layer 3 forwarding decisions between subnets and isolate broadcast domains.

Step-by-Step Solution

1
Identify the primary role of the Core Switch
Matched to high-speed backbone transport with minimal overhead.
Core devices prioritize raw switching speed and low latency across the network backbone.
2
Identify the primary role of the Access Switch
Matched to end-user device connectivity and edge port security.
Access switches represent the entry point for endpoint devices and enforce boundary policies.
3
Identify the primary role of the Lightweight Access Point (LAP)
Matched to handling real-time 802.11 wireless traffic offloading control to a WLC.
Split-MAC architecture assigns real-time wireless tasks to the LAP and management to the WLC.
4
Identify the primary role of the Enterprise Router
Matched to inter-subnet path determination and broadcast domain isolation.
Routers operate at Layer 3 to connect distinct networks and block Layer 2 broadcast frames.

Key Concept

Hierarchical network layers and specialized device roles (Core, Access, LAP split-MAC, and Layer 3 Routing).
Question 206Question →

A network engineer is analyzing performance degradation for an application that streams telemetry data across an enterprise network. The developer considers modifying the application to add custom sequencing and selective retransmission logic over UDP rather than switching to TCP. Which structural characteristic of the UDP header explains why native sequence tracking and acknowledgment cannot occur at the transport layer when using standard UDP?

Show answer & explanation

Answer: The UDP header has a fixed 8-byte length containing only Source Port, Destination Port, Length, and Checksum fields, omitting sequence and acknowledgment fields entirely.

Answer

The UDP header has a fixed 8-byte length containing only Source Port, Destination Port, Length, and Checksum fields, omitting sequence and acknowledgment fields entirely.
UDP features a minimal, fixed 8-byte header comprising four 16-bit fields: Source Port, Destination Port, Length, and Checksum. Because UDP lacks fields for Sequence Numbers, Acknowledgment Numbers, and Window Size (which are present in the 20-byte minimum TCP header), native transport-layer sequencing, flow control, and retransmission mechanisms are impossible without upper-layer application logic.

Step-by-Step Solution

1
Examine the structural composition of the UDP header.
Identify that UDP consists of four 16-bit (2-byte) fields totaling 8 bytes: Source Port, Destination Port, Length, and Checksum.
Understanding the fixed 8-byte UDP header format clarifies which functions the protocol can and cannot natively support.
2
Compare UDP header capabilities against TCP reliability mechanisms.
Note that TCP includes explicit 32-bit Sequence Number and 32-bit Acknowledgment Number fields in its 20-byte baseline header, whereas UDP lacks these fields.
Sequence tracking, reorder detection, and flow control require dedicated protocol header fields that exist in TCP but are absent in UDP.
3
Deduce why sequence tracking must be implemented at the application layer when using UDP.
Because UDP operates statelessly without built-in sequence or ACK fields, any reliability or sequencing features must be handled above Layer 4 inside the application payload.
Confirms why the application must provide custom sequence numbers if it uses UDP for transport.

Key Concept

TCP vs UDP Header Overhead and Protocol Mechanics
Estimated Time:1m 30s
Question 207Question →

A network administrator is designing a virtualized data center environment to host high-performance enterprise applications. The implementation requires an architecture where the virtualization software interacts directly with the server physical hardware, avoiding the latency and overhead of a host operating system. Which component fulfills this requirement?

Show answer & explanation

Answer: A Type 1 hypervisor installed directly on bare-metal hardware

Answer

A Type 1 hypervisor installed directly on bare-metal hardware
The correct answer identifies the Type 1 hypervisor (bare-metal), which installs directly on the host server hardware. This architecture allows direct control of CPU, memory, and storage resources without passing requests through a primary operating system layer.

Step-by-Step Solution

1
Analyze the operational requirements specified in the scenario
The requirement calls for a virtualization management layer running directly on physical hardware without an intermediate host OS footprint.
Eliminating the host operating system reduces hardware resource overhead, latency, and security attack vectors.
2
Compare hypervisor architectural types against the requirement
Type 1 hypervisors execute directly on physical server hardware (bare-metal), whereas Type 2 hypervisors execute as applications on top of a general-purpose host OS.
Direct hardware interaction defines the bare-metal nature of Type 1 hypervisors.
3
Select the component matching the bare-metal architecture role
The Type 1 hypervisor is the correct component.
It directly satisfies the requirement for bare-metal performance and virtualization management.

Key Concept

Type 1 vs. Type 2 Hypervisors in Network and Server Virtualization
Question 208Question →

An administrator is reviewing IPv6 address scopes and interface identifier generation rules. Which of the following statements correctly describe standard IPv6 address prefixes and EUI-64 requirements? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: The IPv6 prefix fe80::/10 is designated for link-local unicast addressing on a local segment.; The IPv6 prefix ff00::/8 is reserved for multicast addresses.

Answer

The statement identifying fe80::/10 as link-local unicast addressing and the statement identifying ff00::/8 as multicast addressing are both correct.
The prefix fe80::/10 is defined in RFC 4291 for link-local unicast addressing, while ff00::/8 is reserved for IPv6 multicast traffic.

Step-by-Step Solution

1
Evaluate the reserved IPv6 scope prefixes.
fe80::/10 is confirmed as Link-Local and ff00::/8 is confirmed as Multicast.
Standard IPv6 RFC specifications assign fe80::/10 to link-local unicast communication and ff00::/8 to multicast.
2
Evaluate EUI-64 identifier construction.
The statement claiming FFFF insertion without bit flipping is false.
EUI-64 requires inserting FFFE into the MAC address midpoint and inverting the 7th bit (U/L bit).
3
Evaluate IPv6 broadcast and subnet host calculations.
The statement claiming two host addresses are subtracted for network and broadcast IDs is false.
IPv6 eliminates traditional subnet broadcast addresses in favor of multicast.

Key Concept

IPv6 Address Scopes and EUI-64 Identifier Rules
Question 209Question →

A network architect is designing a high-availability infrastructure for an enterprise campus network. The design replaces legacy distribution switches with Layer 3 switches and deploys a bare-metal Type 1 hypervisor host to support core virtualized services such as DNS, DHCP, and IPAM. Which of the following statements correctly describe the operational roles and traffic handling behavior of these components? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: The Layer 3 switch terminates VLAN broadcast domains at the distribution layer by performing IP routing between IP subnets.; The Type 1 hypervisor runs directly on the bare-metal host hardware to allocate hardware resources to virtualized infrastructure services without a host operating system layer.

Answer

The correct statements are that the Layer 3 switch terminates VLAN broadcast domains at the distribution layer by routing between IP subnets, and the Type 1 hypervisor runs directly on bare-metal hardware without a host operating system.
Layer 3 switches establish Layer 3 boundaries between subnets, terminating broadcast domains for each VLAN while routing traffic between them. Type 1 hypervisors run directly on the physical hardware (bare metal), providing high performance and direct resource management for server virtual machines.

Step-by-Step Solution

1
Analyze Layer 3 switch operational roles.
Layer 3 switches isolate broadcast domains per VLAN and perform inter-VLAN routing at Layer 3.
Layer 2 switches isolate collision domains but forward broadcast frames within a VLAN. Adding Layer 3 capabilities creates distinct broadcast domain boundaries per subnet.
2
Evaluate hypervisor architecture types.
Type 1 hypervisors operate directly on physical bare metal, while Type 2 hypervisors run inside a conventional host operating system.
Bare-metal deployment minimizes latency and resource overhead for enterprise virtualized network services.

Key Concept

Layer 3 Switch and Hypervisor Component Roles
Estimated Time:2m 0s
Question 210Question →

A network administrator is evaluating virtualization technologies for a new enterprise data center deployment. Which two statements correctly describe the characteristics of a Type 1 hypervisor? (Select two.)

Select all that apply

Show answer & explanation

Answer: It installs directly on physical server hardware without requiring an underlying host operating system.; It delivers lower latency and higher throughput performance compared to hosted hypervisors.

Answer

Type 1 hypervisors install directly on bare-metal server hardware without a host operating system and provide lower latency and superior performance for virtual machines.
Type 1 hypervisors install directly on physical hardware (bare-metal) without requiring a pre-existing host operating system. Because they bypass intermediate host OS software layers, they offer lower latency and higher performance, making them ideal for enterprise virtualized environments.

Step-by-Step Solution

1
Analyze hypervisor architectural placement.
Recognize that Type 1 hypervisors reside directly on server hardware (bare-metal) rather than running on top of an existing operating system.
Architectural placement distinguishes bare-metal Type 1 hypervisors from hosted Type 2 hypervisors.
2
Evaluate operational efficiency and latency characteristics.
Determine that eliminating host operating system abstraction layers minimizes latency and maximizes processing throughput.
Direct hardware control streamlines CPU scheduling and memory management for virtualized enterprise workloads.

Key Concept

Type 1 Bare-Metal Hypervisor Architecture
Question 211Question →

A network administrator is inspecting a Cisco IOS switch interface connected to a server. The output of the `show interfaces GigabitEthernet 0/1` command reveals a high number of Frame Check Sequence (FCS) errors and late collisions. Which two physical or data-link layer issues are the most likely causes of these specific interface counter statistics? (Select two.)

Select all that apply

Show answer & explanation

Answer: Physical cable damage, excessive electromagnetic interference, or faulty connectors corrupting bits in transit.; A duplex mismatch between the switch port and the connected device, causing the half-duplex side to transmit while receiving data after the slot time.

Answer

Physical cable damage or interference causing bit corruption (FCS errors) and a duplex mismatch causing collision detection after the 64-byte window (late collisions).
Physical cable damage or electrical interference causes bits to flip in transit, resulting in Frame Check Sequence (FCS) checksum mismatches. A duplex mismatch causes the half-duplex end to attempt transmission while the full-duplex end sends traffic, producing late collisions after the initial 64-byte collision window.

Step-by-Step Solution

1
Analyze the FCS error counter symptom
Identify physical layer causes such as bad cabling, interference, or faulty interface hardware
FCS errors indicate that frames arrived corrupted at the receiving interface.
2
Analyze the late collision counter symptom
Identify duplex misconfigurations or excessive cable lengths
Late collisions occur when a collision is detected after the 512-bit (64-byte) slot time, characteristic of half-duplex operating while the peer transmits full-duplex.

Key Concept

Interface counter troubleshooting for FCS errors and late collisions
Question 212Question →

A network technician is simplifying an IPv6 address for documentation purposes by compressing leading zeros and contiguous fields of zeros. Which format represents the correctly compressed version of 2001:0db8:0000:0000:0000:0000:0000:0001?

Show answer & explanation

Answer: 2001:db8::1

Answer

The correctly compressed IPv6 address is 2001:db8::1.
The address 2001:db8::1 strictly follows RFC IPv6 address compression rules by omitting the leading zero in '0db8' to become 'db8' and replacing all consecutive fields of zeros (:0000:0000:0000:0000:0000:) with a single double colon (::).

Step-by-Step Solution

1
Omit leading zeros in each 16-bit hexadecimal field (hextet).
The second hextet 0db8 becomes db8, and 0001 becomes 1.
Leading zeros are redundant and should be removed per standard RFC 4291 / RFC 5952 formatting guidelines.
2
Replace the single contiguous run of consecutive all-zero hextets with a double colon (::).
The five zero hextets (:0000:0000:0000:0000:0000:) are replaced by a single double colon (::).
The double colon represents one or more contiguous fields of zeros to shorten the address.
3
Combine the resulting hextets into the final simplified address.
The address compresses to 2001:db8::1.
This yields a valid, unambiguous 128-bit IPv6 address format.

Key Concept

IPv6 Address Compression Rules (RFC 4291 and RFC 5952)
Question 213Question →

An enterprise network administrator is deploying a virtualized server environment using a Type 1 hypervisor to host multiple virtual machines across separate VLANs. Which two statements accurately describe the operational mechanics of the virtual switch (vSwitch) and hypervisor architecture in this scenario? (Select two.)

Select all that apply

Show answer & explanation

Answer: Intra-host traffic between virtual machines on the same VLAN is switched internally by the vSwitch without egressing through physical network interface cards.; The vSwitch can perform IEEE 802.1Q VLAN tagging on virtual machine frames before sending them across physical uplinks to trunk ports on the physical switch.

Answer

Intra-host traffic between virtual machines on the same VLAN is switched internally by the vSwitch without egressing through physical network interface cards, AND the vSwitch can perform IEEE 802.1Q VLAN tagging on virtual machine frames before sending them across physical uplinks to trunk ports on the physical switch.
In Type 1 hypervisor architectures, intra-host traffic between co-located virtual machines on the same virtual switch stays within host software memory without exiting to the physical network interface cards. Additionally, virtual switches can tag frames with IEEE 802.1Q VLAN headers before sending them over physical uplink interfaces connected to physical switch trunk ports.

Step-by-Step Solution

1
Analyze the architecture of a Type 1 hypervisor and vSwitch frame forwarding.
A Type 1 hypervisor operates directly on bare-metal hardware. The vSwitch handles Layer 2 frame forwarding within the hypervisor software kernel.
Understanding bare-metal execution clarifies that there is no general-purpose host OS beneath the hypervisor to handle network I/O.
2
Evaluate intra-host virtual machine traffic paths.
Frames sent between VMs connected to the same vSwitch and VLAN are switched directly within server memory.
Frames only hit physical network interface card (pNIC) uplinks when destined for external network devices.
3
Determine VLAN tagging capabilities and loop prevention mechanics.
vSwitches encapsulate frames with 802.1Q tags for trunk traffic and rely on non-STP forwarding rules (such as prohibiting forwarding between physical uplinks) to avoid loops.
Standard vSwitches do not process or generate Spanning Tree Protocol BPDUs.

Key Concept

Type 1 Hypervisor and Virtual Switch (vSwitch) Frame Forwarding Mechanics
Estimated Time:2m 0s
Question 214Question →

A network technician observes an automatically configured Link-Local IPv6 address of FE80::0250:56FF:FE9A:BCDE on a server's network interface using SLAAC and EUI-64. Based on standard EUI-64 interface identifier rules, which two statements regarding this interface's physical MAC address and link-local generation process are correct? (Select two.)

Select all that apply

Show answer & explanation

Answer: The original burned-in MAC address of the interface is 00:50:56:9A:BC:DE.; The 7th bit (Universal/Local bit) of the MAC address's first byte was inverted from 0 to 1 during EUI-64 processing to indicate a modified interface ID.

Answer

The original burned-in MAC address of the interface is 00:50:56:9A:BC:DE, and the 7th bit (Universal/Local bit) of the MAC address's first byte was inverted from 0 to 1 during EUI-64 processing.
The link-local IPv6 address FE80::0250:56FF:FE9A:BCDE uses an EUI-64 interface identifier created from a 48-bit MAC address. EUI-64 inserts the hexadecimal value FFFE into the middle of the MAC address (between the 3rd and 4th bytes) and flips the 7th bit of the first byte (the Universal/Local bit). Reversing this process on 0250:56FF:FE9A:BCDE requires removing FFFE to leave 02:50:56:9A:BC:DE, and flipping the 7th bit of the first byte 02 (0000 0010 in binary) back to 0, which gives 00 (0000 0000 in binary). Thus, the original MAC address is 00:50:56:9A:BC:DE, and the 7th bit was inverted from 0 to 1 during the process.

Step-by-Step Solution

1
Analyze the 64-bit interface identifier portion of the IPv6 address.
Interface ID is 0250:56FF:FE9A:BCDE.
The link-local prefix is FE80::/64, leaving the last 64 bits as the EUI-64 identifier.
2
Identify and remove the inserted midpoint 16-bit hex sequence.
Removing FFFE from 0250:56FF:FE9A:BCDE yields the 48-bit sequence 02:50:56:9A:BC:DE.
EUI-64 generation inserts FFFE between the 24-bit OUI and the 24-bit device identifier.
3
Invert the 7th bit (Universal/Local bit) of the first byte.
First byte 02 (binary 0000 0010) inverted at the 7th bit becomes 00 (binary 0000 0000). The resulting original MAC address is 00:50:56:9A:BC:DE.
IEEE EUI-64 rule requires flipping the 7th bit (0 for universal scope becomes 1 for local interface ID uniqueness).

Key Concept

EUI-64 Interface Identifier Derivation
Question 215Question →

A network engineer is inspecting packet capture data from a host running a dual-protocol network diagnostic service. The service transmits real-time telemetry using a connectionless transport protocol while concurrently maintaining a stateful management session using a connection-oriented transport protocol. When evaluating the transport layer encapsulation of these two flows, which structural characteristic correctly distinguishes the connectionless transport header from the connection-oriented transport header?

Show answer & explanation

Answer: The connectionless header maintains a fixed 8-byte structure consisting strictly of Source Port, Destination Port, Length, and Checksum fields.

Answer

The connectionless transport header (UDP) maintains a minimalist, fixed 8-byte header structure containing only Source Port, Destination Port, Length, and Checksum fields, whereas connection-oriented TCP headers require at least 20 bytes to accommodate sequencing, acknowledgments, window size, and control flags.
The correct answer accurately describes the standard UDP header structure. UDP operates as a lightweight, connectionless protocol with a fixed 8-byte header overhead. It comprises exactly four fields, each 16 bits (2 bytes) in length: Source Port, Destination Port, Length, and Checksum. This minimal structure allows for rapid transmission with minimal latency and process overhead.

Step-by-Step Solution

1
Identify the protocols representing connectionless and connection-oriented transport layer communications.
UDP represents connectionless communications, and TCP represents connection-oriented communications.
Transport layer protocol comparison requires mapping UDP to connectionless datagram delivery and TCP to stateful stream delivery.
2
Analyze the header structure of the UDP (connectionless) transport protocol.
UDP headers contain four 16-bit (2-byte) fields totaling 8 bytes: Source Port (16 bits), Destination Port (16 bits), Length (16 bits), and Checksum (16 bits).
UDP is designed for low overhead without connection setup, state tracking, sequence numbers, or windowing fields.
3
Compare UDP header characteristics with TCP header characteristics to evaluate distractor validity.
TCP headers are 20 bytes minimum (up to 60 bytes with options) and contain Sequence Numbers, Acknowledgment Numbers, Data Offset, Reserved bits, Control Flags, Window Size, Checksum, and Urgent Pointer.
Distractors incorrectly attribute TCP features, IP-layer offloading, or variable header truncation to UDP.

Key Concept

Transport Layer TCP vs UDP Header Structure and Protocol Overhead
Question 216Question →

A network engineer installs VMware Workstation on a laptop running Windows 11 to set up a test environment with multiple guest operating systems. Which architecture describes this virtualization deployment?

Show answer & explanation

Answer: A Type 2 hypervisor running on top of an existing host operating system

Answer

A Type 2 hypervisor running on top of an existing host operating system
VMware Workstation installed on a Windows 11 laptop is a standard example of a Type 2 hypervisor. Type 2 (hosted) hypervisors run as application software on top of a host operating system, which handles memory, CPU, and device management.

Step-by-Step Solution

1
Analyze the installation context described in the scenario
The hypervisor (VMware Workstation) is being installed inside a pre-existing desktop operating system (Windows 11).
Understanding where the hypervisor executes relative to the physical hardware determines its virtualization class.
2
Classify the hypervisor type based on hardware interaction
Hypervisors that run on top of a host operating system are categorized as Type 2 (hosted) hypervisors.
Type 2 hypervisors rely on the host OS to manage hardware interactions and system devices.

Key Concept

Type 2 (Hosted) Hypervisor Architecture
Question 217Question →

A network administrator is evaluating the core infrastructure components for a new branch office deployment, including a router, an unsegmented Layer 2 switch, a Next-Generation Firewall (NGFW), and virtualization servers. Which two statements accurately describe the operational roles and domain boundaries of these network components?

Select all that apply

Show answer & explanation

Answer: A router isolates broadcast domains at each configured Layer 3 interface.; A Next-Generation Firewall performs stateful inspection and application-level control up through Layer 7 of the OSI model.

Answer

A router isolates broadcast domains at each configured Layer 3 interface, and a Next-Generation Firewall performs stateful inspection and application-level control up through Layer 7.
Routers operate at Layer 3 and prevent Layer 2 broadcast traffic from propagating across interfaces, creating distinct broadcast domains. Next-Generation Firewalls expand upon traditional packet filtering by integrating deep packet inspection and application awareness up through Layer 7 of the OSI model.

Step-by-Step Solution

1
Analyze the forwarding domain boundaries of routers versus Layer 2 switches.
Routers operate at Layer 3 and stop Layer 2 broadcasts from passing between interfaces. Layer 2 switches split collision domains per port but maintain a single shared broadcast domain by default.
Understanding domain separation is fundamental to network architecture.
2
Evaluate the functional scope of Next-Generation Firewalls (NGFW).
NGFWs inspect traffic up to Layer 7 (Application layer) to provide deep packet inspection and threat prevention.
Standard firewalls filter traffic up to Layer 4, while NGFWs inspect application payloads.
3
Differentiate hypervisor architecture classifications.
Type 1 hypervisors run directly on bare metal, while Type 2 hypervisors rely on a host operating system.
Proper classification prevents misidentifying virtualization platform architecture.

Key Concept

Network Component Operational Roles and Domain Boundaries
Estimated Time:1m 30s
Question 218Question →

A Cisco Catalyst switch receives an ingress Ethernet frame tagged for VLAN 10 on port GigabitEthernet0/1. The frame contains a source MAC address of 0010.AB12.3456 and a destination MAC address of 0020.CD34.5678. Prior to receiving this frame, the switch MAC address table contains a dynamic entry listing 0010.AB12.3456 on port GigabitEthernet0/2 in VLAN 10, and a static entry listing 0020.CD34.5678 on port GigabitEthernet0/3 in VLAN 10. Port GigabitEthernet0/4 is an active access port configured in VLAN 20, while port GigabitEthernet0/5 is an active 802.1Q trunk port allowing VLANs 10 and 20. Which two actions does the switch perform upon processing this frame?

Select all that apply

Show answer & explanation

Answer: It updates the MAC address table entry for 0010.AB12.3456 to associate it with port GigabitEthernet0/1 in VLAN 10.; It forwards the frame exclusively out port GigabitEthernet0/3.

Answer

The switch updates its MAC address table entry for source MAC 0010.AB12.3456 to port GigabitEthernet0/1 in VLAN 10, and forwards the frame directly out port GigabitEthernet0/3.
When an Ethernet switch receives a frame, it performs learning and forwarding independently. First, it reads the source MAC address 0010.AB12.3456 and updates the MAC address table entry to map to port GigabitEthernet0/1 in VLAN 10. Second, it looks up the destination MAC address 0020.CD34.5678 in the MAC table for VLAN 10. Because a matching entry exists for GigabitEthernet0/3, the switch forwards the frame directly out GigabitEthernet0/3 without flooding.

Step-by-Step Solution

1
Analyze ingress port processing and source MAC learning logic.
The source MAC address 0010.AB12.3456 arrived on GigabitEthernet0/1 in VLAN 10. The switch updates the existing table mapping for 0010.AB12.3456 from GigabitEthernet0/2 to GigabitEthernet0/1.
Layer 2 switches evaluate every incoming frame's source MAC address to build and maintain up-to-date MAC table port mappings.
2
Analyze egress port forwarding logic based on destination MAC lookup.
The destination MAC 0020.CD34.5678 is matched in the MAC table for VLAN 10 pointing to GigabitEthernet0/3.
When a match is found in the MAC table for the frame's destination address in the matching VLAN, known unicast forwarding directs the frame solely to that designated egress interface.

Key Concept

Layer 2 MAC address table dynamic learning, station movement handling, and known unicast forwarding logic.
Question 219Question →

A network engineer is configuring Variable Length Subnet Masking (VLSM) on a router interface for two internal enterprise subnets using the base block 172.16.10.0/24172.16.10.0/24. Subnet A must support at least 60 usable hosts, while Subnet B must support at least 28 usable hosts. Subnet A is allocated first starting at 172.16.10.0172.16.10.0 using the smallest prefix length that satisfies its requirement. Subnet B is allocated immediately following Subnet A using its smallest required prefix length. Which two statements regarding this subnetting layout are correct? (Select two.)

Select all that apply

Show answer & explanation

Answer: The broadcast address for Subnet A is 172.16.10.63172.16.10.63.; The IP address 172.16.10.94172.16.10.94 is a valid, usable host address in Subnet B.

Answer

The broadcast address for Subnet A is 172.16.10.63172.16.10.63, and the IP address 172.16.10.94172.16.10.94 is a valid, usable host address in Subnet B.
For Subnet A, 60 hosts require 6 host bits (26−2=622^6 - 2 = 62 usable), yielding a /26/26 mask. Starting at 172.16.10.0/26172.16.10.0/26, the block ends at broadcast address 172.16.10.63172.16.10.63. Subnet B starts at the next available address, 172.16.10.64172.16.10.64, and requires 5 host bits (25−2=302^5 - 2 = 30 usable) for 28 hosts, giving a /27/27 mask (172.16.10.64/27172.16.10.64/27). The valid usable host range for Subnet B is 172.16.10.65172.16.10.65 through 172.16.10.94172.16.10.94, making 172.16.10.94172.16.10.94 a valid usable host address.

Step-by-Step Solution

1
Calculate prefix length and host range for Subnet A.
Subnet A prefix is /26/26 (255.255.255.192255.255.255.192), spanning 172.16.10.0/26172.16.10.0/26. Network ID: 172.16.10.0172.16.10.0, Usable range: 172.16.10.1172.16.10.1–172.16.10.62172.16.10.62, Broadcast: 172.16.10.63172.16.10.63.
60 host requirement needs 2h−2≥60  ⟹  h=62^h - 2 \ge 60 \implies h=6 bits. 32−6=2632 - 6 = 26.
2
Determine the starting boundary and prefix length for Subnet B.
Subnet B starts at 172.16.10.64172.16.10.64 with prefix /27/27 (255.255.255.224255.255.255.224). Network ID: 172.16.10.64172.16.10.64, Usable range: 172.16.10.65172.16.10.65–172.16.10.94172.16.10.94, Broadcast: 172.16.10.95172.16.10.95.
Next contiguous subnet after 172.16.10.63172.16.10.63 is 172.16.10.64172.16.10.64. 28 host requirement needs 2h−2≥28  ⟹  h=52^h - 2 \ge 28 \implies h=5 bits. 32−5=2732 - 5 = 27.
3
Evaluate the choices against calculated subnet boundaries and RFC 1918 rules.
Identified 172.16.10.63172.16.10.63 as Subnet A's broadcast address and 172.16.10.94172.16.10.94 as Subnet B's last usable host IP.
Network addresses (172.16.10.64172.16.10.64) cannot be assigned to hosts, and 172.32.x.x172.32.x.x falls outside RFC 1918 private space.

Key Concept

VLSM Subnet Sizing and Boundary Calculation
Question 220Question →

A network engineer is configuring an internal WLAN management interface on an enterprise wireless LAN controller. To comply with corporate security policy, the interface must be assigned an IPv4 address from the RFC 1918 private address space so that it is non-routable over the public Internet. Which of the following IPv4 addresses meets this requirement?

Show answer & explanation

Answer: 172.22.100.15

Answer

172.22.100.15 is the only address within the RFC 1918 private IPv4 specification.
The IPv4 address 172.22.100.15 belongs to the 172.16.0.0/12 block (range 172.16.0.0 – 172.31.255.255), which is explicitly reserved by RFC 1918 for private internal networking.

Step-by-Step Solution

1
Recall the defined RFC 1918 private IPv4 address blocks.
Class A: 10.0.0.0 to 10.255.255.255 (10.0.0.0/8); Class B: 172.16.0.0 to 172.31.255.255 (172.16.0.0/12); Class C: 192.168.0.0 to 192.168.255.255 (192.168.0.0/16).
RFC 1918 sets aside specific address ranges for internal private enterprise networks.
2
Compare each option against the RFC 1918 range limits.
172.22.100.15 has a second octet of 22, which is between 16 and 31.
Any address in the 172.16.0.0/12 block is a valid private IPv4 address.

Key Concept

RFC 1918 Private IPv4 Address Ranges
Estimated Time:1m 0s
PreviousPage 11 / 20Next
Network Fundamentals Practice Questions — Cisco CCNA — Page 11 | Examkin