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Networking Concepts

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Question 141Question →

A web server receives an incoming Ethernet frame carrying an HTTPS request over TLS/TCP. Arrange the following de-encapsulation steps performed by the receiving host network stack in the correct chronological sequence from first (bottom of stack) to last (top of stack).

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Answer

The correct sequence of de-encapsulation is: 1) NIC validates FCS and strips Ethernet header/trailer (Layer 2 frame to Layer 3 packet); 2) Network layer verifies destination IP and strips IPv4 header (Layer 3 packet to Layer 4 segment); 3) Transport layer processes port numbers and strips TCP header (Layer 4 segment to upper-layer data); 4) TLS layer decrypts the encrypted payload (Layer 5/6 Presentation layer processing); 5) Web server process parses the HTTP request (Layer 7 Application layer).
De-encapsulation operates strictly bottom-up (Layer 2 frame → Layer 3 packet → Layer 4 segment → Layer 5/6 decrypted session → Layer 7 application data). At each step, a device inspects its specific layer control header, verifies checksums/integrity, strips the header/trailer, and uses higher-layer protocol indicators (EtherType, IP Protocol field, TCP Port) to pass the remaining PDU up to the next layer.

Step-by-Step Solution

1
Identify Data Link Layer (Layer 2) de-encapsulation
Ethernet frame header and trailer (FCS) are stripped by the NIC upon integrity verification, leaving an IP packet.
De-encapsulation always starts at the lowest layer (Layer 1/2) when receiving signals off the physical wire.
2
Identify Network Layer (Layer 3) de-encapsulation
Destination IP address is checked and the IPv4 header is stripped, leaving a TCP segment.
Layer 3 processing evaluates IP headers and uses the Protocol field (TCP=6) to direct the packet payload to the correct Layer 4 protocol engine.
3
Identify Transport Layer (Layer 4) de-encapsulation
TCP destination port (e.g., 443) is evaluated, sequence numbers are reassembled, and the TCP header is stripped.
Layer 4 uses port numbers to multiplex data streams to specific application sockets.
4
Identify Presentation/Session Layer (Layer 5/6) security unwrapping
TLS cryptographic records are decrypted to reveal the raw HTTP request text.
Decryption must occur after Layer 4 TCP segment reassembly completes, but before the HTTP application daemon can interpret the request syntax.
5
Identify Application Layer (Layer 7) execution
The web server application reads the raw HTTP GET request.
The top of the OSI stack processes pure application payload free of all transport, network, framing headers, and encryption layers.

Key Concept

Inbound De-encapsulation Sequence and PDU Unwrapping
Question 142Question →

An enterprise organization with 8 branch locations is migrating its site-to-site connectivity from a legacy dedicated full-mesh physical WAN to a Dynamic Multipoint VPN (DMVPN) deployment featuring redundant dual central hubs over commercial internet links. Which of the following statements accurately describe the topological changes and operational characteristics resulting from this migration? (Select TWO.)

Select all that apply

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Answer: The physical topology shifts to a dual-hub-and-spoke model reducing static WAN connections from 28 to 16, while the logical topology can dynamically establish direct spoke-to-spoke tunnels on demand.; The virtual overlay operates logically as a non-broadcast multi-access (NBMA) network, requiring encapsulation protocols like Multipoint GRE to transport multicast routing protocol traffic across internet transports.

Answer

The physical topology shifts to a dual-hub-and-spoke model reducing static WAN connections from 28 to 16 while supporting dynamic logical spoke-to-spoke tunnels, and the virtual overlay functions as an NBMA network requiring mGRE to carry multicast and broadcast traffic.
Migrating from a static full-mesh WAN to a dual-hub DMVPN architecture changes the physical connection count from 28 static links (calculated via N(N−1)2\frac{N(N-1)}{2} for 8 nodes) down to 16 static hub links (8 sites×2 hubs8 \text{ sites} \times 2 \text{ hubs}). Furthermore, public internet transport functions logically as an NBMA network, requiring encapsulation like Multipoint GRE (mGRE) to handle broadcast and multicast traffic for dynamic routing protocols.

Step-by-Step Solution

1
Calculate physical WAN link requirements for full mesh versus dual hub-and-spoke.
Full mesh for N=8N=8 sites requires 8(8−1)2=28\frac{8(8-1)}{2} = 28 physical links. Dual-hub topology requires each of the 8 sites to connect to 2 hubs, totaling 8×2=168 \times 2 = 16 static hub-spoke links.
Topological transition reduces static provisioning overhead while using dynamic overlay capabilities.
2
Analyze logical overlay capabilities of DMVPN.
DMVPN combines Multipoint GRE (mGRE) and Next Hop Resolution Protocol (NHRP) to dynamically establish temporary spoke-to-spoke tunnels across an NBMA internet transport.
This allows direct spoke communication without static full-mesh tunnel configuration or permanent routing through central hubs.

Key Concept

Physical vs. Logical Topologies and DMVPN Overlay Mechanics
Question 143Question →

Match each IPv6 address type or field name on the left with its corresponding prefix or functional description on the right.

Click a left item, then click its matching right item

Items

Global Unicast Address
Link-Local Address
Multicast Address
Loopback Address

Matches

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Answer

Global Unicast Address matches 2000::/3, Link-Local Address matches fe80::/10, Multicast Address matches ff00::/8, and Loopback Address matches ::1/128.
Global Unicast addresses start with 2000::/3, Link-Local addresses start with fe80::/10, Multicast addresses start with ff00::/8, and the Loopback address is represented by ::1/128.

Step-by-Step Solution

1
Identify the designated prefix block for each IPv6 address scope.
Global Unicast uses 2000::/3, Link-Local uses fe80::/10, Multicast uses ff00::/8, and Loopback uses ::1/128.
IPv6 standard address allocation defines specific prefix ranges to separate unicast, multicast, link-restricted, and host-internal operations.

Key Concept

IPv6 Address Scopes and Prefixes
Question 144Question →

Arrange the following Protocol Data Units (PDUs) in the correct sequential order as data moves down the Open Systems Interconnection (OSI) model during the encapsulation process, starting from the application layer down to the physical transmission medium.

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Answer

The correct encapsulation sequence from top to bottom is: Data → Segment → Packet → Frame → Bits.
During encapsulation, data moves down the OSI model from Layer 7 to Layer 1. The PDU sequence is Data at the upper layers (Application, Presentation, Session), Segment at the Transport layer (Layer 4), Packet at the Network layer (Layer 3), Frame at the Data Link layer (Layer 2), and Bits at the Physical layer (Layer 1).

Step-by-Step Solution

1
Identify the starting point of data creation in the OSI model.
Information originates at the Application, Presentation, and Session layers as raw Data.
Encapsulation begins at the upper layers before being prepared for network transport.
2
Determine the PDU created at the Transport layer (Layer 4).
Data is segmented and wrapped with transport protocol headers to create Segments (or Datagrams).
Layer 4 manages end-to-end communication and port numbers.
3
Determine the PDU created at the Network layer (Layer 3).
Segments are encapsulated with IP headers containing source and destination IP addresses to create Packets.
Layer 3 handles logical addressing and path selection across networks.
4
Determine the PDU created at the Data Link layer (Layer 2).
Packets are encapsulated with MAC headers and frame check sequence trailers to create Frames.
Layer 2 facilitates node-to-node transfer on the local network segment.
5
Determine the final state at the Physical layer (Layer 1).
Frames are encoded into raw Bits for transmission across physical medium.
Layer 1 transmits binary signals over copper cables, fiber optics, or wireless mediums.

Key Concept

OSI Model Data Encapsulation and PDU Sequence
Question 145Question →

Match each switching concept on the left with its correct operational description on the right.

Click a left item, then click its matching right item

Items

802.1Q Tagging
Native VLAN
Access Port
Trunk Port

Matches

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Answer

802.1Q Tagging matches the standard 4-byte header insertion protocol; Native VLAN matches handling untagged traffic on trunk links; Access Port matches single-VLAN host connection mode; Trunk Port matches multi-VLAN inter-switch transport mode.
Each switching term correctly maps to its foundational Layer 2 behavior: 802.1Q tagging defines the 4-byte frame header, Native VLAN carries untagged frames over trunk links, Access ports assign an interface to a single VLAN for endpoints, and Trunk ports carry traffic across multiple VLANs between switches.

Step-by-Step Solution

1
Identify the protocol used for VLAN encapsulation across trunk links.
IEEE 802.1Q inserts a 4-byte field into the Ethernet frame header containing the VLAN ID.
802.1Q is the industry standard framing mechanism for preserving VLAN identity across switches.
2
Determine how untagged frames are processed on trunk interfaces.
Frames without an 802.1Q tag header passing over a trunk link are automatically assigned to the Native VLAN.
The Native VLAN provides backward compatibility for untagged legacy traffic and control protocols.
3
Distinguish between Access Port and Trunk Port operations.
Access ports strip tags and belong strictly to a single VLAN for end hosts, whereas Trunk ports retain tags to transport multiple VLANs simultaneously.
End-user devices do not inspect 802.1Q tags, so access ports deliver untagged frames to hosts while trunk ports multiplex tagged traffic between networking devices.

Key Concept

Switch Port Roles and IEEE 802.1Q VLAN Tagging
Question 146Question →

A network administrator connects two workstations, Host A and Host B, to access ports on the same physical Layer 2 switch. Host A is configured on VLAN 10 (192.168.10.0/24), while Host B is configured on VLAN 20 (192.168.20.0/24). Host A transmits an ARP broadcast frame to determine an IP address on its local network segment. How will the switch handle this broadcast frame?

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Answer: The switch floods the broadcast frame out all ports assigned to VLAN 10 except the receiving port, preventing Host B on VLAN 20 from receiving it.

Answer

The switch floods the broadcast frame out all ports assigned to VLAN 10 except the receiving port, preventing Host B on VLAN 20 from receiving it.
A Virtual Local Area Network (VLAN) creates a logical boundary at Layer 2. Broadcast frames transmitted by a device are flooded only to interfaces within the exact same VLAN membership (excluding the receiving interface). Consequently, Host B on VLAN 20 is shielded from VLAN 10 broadcast traffic.

Step-by-Step Solution

1
Identify the frame type and ingress interface configuration.
The incoming frame is an Ethernet broadcast frame (destination MAC FF:FF:FF:FF:FF:FF) originating from a host assigned to VLAN 10.
VLAN memberships assigned to switch interfaces define the boundary for Layer 2 broadcast containment.
2
Determine Layer 2 switch forwarding behavior for broadcast traffic within a VLAN.
The switch replicates the broadcast frame to all operational interfaces that belong to VLAN 10, excluding the port where the frame entered.
VLANs act as independent logical switches; frames in one VLAN are isolated from interfaces assigned to other VLANs.
3
Evaluate the impact on hosts in different VLANs.
Host B, residing on an access port configured for VLAN 20, does not receive the broadcast frame.
Without a Layer 3 routing mechanism (such as a router or Layer 3 switch), traffic cannot cross VLAN boundaries.

Key Concept

VLAN Broadcast Domain Isolation
Estimated Time:1m 15s
Question 147Question →

A network technician needs to separate broadcast traffic between two departments connected to the same physical switch. Which technology should the technician implement on the switch to isolate the broadcast domains?

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Answer: Virtual Local Area Networks (VLANs)

Answer

Virtual Local Area Networks (VLANs) logically segment a physical Layer 2 switch into distinct broadcast domains.
Virtual Local Area Networks (VLANs) allow network administrators to partition a single physical switch into multiple logical broadcast domains, keeping broadcast traffic isolated within each designated group of ports.

Step-by-Step Solution

1
Analyze the network requirement
The requirement calls for isolating broadcast traffic between different host groups connected to the same physical switch.
By default, all ports on an unconfigured Layer 2 switch belong to a single shared broadcast domain.
2
Identify the appropriate Layer 2 feature
Configuring Virtual Local Area Networks (VLANs) groups switch ports into logical sub-networks.
VLANs restrict broadcast frames to only those switch ports assigned to the same VLAN ID.

Key Concept

VLAN Broadcast Domain Isolation
Estimated Time:45s
Question 148Question →

A network engineer is allocated the IPv4 address block 10.240.16.0/2010.240.16.0/20 to create subnets for distinct department segments. Each segment requires a minimum of 120120 usable host IP addresses. What is the maximum number of subnets of this minimum required size that can be created from the given block?

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Answer: 32

Answer

The maximum number of subnets of the minimum required size that can be created is 32.
To support a minimum of 120 hosts, 7 host bits are required (27−2=126≥1202^7 - 2 = 126 \ge 120). A subnet with 7 host bits has a prefix length of /25/25 (32−7=2532 - 7 = 25). Partitioning a /20/20 block into /25/25 subnets borrows 5 bits (25−20=525 - 20 = 5), resulting in 25=322^5 = 32 subnets.

Step-by-Step Solution

1
Calculate the host bits required for the minimum host capacity requirement
7 host bits are required because 27−2=126≥1202^7 - 2 = 126 \ge 120.
Using 6 host bits only provides 26−2=622^6 - 2 = 62 usable addresses, which fails to meet the 120 host requirement.
2
Calculate the required subnet mask prefix length
The prefix length is /25/25, calculated as 32−7=2532 - 7 = 25.
Subtracting host bits from the 32 total IPv4 address bits determines the network prefix.
3
Calculate the total subnets formed by splitting a /20/20 block into /25/25 subnets
225−20=25=322^{25 - 20} = 2^5 = 32 subnets.
The number of subnets created is 2 raised to the power of the borrowed subnet bits.

Key Concept

IPv4 Subnetting and Host Capacity Calculation
Question 149Question →

A network engineer is configuring a core router that receives routes to the destination network 10.100.20.0/2410.100.20.0/24 from two distinct dynamic routing sources simultaneously:

- An OSPF advertisement with a calculated path metric of 4545
- An Internal EIGRP (iEIGRP) advertisement with a calculated path metric of 2,560,0002,560,000

Which path will the router install into its active IPv4 routing table for the 10.100.20.0/2410.100.20.0/24 network, and what is the primary reason for this selection?

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Answer: The Internal EIGRP path, because Internal EIGRP has a lower administrative distance (9090) than OSPF (110110).

Answer

The router installs the Internal EIGRP path because Internal EIGRP has a lower Administrative Distance (90) than OSPF (110). Metric values are protocol-specific and only evaluated after administrative distance comparison.
When a router receives multiple routes for the exact same network prefix from different routing protocols, it compares their Administrative Distance (AD). Internal EIGRP has a lower default AD (90) than OSPF (110). Because lower AD values are preferred, the router selects and installs the Internal EIGRP route into the routing table. Metric values are never directly compared between different dynamic routing protocols.

Step-by-Step Solution

1
Identify the destination prefix and routing sources
Both routing protocols offer a route to the exact same prefix (10.100.20.0/2410.100.20.0/24).
When prefix lengths are equal, the router compares Administrative Distance (AD) to select the trustworthy routing source.
2
Compare Administrative Distance (AD) values of the competing protocols
Internal EIGRP has a default AD of 9090, while OSPF has a default AD of 110110.
Lower AD values indicate higher trustworthiness. Since 90<11090 < 110, Internal EIGRP is selected.
3
Determine why metric comparison is bypassed
Metric values (4545 vs 2,560,0002,560,000) are ignored because metrics cannot be compared across different routing protocols.
Different dynamic routing protocols use incompatible metric formulas (e.g., cost based on bandwidth vs. composite metric based on bandwidth and delay).

Key Concept

Administrative Distance vs. Routing Metric in Path Selection
Question 150Question →

Match each network scope classification on the left with its correct geographical footprint and primary application on the right.

Click a left item, then click its matching right item

Items

Personal Area Network (PAN)
Storage Area Network (SAN)
Metropolitan Area Network (MAN)
Campus Area Network (CAN)

Matches

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Answer

Personal Area Network (PAN) matches short-range personal device connections around a user; Storage Area Network (SAN) matches dedicated high-speed block storage access for data center servers; Metropolitan Area Network (MAN) matches city-wide municipal coverage; Campus Area Network (CAN) matches interconnecting LANs across adjacent buildings on a shared site.
Each network type matches its unique operational scale: PANs handle personal device range (<10m), SANs handle dedicated block storage for servers, MANs cover municipal-scale cities, and CANs connect multi-building private properties like universities.

Step-by-Step Solution

1
Determine the scale and purpose of Personal Area Network (PAN).
PAN covers small personal spaces (up to 10 meters) for peripheral pairing.
PAN technology (such as Bluetooth) is designed for short-range personal device communication.
2
Determine the scale and purpose of Storage Area Network (SAN).
SAN is a specialized data center network providing block-level access to disk arrays.
SAN offloads storage traffic from the main user LAN onto a high-performance network fabric.
3
Determine the scale and purpose of Metropolitan Area Network (MAN).
MAN spans a city-sized geographical region.
MAN connects disparate facilities situated across a municipality.
4
Determine the scale and purpose of Campus Area Network (CAN).
CAN links multiple local area networks across contiguous buildings on a shared campus.
CAN is larger than a standard LAN but smaller than a city-wide MAN, typically owned and managed by a single institution.

Key Concept

Classification of network types based on geographic coverage and architectural function.
Estimated Time:1m 0s
Question 151Question →

A network technician is using a packet capture utility to inspect outbound host traffic during an HTTPS session initiation. Arrange the following data encapsulation steps in the correct chronological sequence from the initial application request generation down to media transmission.

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Answer

The correct order follows the top-down OSI encapsulation model: Application data formatting (Upper layers), TCP segment creation with port numbers (Transport layer), IP packet creation with logical addresses (Network layer), Ethernet frame creation with MAC addresses (Data Link layer), and bit signal conversion (Physical layer).
During outbound network transmission, data is processed sequentially from Layer 7 (Application) down to Layer 1 (Physical). Application data is first generated, then encapsulated into a Transport layer segment (TCP/UDP ports), further encapsulated into a Network layer packet (IP addresses), wrapped into a Data Link layer frame (MAC addresses and FCS trailer), and finally converted into physical signals (bits) at the Physical layer.

Step-by-Step Solution

1
Identify the starting point of outbound traffic flow
Outbound transmission begins at the highest layers with data payload creation.
The client application creates the original request message before network headers are appended.
2
Determine the Layer 4 Transport encapsulation
TCP port header is added to form a segment.
Transport protocols (Layer 4) manage host-to-host session multiplexing using port numbers.
3
Determine the Layer 3 Network encapsulation
IP header is added to form a packet.
Network layer (Layer 3) handles logical host identification and path determination across network boundaries.
4
Determine the Layer 2 Data Link encapsulation
Ethernet MAC header and FCS trailer are added to form a frame.
Data Link layer (Layer 2) prepares data for physical link traversal by assigning physical hardware addresses.
5
Determine the Layer 1 Physical transmission step
Digital data is serialized and converted to signal pulses (bits).
Physical layer (Layer 1) deals directly with signal encoding across physical media.

Key Concept

OSI Model Encapsulation Sequence
Estimated Time:1m 30s
Question 152Question →

An enterprise network administrator is troubleshooting directory service integration between an internal web proxy and an Active Directory Domain Controller. Authentication using implicit SSL encryption (LDAPS) over TCP port 636 functions correctly. However, automated directory synchronization tasks configured on the proxy to use explicit TLS encryption (STARTTLS) fail to establish a secure connection. Network packet captures confirm that the proxy initiates communication over the default unencrypted LDAP port before issuing the STARTTLS upgrade command, but traffic is blocked by an intermediate firewall ACL. Which of the following transport protocols and destination port combinations must be permitted on the firewall to allow STARTTLS directory synchronization to succeed?

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Answer: TCP port 389

Answer

TCP port 389
Explicit TLS (STARTTLS) allows a client to establish an initial connection over the standard unencrypted port (TCP port 389 for LDAP) and issue a command to upgrade the active channel to TLS encryption. For STARTTLS negotiation to succeed, the firewall ACL must permit TCP port 389.

Step-by-Step Solution

1
Analyze the protocol handshake mechanism for explicit TLS (STARTTLS).
Recognize that STARTTLS is an explicit security extension that begins session negotiation over the default unencrypted service port before initiating a TLS upgrade.
Explicit TLS requires initial access to the unencrypted service port to send the protocol upgrade request.
2
Identify the standard port number associated with unencrypted directory services.
Determine that standard LDAP operates on port 389.
Lightweight Directory Access Protocol (LDAP) uses port 389 as its standard listener port.
3
Evaluate transport layer requirements for TLS session negotiation.
Select TCP over UDP for session state establishment.
TLS state handshakes and data integrity enforcement require the connection-oriented reliability offered by TCP.

Key Concept

Explicit TLS (STARTTLS) vs Implicit TLS Port Behavior
Question 153Question →

A network engineer is configuring a static route on a core router using the uncompressed IPv6 next-hop address 2001:0db8:0000:0000:0008:0000:0000:0001. According to RFC 5952 IPv6 addressing standards, which of the following represents the correct compressed form of this address?

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Answer: 2001:db8::8:0:0:1

Answer

The address 2001:db8::8:0:0:1 is the correct compressed form.
The correct representation, 2001:db8::8:0:0:1, suppresses leading zeros in each hexadecimal word and compresses the first sequence of contiguous zero blocks using ::. Per RFC 5952, when two zero sequences have the same length, the first sequence must be compressed.

Step-by-Step Solution

1
Suppress leading zeros in each 16-bit hexadecimal field.
2001:0db8 becomes 2001:db8, 0008 becomes 8, and 0001 becomes 1.
Leading zeros within any 4-digit hexadecimal group are optional and omitted for brevity.
2
Identify contiguous blocks of all-zero fields.
There are two separate sequences of two consecutive zero blocks: blocks 3-4 and blocks 6-7.
IPv6 allows replacing contiguous zero blocks with a double-colon (::).
3
Apply the double-colon (::) according to RFC 5952 rules.
The first sequence of two zero blocks is replaced by ::, yielding 2001:db8::8:0:0:1.
The double-colon can only be used once per address. When zero sequences are of equal length, standard conventions require compressing the first occurrence.

Key Concept

IPv6 Address Compression Rules (RFC 5952)
Estimated Time:1m 0s
Question 154Question →

A network security analyst is monitoring how an inbound HTTPS session is processed by a host system during decapsulation. Arrange the following data processing actions in the correct sequential order as the received data moves up the OSI reference model, starting from Layer 1 (Physical) to Layer 7 (Application).

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Answer

The correct sequence follows the bottom-up decapsulation process of the OSI model: Physical (Layer 1) → Data Link (Layer 2) → Network (Layer 3) → Transport (Layer 4) → Presentation (Layer 6) → Application (Layer 7).
Decapsulation proceeds from Layer 1 up to Layer 7. Incoming physical signals are first converted to bits (Physical / L1), framed and verified using MAC/FCS (Data Link / L2), processed for IP routing and TTL (Network / L3), reassembled by TCP sequence and port numbers (Transport / L4), decrypted from TLS wrappers (Presentation / L6), and finally processed as an HTTP payload by the application (Application / L7).

Step-by-Step Solution

1
Identify Layer 1 operation
Converting physical signals (electrical pulses) into binary bits occurs at the Physical layer (Layer 1).
Layer 1 is responsible for the physical interface, encoding, and raw bit transmission.
2
Identify Layer 2 operation
Checking the Frame Check Sequence (FCS) and verifying the MAC address occurs at the Data Link layer (Layer 2).
Layer 2 encapsulates network packets into frames containing hardware MAC addresses and CRC error checking.
3
Identify Layer 3 operation
Inspecting IP headers, checking TTL, and handling logical address routing occurs at the Network layer (Layer 3).
Layer 3 operates on packets and processes IP addresses and routing fields.
4
Identify Layer 4 operation
TCP segment reassembly using sequence numbers and reading port numbers occurs at the Transport layer (Layer 4).
Layer 4 provides flow control, connection management, port multiplexing, and reliable segment reassembly.
5
Identify Layer 6 operation
Decrypting TLS/SSL wrappers and formatting data occurs at the Presentation layer (Layer 6).
Layer 6 manages syntax formatting, encryption/decryption, and data compression.
6
Identify Layer 7 operation
Processing the HTTP GET request by the web application service occurs at the Application layer (Layer 7).
Layer 7 directly interfaces with end-user software applications and protocol payloads.

Key Concept

OSI Model Layer Decapsulation Order and Layer Functions
Question 155Question →

Match each IPv6 base header field on the left with its correct functional description on the right.

Click a left item, then click its matching right item

Items

Traffic Class
Flow Label
Next Header
Hop Limit

Matches

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Answer

Traffic Class matches with identifying packet priority and QoS markings; Flow Label matches with maintaining the same path for a specific sequence of packets; Next Header matches with specifying upper-layer protocol or extension header; Hop Limit matches with decrementing at each hop to prevent loops.
Each fundamental IPv6 header field maps to a specific purpose: Traffic Class provides QoS markings; Flow Label maintains packet flow path consistency; Next Header identifies encapsulated protocols or extension headers; Hop Limit prevents network routing loops by decrementing per hop.

Step-by-Step Solution

1
Identify the primary role of the Traffic Class header field.
Determine that Traffic Class manages packet priority and QoS, mirroring IPv4 Differentiated Services.
Traffic Class handles packet prioritization in IPv6 header structures.
2
Identify the primary role of the Flow Label header field.
Determine that Flow Label tags specific packet streams for consistent router path delivery.
Flow Label facilitates real-time data flow handling across intermediate routers.
3
Identify the primary role of the Next Header field.
Determine that Next Header specifies payload protocols or links to optional extension headers.
Next Header replaces IPv4's Protocol field and enables modular extension header chaining.
4
Identify the primary role of the Hop Limit field.
Determine that Hop Limit decrements per layer 3 hop to mitigate routing loops.
Hop Limit replaces IPv4 TTL while serving the exact same loop-prevention mechanism.

Key Concept

IPv6 Base Header Fields and Functions
Question 156Question →

An enterprise architecture team is redesigning the physical WAN connections between 55 regional facilities. Currently, all 55 sites are interconnected using a fully meshed topology of dedicated point-to-point links to guarantee fault tolerance. To reduce recurring leased-line costs, management requests transitioning to a partial-mesh design that decommissions as many links as possible while guaranteeing that every site still retains at least two physically independent transmission paths to reach any other site. What is the maximum number of point-to-point links that can be decommissioned from the original network design?

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Answer: 5 links

Answer

5 links can be decommissioned from the original full-mesh deployment.
For a network of 55 sites (n=5n=5), a full-mesh topology requires n(n−1)2=5×42=10\frac{n(n-1)}{2} = \frac{5 \times 4}{2} = 10 links. To guarantee that every site maintains at least two independent pathways to every other site, the minimum viable partial-mesh layout is a ring topology requiring 55 links. Subtracting the 55 required links from the original 1010 links allows maximum decommissioning of 55 links.

Step-by-Step Solution

1
Calculate total point-to-point links in the existing full-mesh topology
Using the formula L=n(n−1)2L = \frac{n(n-1)}{2} where n=5n = 5: L=5×42=10L = \frac{5 \times 4}{2} = 10 physical links.
Full mesh requires a direct point-to-point cable run between every unique pair of nodes.
2
Determine the minimum number of links needed to satisfy the dual-path redundancy constraint
A minimum of 55 links configured in a closed ring (partial mesh) topology.
In a ring topology with n=5n = 5 nodes, every node has a degree of 2 (connected to two neighbors), providing two distinct physical paths (clockwise and counter-clockwise) to reach any destination node.
3
Subtract the required partial-mesh link count from the original full-mesh link count
10 links (full mesh)−5 links (ring partial mesh)=5 links decommissioned10 \text{ links (full mesh)} - 5 \text{ links (ring partial mesh)} = 5 \text{ links decommissioned}.
This yields the maximum number of links that can be removed while preserving fault tolerance constraints.

Key Concept

Full Mesh vs Partial Mesh Link Calculations
Question 157Question →

A network administrator is manually configuring a static global unicast IPv6 address on a core server interface. The administrator enters the address string `2001:db8:100:0::45::1/64` into the operating system interface configuration file. Upon applying the configuration, the system rejects the input with a syntax error. Which of the following technical explanations identifies why this IPv6 address representation is invalid?

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Answer: The double-colon shorthand notation (`::`) is included twice in a single address string, which prevents the parser from unambiguously calculating the total number of 16-bit zero blocks represented by each occurrence.

Answer

The double-colon shorthand notation (`::`) is included twice in a single address string, which prevents the parser from unambiguously calculating the total number of 16-bit zero blocks represented by each occurrence.
The double-colon (`::`) zero compression rule allows shortening contiguous 16-bit fields of zeros. However, using `::` more than once in a single address string makes it mathematically impossible to determine how many 16-bit zero blocks each `::` replaces, leading to syntax rejection by network operating systems.

Step-by-Step Solution

1
Analyze standard IPv6 address structure rules.
An IPv6 address comprises eight 16-bit hexadecimal fields (128 bits total), separated by colons.
To shorten long addresses, contiguous blocks of zero-value fields can be compressed using a double-colon (`::`).
2
Evaluate the string `2001:db8:100:0::45::1` for shorthand compliance.
The string contains two separate instances of `::`.
Multiple `::` occurrences lead to mathematical ambiguity regarding how many 16-bit zero blocks are assigned to the first versus second double-colon.
3
Determine the required syntax resolution.
The double-colon must be used only once (e.g., `2001:db8:100:0:0:45::1` or `2001:db8:100:0::45:0:1`).
This guarantees that an automated parser can reliably pad the missing zero fields to reconstruct the exact 128-bit binary address.

Key Concept

IPv6 Address Zero Compression Rules
Question 158Question →

A network administrator is configuring stateless inbound and outbound access control lists (ACLs) on an enterprise edge firewall connecting a remote branch to the corporate datacenter. Corporate security policy mandates that all administrative remote access and directory authentication services must be fully encrypted in transit, strictly prohibiting cleartext credentials or unencrypted management protocols. During post-deployment testing, packet captures show that administrative CLI management sessions and secure directory queries are being dropped by the firewall. Which of the following transport layer protocol and destination port combinations MUST be explicitly allowed in the firewall rules to restore these required secure services? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: TCP port 22; TCP port 636

Answer

The firewall must permit TCP port 22 for secure SSH terminal access and TCP port 636 for encrypted LDAPS directory service authentication.
The correct answers represent the standard transport protocol and port assignments for secure administrative access and encrypted directory services. Secure Shell (SSH) uses TCP port 22 to establish encrypted terminal sessions. Lightweight Directory Access Protocol Secure (LDAPS) uses TCP port 636 to encapsulate directory service communication within an SSL/TLS tunnel, fulfilling the corporate requirement for encrypted transit.

Step-by-Step Solution

1
Identify the protocol requirements from the scenario constraints.
The scenario requires encrypted remote CLI management and encrypted directory service authentication.
Security policy explicitly forbids cleartext protocols such as Telnet or standard LDAP.
2
Determine the transport protocol and default port number for secure remote CLI management.
SSH (Secure Shell) provides encrypted CLI administration over TCP port 22.
SSH requires connection-oriented TCP to maintain reliable session control and stream encryption.
3
Determine the transport protocol and default port number for encrypted directory authentication.
LDAPS (LDAP over TLS/SSL) operates over TCP port 636.
Standard LDAP over TCP port 389 sends credentials in cleartext; TLS wrapper on port 636 encrypts the entire LDAP session payload.

Key Concept

Identification of default port numbers and transport layer choices for secure management and directory protocols (SSH on TCP 22 vs LDAPS on TCP 636).
Question 159Question →

An network administrator is deploying Virtual Local Area Networks (VLANs) on a newly installed Layer 2 enterprise switch. Which of the following statements accurately describe key operational characteristics of VLANs? (Select TWO).

Select all that apply

Show answer & explanation

Answer: VLANs divide a single physical switch into multiple separate broadcast domains.; Hosts assigned to different VLANs on the same physical switch require a Layer 3 routing function to communicate.

Answer

VLANs divide a single physical switch into multiple separate broadcast domains, and hosts assigned to different VLANs on the same switch require a Layer 3 routing function to communicate.
VLANs logically partition a physical Layer 2 switch into independent broadcast domains. Because each VLAN is isolated at Layer 2, any communication between hosts located on different VLANs requires inter-VLAN routing performed by a Layer 3 device.

Step-by-Step Solution

1
Analyze Layer 2 segmentation properties of VLANs
Creating VLANs breaks a physical switch into distinct logical segments, where each segment acts as its own independent broadcast domain.
Broadcast traffic generated on one VLAN is restricted solely to ports assigned to that specific VLAN.
2
Determine requirements for inter-VLAN communication
Traffic destined for a host in another VLAN must pass through a Layer 3 device (such as a router or Layer 3 switch).
Layer 2 switches do not forward frames between different broadcast domains without IP routing capabilities.

Key Concept

VLAN Broadcast Isolation and Inter-VLAN Routing Requirements
Estimated Time:45s
Question 160Question →

A network administrator is mapping network application requirements to transport layer protocol features. Match each operational requirement or characteristic on the left with its corresponding protocol implementation detail on the right.

Click a left item, then click its matching right item

Items

Three-way handshake establishment
Low-latency transmission without delivery guarantees
Dynamic flow control via sliding windowing
Stateless subnet-wide broadcast support

Matches

Show answer & explanation

Answer

Three-way handshake establishment matches establishing a session using SYN, SYN-ACK, and ACK flags. Low-latency transmission matches using connectionless UDP datagrams with minimal header overhead. Dynamic flow control matches adjusting transmission rates via sliding windowing. Stateless broadcast support matches leveraging connectionless transport to distribute datagrams without tracking.
TCP is connection-oriented and utilizes sequence numbers, three-way handshakes (SYN, SYN-ACK, ACK), and sliding windows for reliability and flow control. UDP is connectionless, prioritizing low latency and low overhead while supporting broadcast modes.

Step-by-Step Solution

1
Analyze connection-oriented protocol characteristics.
Identified TCP mechanisms: SYN/SYN-ACK/ACK handshake for session initialization and sliding windowing for flow control.
TCP maintains session state and manages sender rate dynamically to guarantee reliable, ordered data delivery.
2
Analyze connectionless protocol characteristics.
Identified UDP mechanisms: lightweight 8-byte headers for time-sensitive traffic and broadcast support without connection establishment.
UDP lacks handshakes, state tracking, and flow control mechanisms, making it ideal for low-overhead or one-to-many communication.
3
Map left-side requirements to right-side implementations.
Paired left_1 to right_1, left_2 to right_2, left_3 to right_3, and left_4 to right_4.
Each feature corresponds precisely to the unique operational mechanisms of TCP or UDP.

Key Concept

Connection-Oriented (TCP) vs Connectionless (UDP) Protocol Operational Traits
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