Networking Concepts

538 questions

All types538 Multiple choice279 Multiple select117 Numeric19 Matching95 Ordering28

An enterprise edge router receives advertisements for the destination prefix $10.20.0.0/16$ from three distinct routing sources: an eBGP session (Administrative Distance $20$ ), an OSPF area (Administrative Distance $110$ ), and a floating static route configured with an Administrative Distance of $130$ . Currently, the router installs the eBGP route into its routing table. The network operations team requires all traffic bound for $10.20.0.0/16$ to prefer the internal OSPF path under normal operating conditions, while maintaining the eBGP path as an automatic secondary backup if OSPF fails. Which configuration change directly achieves this objective without modifying network subnets or prefixes?

Increase the OSPF Administrative Distance for the prefix to a value greater than 130.

Lower the OSPF cost metric for the

10.20.0.0/16

route to 0.

Increase the eBGP Administrative Distance for the route above 110 but below 130.

Change the transport layer port number used by OSPF to match BGP port 179.

Show answer & explanation

Answer: Increase the eBGP Administrative Distance for the route above 110 but below 130.

Answer

Increase the eBGP Administrative Distance for the route to a value greater than OSPF (110) but less than the floating static route (130).

When a router learns the exact same destination prefix through multiple routing sources, it evaluates Administrative Distance (AD) first. Lower AD values indicate higher trust. By increasing the eBGP AD for the route to a value between 111 and 129, OSPF (AD 110) becomes the primary active route in the routing table, while eBGP (e.g., AD 120) remains available as the secondary backup before the floating static route (AD 130).

Step-by-Step Solution

Analyze how routers select paths among multiple routing sources.

When a router receives identical prefix routes (

10.20.0.0/16

) from different protocols, it compares their Administrative Distance (AD). The route with the lowest AD is installed in the Routing Information Base (RIB).

Administrative Distance indicates the trustworthiness of a routing source.

Evaluate the current AD hierarchy.

Current order: eBGP (

20

) < OSPF (

110

) < Floating Static (

130

). Hence, eBGP is active.

Lower AD values take precedence over higher AD values.

Adjust AD values to achieve the desired failover hierarchy (Primary: OSPF, Secondary: eBGP, Tertiary: Floating Static).

Increasing eBGP's AD to a value between

111

and

129

(e.g.,

120

) causes OSPF (

110

) to become the primary route, followed by eBGP (

120

), and lastly the floating static route (

130

This establishes the exact failover ordering specified in the requirements without altering prefix lengths.

Key Concept

Administrative Distance Hierarchy and Multi-Protocol Path Selection

Question 342Question →

While inspecting a packet capture file on a network monitoring station, a system administrator isolates an outbound HTTPS request generated by a client host. Which of the following correctly describes the encapsulation sequence and resulting Protocol Data Units (PDUs) as the application data travels down the protocol stack from Layer 4 to Layer 2?

The Layer 4 protocol appends a TCP header to form a Segment, which is encapsulated by Layer 3 into an IP Packet, and finally encapsulated by Layer 2 with a header and trailer to form a Frame.

The Layer 4 protocol appends an IP header to form a Packet, which is encapsulated by Layer 3 into a TCP Segment, and finally encapsulated by Layer 2 to form a Frame.

The Layer 4 protocol appends a TCP header to form a Frame, which is encapsulated by Layer 3 into a Packet, and finally encapsulated by Layer 2 to form a Segment.

The Layer 4 protocol appends a UDP header to form a Packet, which is encapsulated by Layer 2 into an Ethernet Frame, bypassing Layer 3 IP encapsulation entirely.

Show answer & explanation

Answer: The Layer 4 protocol appends a TCP header to form a Segment, which is encapsulated by Layer 3 into an IP Packet, and finally encapsulated by Layer 2 with a header and trailer to form a Frame.

Answer

The correct sequence begins with the Transport layer (Layer 4) appending a TCP header to the application payload to create a Segment. Next, the Network layer (Layer 3) adds an IP header to create a Packet. Finally, the Data Link layer (Layer 2) wraps the packet with an Ethernet header and frame check sequence trailer to create a Frame.

During outbound network transmission, data flows downward through the protocol stack. Layer 4 (Transport) appends a TCP header to application data to produce a Segment. Layer 3 (Network) wraps the Segment inside an IP header to form a Packet. Layer 2 (Data Link) wraps the Packet with an Ethernet header and trailer to form a Frame ready for physical transmission.

Step-by-Step Solution

Identify Layer 4 (Transport Layer) processing

Application data receives a TCP header containing source and destination port numbers, creating a Transport layer Protocol Data Unit (PDU) known as a Segment.

HTTPS traffic utilizes TCP at Layer 4 to ensure reliable, connection-oriented delivery.

Identify Layer 3 (Network Layer) processing

The entire TCP Segment becomes the payload for Layer 3, where an IP header containing source and destination IP addresses is attached, creating a Packet.

Layer 3 encapsulation is required to enable logical routing across subnets.

Identify Layer 2 (Data Link Layer) processing

The IP Packet is encapsulated inside a Layer 2 header (containing source/destination MAC addresses) and a trailer (containing a Frame Check Sequence), creating a Frame.

Layer 2 framing prepares data for physical transmission across local network media.

Key Concept

Data Encapsulation and PDU Hierarchy

Estimated Time:1m 0s

Question 343Question →

A network engineer is designing a core infrastructure interconnecting five datacenter switches. The design must provide maximum fault tolerance and link redundancy without relying on an intermediate core hub. Which TWO of the following statements correctly describe the physical requirements and operational characteristics of implementing a full-mesh topology in this scenario?

Select all that apply

Connecting all five core switches in this topology requires a total of

10

dedicated physical point-to-point links.

Every switch maintains a direct physical connection to every other switch, eliminating single points of failure between core nodes.

Establishing physical mesh connections between switches automatically eliminates the need for Layer 3 routing by merging all connected switches into a single broadcast domain.

Adding redundant mesh links automatically doubles interface throughput by forcing all ports into full-duplex mode without administrative configuration.

Show answer & explanation

Answer: Connecting all five core switches in this topology requires a total of $10$ dedicated physical point-to-point links.; Every switch maintains a direct physical connection to every other switch, eliminating single points of failure between core nodes.

Answer

Connecting all five core switches requires a total of 10 dedicated physical links, and every switch maintains a direct point-to-point connection to every other switch.

In a full-mesh topology consisting of 5 switches, the link calculation formula

\frac{n(n-1)}{2}

specifies exactly 10 dedicated physical links. Additionally, the defining characteristic of a full-mesh topology is that every node has a direct connection to every other node, providing complete path redundancy.

Step-by-Step Solution

Calculate the required physical link count for a 5-node full-mesh topology

Using

n = 5

\frac{n(n-1)}{2}

yields

\frac{5 \times 4}{2} = 10

links.

A full-mesh topology requires a dedicated connection between every unique pair of devices.

Evaluate the architectural redundancy of full mesh

Direct connections exist between all nodes.

This guarantees that no individual switch or cable failure will isolate any remaining device.

Identify misconceptions regarding Layer 3 boundaries and duplex modes

Physical mesh cabling does not replace Layer 3 routing nor automatically reconfigure port duplex settings.

Physical layout dictates path availability, whereas broadcast domain boundaries and port configurations operate at Layer 2/3 and interface configuration levels.

Key Concept

Full-Mesh Network Topology Link Calculations and Redundancy Characteristics

Question 344Question →

A network administrator is establishing stateless inbound Access Control List (ACL) rules on an internal gateway router to allow remote network infrastructure monitoring and log aggregation. The management policy mandates that managed switches send encrypted log events via Secure Syslog (Syslog-TLS) to a central server while concurrently synchronizing system clocks with a dedicated internal time server. Which combination of transport layer protocols and destination port numbers must be permitted on the router to support both Secure Syslog and Network Time Protocol (NTP) traffic?

TCP port 6514 for Secure Syslog and UDP port 123 for NTP

UDP port 514 for Secure Syslog and TCP port 123 for NTP

TCP port 514 for Secure Syslog and UDP port 512 for NTP

IP protocol 514 for Secure Syslog and IP protocol 123 for NTP at Layer 3

Show answer & explanation

Answer: TCP port 6514 for Secure Syslog and UDP port 123 for NTP

Answer

The correct rule configuration permits TCP port 6514 for Secure Syslog and UDP port 123 for NTP.

Secure Syslog (RFC 5425) specifies TCP port 6514 for transport layer security (TLS) log transmission to ensure confidentiality and reliable delivery. Network Time Protocol (NTP) utilizes UDP port 123 to handle low-latency timestamp exchanges across network devices.

Step-by-Step Solution

Identify the transport protocol and port number required for Secure Syslog (Syslog-TLS).

Standard unencrypted Syslog uses UDP port 514, while Secure Syslog over TLS mandates TCP port 6514 to establish encrypted transport sessions.

Encryption and session reliability require connection-oriented TCP at Layer 4.

Identify the transport protocol and port number required for Network Time Protocol (NTP).

NTP operates exclusively over UDP port 123.

Time synchronization requires minimal latency overhead provided by connectionless UDP.

Synthesize the ACL requirements to combine both port definitions.

The firewall rule must permit TCP port 6514 for Secure Syslog and UDP port 123 for NTP.

Matching both service specifications ensures secure log collection and accurate time synchronization across subnets.

Key Concept

Port and transport layer protocol mappings for Secure Syslog (TCP 6514) and NTP (UDP 123)

Question 345Question →

A Layer 2 switch receives an Ethernet frame on Port 2, which is an access port assigned to VLAN 15. The source MAC address $0011.AA22.BB33$ is not currently listed in the switch's CAM table. The destination MAC address $0033.CC44.DD55$ is already present in the CAM table and associated with Port 7, an access port also assigned to VLAN 15. Port 10 is an access port assigned to VLAN 30, and Port 20 is configured as an 802.1Q trunk port carrying VLAN 15 and VLAN 30.

Which TWO of the following actions will the switch perform upon processing this frame? (Select TWO.)

Select all that apply

The switch enters the source MAC address 0011.AA22.BB33 into its CAM table, associating it with Port 2 and VLAN 15.

The switch forwards the frame directly out Port 7 as an untagged frame.

The switch floods the frame out all active interfaces in all VLANs because the source address was not previously recorded.

The switch drops the frame because frames entering access ports assigned to non-default VLANs must contain an 802.1Q tag.

The switch replicates the frame and forwards a tagged copy out trunk Port 20.

Show answer & explanation

Answer: The switch enters the source MAC address 0011.AA22.BB33 into its CAM table, associating it with Port 2 and VLAN 15.; The switch forwards the frame directly out Port 7 as an untagged frame.

Answer

The switch records the source MAC address 0011.AA22.BB33 mapped to Port 2 and VLAN 15 in its CAM table, and forwards the frame as an untagged Ethernet frame directly out Port 7.

When an Ethernet frame arrives at a switch, Layer 2 forwarding operations follow two distinct checks: source MAC learning and destination MAC forwarding. First, the switch checks the source MAC address. Because it is absent from the CAM table, the switch records the address 0011.AA22.BB33 associated with Port 2 in VLAN 15. Second, the switch looks up the destination MAC address 0033.CC44.DD55. Because it exists in the CAM table on Port 7 (which belongs to the same VLAN), the switch performs a microsegmentated unicast transmission directly out Port 7 as an untagged frame.

Step-by-Step Solution

Analyze source MAC address processing (Ingress learning).

The source MAC address 0011.AA22.BB33 is checked against the CAM table. Since it is missing, the switch adds an entry binding 0011.AA22.BB33 to Port 2 within VLAN 15.

Layer 2 switches perform dynamic MAC address learning on incoming frames to maintain an accurate CAM table.

Analyze destination MAC address processing (Egress forwarding).

The destination MAC address 0033.CC44.DD55 is looked up in the CAM table and found on Port 7 in VLAN 15. The switch forwards the frame directly to Port 7.

Known unicast frames are forwarded exclusively out the specific egress interface mapped to the destination MAC address within the matching VLAN.

Determine frame encapsulation format on egress.

The frame is transmitted out Port 7 without an 802.1Q tag.

Access switchports communicate with end hosts using standard untagged Ethernet frames.

Key Concept

Layer 2 Switch Frame Forwarding and Dynamic MAC Learning

Question 346Question →

A core router receives an IP packet destined for host address $192.168.88.77$ . The router's routing table contains the following matching entries:

Type	Destination Subnet	Administrative Distance / Metric	Next-Hop Interface
Static	$192.168.88.0/24$	$[1/0]$	$10.0.0.1$
BGP	$192.168.0.0/16$	$[20/100]$	$10.0.0.2$
OSPF	$192.168.88.64/26$	$[110/15]$	$10.0.0.3$
RIPv2	$192.168.88.72/29$	$[120/3]$	$10.0.0.4$

Based on standard Layer 3 forwarding logic, to which next-hop IP address will the router forward the packet?

10.0.0.4

10.0.0.1

10.0.0.2

10.0.0.3

Show answer & explanation

Answer: $10.0.0.4$

Answer

The router forwards the packet to next-hop

10.0.0.4

based on the Longest Prefix Match rule.

When a router performs a forwarding lookup for a destination IP, it evaluates the prefix length of all matching routes in the routing table. The route with the longest matching prefix mask (

/29

192.168.88.72/29

) is the most specific route to

192.168.88.77

. Longest prefix match takes precedence over administrative distance and metric values during packet forwarding decisions. Therefore, the router forwards traffic to next-hop

10.0.0.4

Step-by-Step Solution

Determine which subnets contain the target IP address

192.168.88.77

The target address matches all four entries:

192.168.88.0/24

192.168.0.0/16

192.168.88.64/26

(range

.64

–

.127

), and

192.168.88.72/29

(range

.72

–

.79

All four routing entries encompass the destination IP address.

Compare the prefix lengths (subnet mask specificity) of all matching routes.

The prefix lengths are

/16

/24

/26

, and

/29

. The longest (most specific) prefix match is

/29

Routers evaluate the Longest Prefix Match (LPM) first when selecting a path from the routing table.

Select the next-hop associated with the longest prefix match.

The route for

192.168.88.72/29

points to next-hop

10.0.0.4

Administrative distance and metric are only used to compare identical prefix destinations from different sources, not routes with different prefix lengths.

Key Concept

Longest Prefix Match (LPM) routing table lookup hierarchy

Estimated Time:2m 0s

Question 347Question →

A network engineer attempts to configure a static route on a enterprise router using the destination IPv6 address 2001:db8:1000:1::0:100::5. Upon executing the command, the router CLI rejects the entry and displays a syntax error. Which of the following best explains why this IPv6 address representation is invalid?

The address uses the double-colon shorthand (::) more than once, creating ambiguity in the total number of zero-padded 16-bit hextets.

The address omits leading zeros within the hextet blocks, which violates RFC 5952 IPv6 compression rules.

The double-colon shorthand (::) is restricted to the start or end of an IPv6 address string and cannot appear in middle blocks.

The hexadecimal characters in the address are written in lowercase letters rather than required uppercase letters.

Show answer & explanation

Answer: The address uses the double-colon shorthand (::) more than once, creating ambiguity in the total number of zero-padded 16-bit hextets.

Answer

The address uses the double-colon shorthand (::) more than once, creating ambiguity in the total number of zero-padded 16-bit hextets.

The IPv6 specification mandates that the double-colon shorthand (::) can only be used ONCE in any single address string. Because an IPv6 address must expand to exactly eight 16-bit hextets (128 bits total), having two instances of :: makes it impossible for the system parser to determine how many zero blocks each :: represents.

Step-by-Step Solution

Analyze the full structural components of an IPv6 address.

An IPv6 address consists of 128 bits divided into eight 16-bit blocks (hextets) separated by colons.

Understanding the 8-hextet baseline allows accurate expansion of compressed IPv6 representations.

Evaluate the provided IPv6 string 2001:db8:1000:1::0:100::5 for shorthand compression compliance.

The string contains two separate instances of the double-colon (::) operator.

The double-colon represents one or more contiguous hextets of all zeros.

Determine the impact of using multiple double-colon operators.

When expanding two double colons, the exact count of zero hextets represented by each double colon cannot be unambiguously calculated.

Because of this mathematical ambiguity, networking software and CLI parsers reject addresses containing more than one double colon as invalid syntax.

Key Concept

IPv6 Address Compression Rules (RFC 5952)

Estimated Time:1m 0s

Question 348Question →

A network administrator is configuring a new subnet assigned a $/26$ CIDR prefix. What is the maximum number of usable host IP addresses that can be assigned to devices on this subnet?

Show answer & explanation

Answer: 62

Answer

The maximum number of usable host IP addresses on a

/26

subnet is 62.

/26

subnet prefix leaves

32 - 26 = 6

bits for host assignment. The total address capacity is

2^6 = 64

IP addresses. Subtracting the network identifier and broadcast address yields

64 - 2 = 62

usable host addresses.

Step-by-Step Solution

Determine host portion bit length

6 bits (

32 - 26 = 6

)

IPv4 addresses consist of 32 bits total. Subtracting the prefix length yields the number of bits allocated for host addresses.

Calculate total address space

64 IP addresses (

2^6 = 64

)

The total number of unique address combinations is

2^h

, where

h

is the number of host bits.

Subtract reserved addresses

62 usable host addresses (

64 - 2 = 62

)

The network address (all host bits 0) and the broadcast address (all host bits 1) cannot be assigned to endpoints.

Key Concept

Calculating usable host count from IPv4 CIDR prefix length

Question 349Question →

During a routine security audit of host communications, a technician inspects raw protocol headers passing through a socket interface. The payload header contains source and destination port multiplexing fields along with window size parameters for flow control, but contains no IP routing headers or hardware physical addresses. At which layer of the OSI model does this data unit function?

Transport Layer

Network Layer

Data Link Layer

Session Layer

Show answer & explanation

Answer: Transport Layer

Answer

Transport Layer

The Transport Layer (Layer 4) of the OSI model manages end-to-end transport services. Key functions at this layer include service multiplexing using port numbers (e.g., TCP/UDP ports), segmentation, error control via checksums, and flow control using mechanisms like sliding window size.

Step-by-Step Solution

Analyze the header fields described in the scenario.

Identified port numbers (source/destination) and window size parameters (flow control).

Port numbers provide service multiplexing, and window sizes regulate data transmission rates.

Map identified protocol functions to the OSI model layers.

Port assignment and flow control are core duties of Layer 4.

The Transport Layer manages end-to-end communication control between end hosts.

Key Concept

OSI Layer 4 (Transport Layer) Functions and Protocol Data Units

Question 350Question →

A network administrator needs to segment the IPv4 network block $10.200.16.0/20$ to accommodate multiple remote office LANs, with the largest subnet requiring support for up to 500 usable host IP addresses. Which of the following statements regarding the subnet design and IP address allocation are correct? (Select TWO.)

Select all that apply

A /23 prefix (subnet mask 255.255.254.0) is the smallest subnet mask that accommodates 500 usable host IP addresses.

The first usable host IP address in the second /23 subnet carved from this block is 10.200.18.1.

A /24 prefix (subnet mask 255.255.255.0) can be assigned to accommodate 500 usable hosts by using variable-length subnet masking.

The broadcast address for the first /23 subnet allocated from this block is 10.200.16.255.

Show answer & explanation

Answer: A /23 prefix (subnet mask 255.255.254.0) is the smallest subnet mask that accommodates 500 usable host IP addresses.; The first usable host IP address in the second /23 subnet carved from this block is 10.200.18.1.

Answer

The correct statements are that a /23 prefix (255.255.254.0) is the smallest prefix capable of supporting 500 usable hosts, and the first usable host IP address in the second /23 subnet is 10.200.18.1.

To support 500 hosts, 9 host bits are necessary (

2^9 - 2 = 510

), resulting in a /23 mask (

255.255.254.0

). The first /23 subnet ranges from

10.200.16.0

10.200.17.255

with broadcast address

10.200.17.255

. The second /23 subnet starts at

10.200.18.0

, making

10.200.18.1

its first usable host address.

Step-by-Step Solution

Determine the required host bits and prefix length for 500 hosts

9 host bits are needed because

2^8 - 2 = 254

(insufficient) and

2^9 - 2 = 510

usable hosts (sufficient). The prefix length is

32 - 9 = /23

(subnet mask

255.255.254.0

Calculating usable hosts requires subtracting 2 (network and broadcast addresses) from

2^h

Calculate subnet boundaries for consecutive /23 subnets starting from 10.200.16.0/20

Subnet 1:

10.200.16.0/23

(Network ID:

10.200.16.0

, First Usable:

10.200.16.1

, Last Usable:

10.200.17.254

, Broadcast:

10.200.17.255

). Subnet 2:

10.200.18.0/23

(Network ID:

10.200.18.0

, First Usable:

10.200.18.1

, Last Usable:

10.200.19.254

, Broadcast:

10.200.19.255

A /23 subnet increments by 2 in the third octet.

Evaluate the choices against the calculated results

The statement specifying /23 as the minimum mask for 500 hosts is correct. The statement identifying 10.200.18.1 as the first usable IP of the second subnet is correct. Statements proposing /24 or 10.200.16.255 as the first subnet broadcast address are incorrect.

Comparing calculated addressing parameters against each statement verifies validity.

Key Concept

IPv4 Subnetting and Host Capacity Calculation

Question 351Question →

An administrator configures two separate VLANs—VLAN $10$ (Sales) and VLAN $20$ (Marketing)—on a single Layer 2 switch. Host computers in both departments are connected to switch access ports assigned to their respective VLAN IDs. Although physical links are up and all devices are correctly assigned IP addresses within their subnets, hosts in VLAN $10$ cannot communicate with hosts in VLAN $20$ . Which of the following best explains why these hosts cannot communicate, and what is required to enable connectivity between them?

VLANs create separate Layer 2 broadcast domains; therefore, a Layer 3 routing device, such as a router or multi-layer switch, is required to forward traffic between them.

Access ports block inter-subnet traffic by default; converting all host access ports into 802.1Q trunk ports will allow untagged communication between VLAN

10

and VLAN

20

The switch drops inter-VLAN frames because the native VLAN on the switch has not been set to VLAN

10

to match the Sales department subnet.

Spanning Tree Protocol (STP) has placed the switch ports into a blocking state due to a mismatch, requiring an adjustment to the STP root bridge priority.

Show answer & explanation

Answer: VLANs create separate Layer 2 broadcast domains; therefore, a Layer 3 routing device, such as a router or multi-layer switch, is required to forward traffic between them.

Answer

VLANs create separate Layer 2 broadcast domains; therefore, a Layer 3 routing device, such as a router or multi-layer switch, is required to forward traffic between them.

Each VLAN configured on a switch acts as a separate Layer 2 broadcast domain. Because Layer 2 switches forward Ethernet frames strictly within the origin VLAN, traffic destined for another VLAN must be forwarded to a Layer 3 device (such as a router or Layer 3 switch) capable of routing packets between IP subnets.

Step-by-Step Solution

Identify the Layer 2 boundary created by Virtual Local Area Networks (VLANs).

Recognize that VLAN

10

and VLAN

20

exist in completely separate broadcast domains on the switch.

By design, Layer 2 switches do not forward frames between different VLANs.

Determine the necessary networking layer component required to bridge distinct broadcast domains.

A Layer 3 routing function (such as a router-on-a-stick or Switched Virtual Interfaces on a Layer 3 switch) must evaluate IP packet headers.

Inter-VLAN communication requires IP routing because Layer 2 switching logic isolates VLAN traffic.

Key Concept

VLAN Isolation and Inter-VLAN Routing Requirements

Question 352Question →

An enterprise edge router receives an IP packet destined for host address $172.16.42.100$ . The router's active routing table contains four candidate routes:

Protocol Source	Destination Subnet	Administrative Distance	Metric	Next Hop
Static	$172.16.0.0/16$	1	0	Null0
eBGP	$172.16.40.0/21$	20	100	$192.168.1.1$
OSPF	$172.16.42.0/24$	110	20	$10.0.0.2$
RIPv2	$172.16.42.96/28$	120	2	$10.0.0.6$

Which route will the router select to forward the packet, and why?

The RIPv2 route (

172.16.42.96/28

), because Longest Prefix Match takes precedence over Administrative Distance.

The Static route (

172.16.0.0/16

), because static routes have the lowest Administrative Distance.

The eBGP route (

172.16.40.0/21

), because exterior gateway protocols are prioritized over interior gateway protocols.

The OSPF route (

172.16.42.0/24

), because it offers the optimal balance between prefix specificity and Administrative Distance.

Show answer & explanation

Answer: The RIPv2 route ( $172.16.42.96/28$ ), because Longest Prefix Match takes precedence over Administrative Distance.

Answer

The router will select the RIPv2 route (

172.16.42.96/28

) because Longest Prefix Match (LPM) is the absolute first criterion used in IP routing decisions, taking precedence over Administrative Distance and metric.

When a router receives a packet, it evaluates all matching routing table entries using the Longest Prefix Match (LPM) rule first. Because

172.16.42.96/28

has the longest subnet mask (/28 = 28 matching bits) among all matching candidate routes, it is chosen immediately. Administrative Distance (AD) and metrics are only evaluated to select a winner when two or more routes advertise identical destination prefix lengths.

Step-by-Step Solution

Determine which routing table entries match the target IP address (

172.16.42.100

All four entries match:

172.16.0.0/16

(range

172.16.0.0 - 172.16.255.255

172.16.40.0/21

(range

172.16.40.0 - 172.16.47.255

172.16.42.0/24

(range

172.16.42.0 - 172.16.42.255

), and

172.16.42.96/28

(range

172.16.42.96 - 172.16.42.111

The router must filter candidate routes to only those subnets that encompass the destination address.

Compare the prefix lengths (subnet masks) of all matching candidate routes.

Prefix lengths are /16, /21, /24, and /28. The longest prefix match is /28.

In IP routing, the most specific route (highest prefix length) is always selected before Administrative Distance or metric are evaluated.

Identify the protocol associated with the longest prefix match (/28).

The /28 route corresponds to the RIPv2 learned entry via next-hop

10.0.0.6

Administrative Distance (120 vs 1 vs 20 vs 110) is ignored because the prefix lengths are not equal.

Key Concept

Longest Prefix Match (LPM) Rule

Question 353Question →

A network technician is configuring a local area network interface using the IPv4 network block $10.20.30.0/26$ . Which of the following statements regarding this subnet configuration are correct? (Select TWO.)

Select all that apply

The subnet mask in dotted-decimal notation is

255.255.255.192

The total number of usable host IP addresses in this subnet is

62

The total number of usable host IP addresses in this subnet is

64

The IP address

10.20.30.63

can be assigned directly to a host workstation interface.

Show answer & explanation

Answer: The subnet mask in dotted-decimal notation is $255.255.255.192$ .; The total number of usable host IP addresses in this subnet is $62$ .

Answer

The correct statements are that the subnet mask in dotted-decimal notation is

255.255.255.192

and the subnet contains

62

usable host IP addresses.

A CIDR prefix of

/26

uses 26 bits for the network portion and 6 bits for the host portion (

32 - 26 = 6

). In dotted-decimal notation, 26 subnet bits translate to

255.255.255.192

. The total number of IP addresses in a

/26

subnet is

2^6 = 64

. Subtracting 2 (one for the network address

10.20.30.0

and one for the broadcast address

10.20.30.63

) leaves exactly 62 usable host IP addresses.

Step-by-Step Solution

Calculate the subnet mask from CIDR notation /26

26 binary 1s followed by 6 binary 0s gives 11111111.11111111.11111111.11000000, which converts to

255.255.255.192

Converting CIDR prefix to dotted-decimal notation is necessary to determine the subnet boundary.

Determine host bits and total IP address capacity

Remaining host bits =

32 - 26 = 6

. Total addresses =

2^6 = 64

Host bits determine the size of the IP block.

Calculate usable host IP addresses

Usable host addresses =

64 - 2 = 62

. Network ID is

10.20.30.0

and Broadcast address is

10.20.30.63

The first address (network ID) and last address (broadcast address) are reserved and cannot be assigned to end devices.

Key Concept

IPv4 Subnet Masking and Usable Host Calculation

Question 354Question →

A network technician is provisioning an isolated subnet for a small cluster of administrative servers using the $/29$ IPv4 prefix. What is the maximum number of usable host IP addresses available on this subnet?

Show answer & explanation

Answer: 6

Answer

The maximum number of usable host IP addresses available on a

/29

subnet is 6.

For a

/29

prefix,

32 - 29 = 3

bits remain for host addressing. The total number of IP addresses in the block is

2^3 = 8

. Because 1 address is reserved for the network ID and 1 address is reserved for the directed broadcast ID, the number of usable host IP addresses is

8 - 2 = 6

Step-by-Step Solution

Calculate the number of host bits available in the CIDR prefix.

32 total IPv4 bits - 29 network bits = 3 host bits.

IPv4 addresses consist of 32 bits total.

Calculate total IP addresses supported by 3 host bits.

2^3 = 8 total IP addresses.

Each host bit doubles the number of possible IP addresses.

Subtract reserved network and broadcast addresses.

8 total IPs - 2 = 6 usable host addresses.

The network address (all host bits 0) and broadcast address (all host bits 1) cannot be assigned to host interfaces.

Key Concept

Calculating usable host IP addresses in IPv4 subnetting by subtracting network and broadcast addresses.

Estimated Time:45s

Question 355Question →

An IT technician is deploying a network infrastructure for a regional facility consisting of four distinct buildings situated on a single contiguous corporate park. The internal networks of each building are connected together using private, high-speed optical fiber cabling owned and managed internally by the organization. Which of the following network scope classifications best describes this setup?

Campus Area Network (CAN)

Metropolitan Area Network (MAN)

Wide Area Network (WAN)

Personal Area Network (PAN)

Show answer & explanation

Answer: Campus Area Network (CAN)

Answer

Campus Area Network (CAN)

A Campus Area Network (CAN) is a network connecting multiple local area networks (LANs) within a contiguous geographical area, such as a corporate office park or university campus, utilizing privately owned cabling and switches.

Step-by-Step Solution

Analyze the physical scope and infrastructure ownership

The network spans multiple buildings within a single corporate park using privately owned infrastructure.

Geographical boundary and ownership differentiate network types.

Map the characteristics to standardized network classifications

A network connecting multiple LANs across a single contiguous property or facility complex is defined as a Campus Area Network (CAN).

CAN is smaller than a MAN/WAN and relies on private infrastructure interconnecting localized LANs.

Key Concept

Campus Area Network (CAN) Characteristics

Estimated Time:1m 0s

Question 356Question →

A network administrator is configuring IP addressing services for a newly created client subnet (VLAN 30) across a routed enterprise network. The centralized DHCP server is located on a different subnet (VLAN 10). Users on VLAN 30 report that their workstations fail to acquire dynamic IP addresses and instead receive APIPA addresses in the range $169.254.0.0/16$ . Which configuration change must be implemented on the default gateway router interface for VLAN 30 to resolve this issue?

Configure an IP helper address (DHCP relay agent) on the VLAN 30 gateway interface pointing to the DHCP server IP address

Configure a DHCP relay agent on the central DHCP server interface on VLAN 10 pointing back to the client subnet

Create an AAAA resource record on the DNS server for the VLAN 30 default gateway interface

Reconfigure the central DHCP server to listen for client lease requests over TCP port 67 instead of UDP port 67

Show answer & explanation

Answer: Configure an IP helper address (DHCP relay agent) on the VLAN 30 gateway interface pointing to the DHCP server IP address

Answer

Configure an IP helper address (DHCP relay agent) on the VLAN 30 gateway interface pointing to the DHCP server IP address

DHCP clients discover servers using local broadcasts. Because Layer 3 routers drop broadcast traffic by default, a DHCP relay agent (configured via an IP helper address on the router interface local to the client subnet) is required to translate client broadcast requests into unicast packets sent directly to the central DHCP server.

Step-by-Step Solution

Analyze client failure symptoms

Workstations receive APIPA addresses (

169.254.x.x

), indicating they cannot reach a DHCP server to obtain an IP lease.

Routers do not forward Layer 2 broadcast packets (such as DHCPDISCOVER messages) across subnet boundaries by default.

Identify the required network service component

A DHCP relay agent (IP helper address) is needed on the client-facing gateway interface.

The relay agent intercepts client broadcast messages on VLAN 30 and encapsulates them into unicast packets directed to the DHCP server on VLAN 10.

Key Concept

DHCP Relay Agent and IP Helper Functionality across Layer 3 Subnet Boundaries

Question 357Question →

A system administrator is configuring perimeter firewall rules for a newly provisioned edge appliance. The appliance requires remote encrypted web-based management and secure file upload capabilities via Secure Shell (SSH). Which TWO of the following port and transport protocol combinations must be opened inbound on the firewall to satisfy these operational requirements? (Select TWO)

Select all that apply

TCP port 443 for HTTPS administrative access

TCP port 22 for SSH-based Secure File Transfer Protocol (SFTP)

UDP port 443 for HTTPS administrative access

TCP port 21 for SSH-based Secure File Transfer Protocol (SFTP)

Show answer & explanation

Answer: TCP port 443 for HTTPS administrative access; TCP port 22 for SSH-based Secure File Transfer Protocol (SFTP)

Answer

The firewall rules must permit inbound traffic on TCP port 443 for HTTPS administrative access and TCP port 22 for SFTP secure file transfer.

TCP port 443 is the standard port reserved for HTTPS traffic, providing encrypted web management capabilities over connection-oriented TCP. TCP port 22 is the standard port for Secure Shell (SSH) services, which includes SFTP for secure file operations.

Step-by-Step Solution

Determine the default transport protocol and port for encrypted web access.

HTTPS uses TCP port 443.

Hypertext Transfer Protocol Secure (HTTPS) requires reliable connection-oriented delivery over TCP port 443.

Determine the default transport protocol and port for secure SSH file management.

SFTP uses TCP port 22.

SSH File Transfer Protocol (SFTP) runs as an subsystem of SSH, utilizing TCP port 22.

Key Concept

Default transport protocols and port assignments for encrypted web and remote management services

Question 358Question →

A network architect is implementing Variable Length Subnet Masking (VLSM) on the assigned address block $172.16.48.0/20$ to provision distinct functional zones in a datacenter. Zone A requires host capacity for at least 500 IP addresses, while Zone B requires host capacity for at least 250 IP addresses. Subnets are allocated sequentially starting from the lowest available network address of the block, with Zone A allocated first, followed immediately by Zone B. What is the broadcast address of the subnet allocated to Zone B?

172.16.50.255

172.16.50.254

172.16.49.255

172.16.51.255

Show answer & explanation

Answer: $172.16.50.255$

Answer

The broadcast address of the subnet allocated to Zone B is

172.16.50.255

To host at least 500 devices in Zone A, 9 host bits are needed (

2^9 - 2 = 510

), resulting in a

/23

subnet (

172.16.48.0

172.16.49.255

). The next available network address for Zone B is

172.16.50.0

. To host 250 devices in Zone B, 8 host bits are required (

2^8 - 2 = 254

), resulting in a

/24

subnet (

172.16.50.0

172.16.50.255

). The broadcast address of this subnet is the final address in its range,

172.16.50.255

Step-by-Step Solution

Determine prefix length and block size for Zone A (500 hosts required)

Host bits needed:

2^h - 2 \ge 500 \implies h = 9

bits (

2^9 - 2 = 510

usable hosts). Subnet mask prefix is

32 - 9 = /23

, which gives a block size of

2^9 = 512

addresses.

Zone A requires at least 500 hosts, so a

/24

(254 usable hosts) is insufficient and a

/23

must be used.

Calculate the address boundary for Zone A

Zone A starts at

172.16.48.0/23

and spans

172.16.48.0

through

172.16.49.255

/23

block starting at

172.16.48.0

increments the third octet by 2 (

48 + 2 = 50

Determine prefix length and block size for Zone B (250 hosts required)

Starting address for Zone B is

172.16.50.0

. Host bits needed:

2^h - 2 \ge 250 \implies h = 8

bits (

2^8 - 2 = 254

usable hosts). Subnet mask prefix is

32 - 8 = /24

, giving a block size of

2^8 = 256

addresses.

250 hosts require 8 host bits, corresponding to a

/24

subnet mask.

Determine the broadcast address of Zone B

The subnet range for Zone B is

172.16.50.0

172.16.50.255

. The broadcast address is

172.16.50.255

The last address in an IP subnet block is reserved as the broadcast address.

Key Concept

Variable Length Subnet Masking (VLSM) calculations, subnet block size determination, and broadcast address identification.

Question 359Question →

A network administrator is configuring switch ports connected to desktop environments where IP phones and workstations share a single physical Ethernet drop. The switchports are configured to handle untagged data traffic on VLAN 10 and tagged voice traffic on VLAN 20. Which of the following statements correctly describe Layer 2 switch forwarding and VLAN broadcast domain behavior in this setup? (Select TWO.)

Select all that apply

Untagged Ethernet frames entering the port are processed internally within VLAN 10.

Broadcast traffic originating on VLAN 10 is constrained to ports belonging to VLAN 10 and will not cross into VLAN 20 at Layer 2.

Hosts on VLAN 10 can communicate directly with hosts on VLAN 20 through standard Layer 2 switch forwarding without a Layer 3 routing device.

The 802.1Q tagging process inserts VLAN identifier metadata directly into the Layer 3 IP header before Layer 2 framing.

If a native VLAN mismatch occurs across an 802.1Q trunk link, all untagged traffic is automatically routed to its intended destination without potential security leakage.

Show answer & explanation

Answer: Untagged Ethernet frames entering the port are processed internally within VLAN 10.; Broadcast traffic originating on VLAN 10 is constrained to ports belonging to VLAN 10 and will not cross into VLAN 20 at Layer 2.

Answer

Untagged Ethernet frames entering the port are processed internally within VLAN 10, and broadcast traffic originating on VLAN 10 is constrained to ports belonging to VLAN 10 and will not cross into VLAN 20 at Layer 2.

VLANs operate by logically partitioning a single physical switch into distinct Layer 2 broadcast domains. Untagged frames received on a port configured for a specific data VLAN are assigned to that VLAN's broadcast domain. Consequently, broadcast frames generated on VLAN 10 are forwarded only to ports assigned to VLAN 10, maintaining full isolation from VLAN 20.

Step-by-Step Solution

Analyze how untagged frames are handled on switch ports configured for voice and data VLANs.

Untagged frames are mapped to the access/data VLAN ID (VLAN 10) for internal Layer 2 forwarding.

Switches assign incoming untagged Ethernet frames to the port's configured access/native VLAN.

Evaluate the broadcast domain boundary enforcement of VLANs.

Broadcast frames (destination MAC FF:FF:FF:FF:FF:FF) on VLAN 10 are replicated only to ports assigned to VLAN 10.

VLANs segment physical switch infrastructure into independent Layer 2 broadcast domains.

Verify inter-VLAN routing requirements.

Traffic moving between VLAN 10 and VLAN 20 requires a Layer 3 routing mechanism.

Layer 2 switches cannot forward frames directly between different VLANs without IP routing.

Key Concept

VLAN Broadcast Domain Segmentation and Layer 2 Frame Processing

Question 360Question →

Match each core IP addressing service or DNS record type to its primary operational function within enterprise network infrastructure.

Click a left item, then click its matching right item

Show answer & explanation

Answer

MX Record maps to directing incoming domain email traffic; Dynamic DNS maps to automatically updating DNS zone records upon IP assignment changes; IPAM maps to centralized tracking and management of IP address spaces; NTP maps to synchronizing system clocks across network devices.

Each addressing service is mapped strictly according to standard protocol definitions: MX records specify mail routing destinations; Dynamic DNS (DDNS) automates DNS record modification when DHCP leases change; IPAM centrally coordinates IP address allocation and scope management; NTP maintains unified time synchronization across all networked devices.

Step-by-Step Solution

Identify the purpose of domain email resolution records.

MX records specify designated mail transfer hostnames for a domain.

Domain name systems rely on specialized MX records rather than standard A records to process inbound mail traffic.

Analyze how variable IP addressing integrates with name resolution.

Dynamic DNS (DDNS) automatically registers changes between DHCP lease grants and DNS resource records.

Hostnames remain resolvable even when workstations receive changing IPv4 or IPv6 dynamic leases.

Determine the administrative system used for network addressing governance.

IPAM provides centralized management of subnets, scopes, and address utilization metrics.

Enterprise networks require IPAM to prevent IP address overlapping and streamline DHCP/DNS infrastructure configuration.

Evaluate time distribution protocol functionality across network infrastructure.

NTP synchronizes client and infrastructure system time over UDP port 123.

Consistent system time is required for accurate log correlation, Kerberos authentication, and certificate validation.

Key Concept

Core IP Addressing Services and DNS Record Functions

Estimated Time:1m 30s

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