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A network engineer is deploying IPv6 across a corporate branch office after receiving a $/56$ IPv6 prefix delegation of $2001:\text{db8}:1234:5600::/56$ from the upstream Internet Service Provider (ISP). Enterprise standards require that all internal local area networks (LANs) be provisioned using standard $/64$ subnets to support Stateless Address Autoconfiguration (SLAAC). How many total $/64$ subnets can the engineer allocate from this delegated prefix, and what is the RFC 5952 compliant IPv6 prefix for the final available subnet?

256 subnets, with the final subnet prefix designated as 2001:db8:1234:56ff::/64

254 subnets, with the final subnet prefix designated as 2001:db8:1234:56fe::/64

16 subnets, with the final subnet prefix designated as 2001:db8:1234:56f0::/64

256 subnets, with the final subnet prefix designated as 2001:db8:1234:5600::ff::/64

Show answer & explanation

Answer: 256 subnets, with the final subnet prefix designated as 2001:db8:1234:56ff::/64

Answer

The correct subnet count is 256 subnets, and the final subnet prefix is 2001:db8:1234:56ff::/64.

Subtracting the ISP delegation prefix length (/56) from the standard host SLAAC prefix length (/64) yields 8 bits for internal subnet division. Calculating

2^8

produces 256 total subnets. The first 56 bits (

2001:\text{db8}:1234:56

) remain fixed, while the remaining 8 bits of the fourth hextet range from 0x00 to 0xFF hex. Thus, the last valid /64 subnet prefix is 2001:db8:1234:56ff::/64.

Step-by-Step Solution

Calculate the number of bits allocated for subnetting.

Subnet bits =

64 - 56 = 8

bits.

The delegated prefix is /56 and target LAN subnets require a /64 prefix length for SLAAC.

Determine total usable subnets from the 8 subnet bits.

2^8 = 256

subnets.

In IPv6 subnetting, there are no reserved all-zeros network or all-ones broadcast subnet restrictions like in IPv4; all combinations are usable.

Calculate the fourth hextet range in hexadecimal.

The fourth hextet is 16 bits long: bits 48-55 are fixed as 0x56, and bits 56-63 range from 0x00 to 0xFF.

Starting at 0x5600, incrementing the 8-bit subnet ID from 0x00 to 0xFF results in a fourth hextet range of 5600 to 56FF.

Format the final subnet prefix according to RFC 5952 compression guidelines.

2001:db8:1234:56ff::/64

Trailing zero hextets in the host portion are replaced by a single double-colon (::).

Key Concept

IPv6 Prefix Subnetting and RFC 5952 Address Formatting

Estimated Time:2m 0s

Question 1622Question →

A network administrator is configuring a host on an enterprise network segment using Stateless Address Autoconfiguration (SLAAC). The local router advertises the IPv6 prefix $2001:\text{db8}:0000:00\text{ab}::/64$ . The host network interface controller has the MAC address `00:50:56:A1:B2:C3`. Assuming the host generates its 64-bit Interface Identifier using the standard EUI-64 mechanism, what is the resulting host IPv6 address expressed in strict compliance with RFC 5952 zero-compression and leading-zero suppression rules?

Show answer & explanation

Answer: 2001:db8:0:ab:250:56ff:fea1:b2c3; 2001:DB8:0:AB:250:56FF:FEA1:B2C3

Answer

2001:db8:0:ab:250:56ff:fea1:b2c3

Generating the SLAAC IPv6 address involves forming the EUI-64 interface ID from MAC address `00:50:56:A1:B2:C3` by inserting `FF:FE` between the third and fourth bytes (`00:50:56:FF:FE:A1:B2:C3`) and inverting the 7th bit of the first byte (`00` becomes `02`), yielding `0250:56ff:fea1:b2c3`. Combining this with prefix `2001:0db8:0000:00ab::/64` gives `2001:0db8:0000:00ab:0250:56ff:fea1:b2c3`. Under RFC 5952, leading zeros in each field are omitted (`0db8` -> `db8`, `0000` -> `0`, `00ab` -> `ab`, `0250` -> `250`). Because there is only one contiguous 16-bit zero field, `::` cannot be applied, resulting in `2001:db8:0:ab:250:56ff:fea1:b2c3`.

Step-by-Step Solution

Split the 48-bit MAC address into two 24-bit halves and insert FF:FE in the middle.

Inserting `FF:FE` into `00:50:56` and `A1:B2:C3` yields `00:50:56:FF:FE:A1:B2:C3` (or formatted as hextets: `0050:56ff:fea1:b2c3`).

EUI-64 expands a 48-bit MAC address to a 64-bit interface identifier by inserting the reserved 16-bit hex sequence `FF-FE`.

Invert the 7th bit (Universal/Local bit) of the first byte of the MAC address.

The first byte `00` in binary is `0000 0000`. Inverting the 7th bit (the U/L bit) gives `0000 0010`, which is `02` in hexadecimal. The modified interface ID becomes `0250:56ff:fea1:b2c3`.

The IEEE EUI-64 standard specifies that the 7th bit of the first octet must be inverted to indicate whether the address is universally unique (0 changed to 1).

Combine the /64 network prefix with the 64-bit EUI-64 interface identifier.

Combining `2001:0db8:0000:00ab` with `0250:56ff:fea1:b2c3` produces the uncompressed 128-bit address `2001:0db8:0000:00ab:0250:56ff:fea1:b2c3`.

The full IPv6 unicast address is formed by concatenating the subnet prefix and host interface ID.

Apply RFC 5952 compression rules strictly.

Suppress leading zeros in all hextets (`0db8` -> `db8`, `0000` -> `0`, `00ab` -> `ab`, `0250` -> `250`). Note that double-colon `::` cannot be used to replace a single zero hextet (`0`). The final compressed address is `2001:db8:0:ab:250:56ff:fea1:b2c3`.

RFC 5952 mandates lower-case hexadecimal digits, leading zero suppression in each 16-bit field, and explicitly forbids using double-colon `::` for a single isolated zero field.

Key Concept

EUI-64 Interface ID Generation & RFC 5952 Address Compression Rules

Question 1623Question →

A network administrator is assigning static IP addresses to virtual machine interfaces within a subnet designated as $10.200.45.64/27$ . Which of the following IP addresses represent valid usable host addresses within this subnet? (Select TWO)

Select all that apply

10.200.45.65

10.200.45.94

10.200.45.64

10.200.45.95

Show answer & explanation

Answer: $10.200.45.65$ ; $10.200.45.94$

Answer

The addresses

10.200.45.65

and

10.200.45.94

are valid usable host addresses.

For a

/27

subnet starting at network ID

10.200.45.64

, the block contains 32 total IP addresses ending at

10.200.45.95

. The first address (

10.200.45.64

) serves as the network ID and the last address (

10.200.45.95

) serves as the subnet broadcast address. Therefore, the assignable usable host IP range spans from

10.200.45.65

10.200.45.94

, making

10.200.45.65

and

10.200.45.94

valid usable host IP addresses.

Step-by-Step Solution

Determine the total block size from the CIDR prefix.

/27

prefix length leaves

32 - 27 = 5

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

2^5 = 32

The CIDR prefix determines the number of bits allocated to the host portion of the address.

Calculate the network ID and broadcast address for the subnet.

The network address is

10.200.45.64

. The broadcast address is

10.200.45.64 + 32 - 1 = 10.200.45.95

The network address is the first IP in the block, and the broadcast address is the final IP in the block.

Identify the range of usable host IP addresses.

Excluding the network ID (

10.200.45.64

) and broadcast address (

10.200.45.95

) yields a usable host range from

10.200.45.65

10.200.45.94

Usable host addresses must exclude the reserved network ID and broadcast addresses.

Key Concept

IPv4 Subnet Boundaries and Usable Host Range Calculation

Estimated Time:1m 30s

Question 1624Question →

An enterprise network infrastructure team is implementing IPv6 addressing across corporate VLANs. The core router is configured to transmit Router Advertisement (RA) messages on the local segment with the Managed Address Configuration flag set to 0 ( $M=0$ ) and the Other Stateful Configuration flag set to 1 ( $O=1$ ). Which of the following best describes how client workstations on this subnet will obtain their IPv6 configuration parameters?

Workstations generate their IPv6 global unicast address autonomously using SLAAC from the advertised prefix, and query a stateless DHCPv6 server for additional parameters such as DNS server addresses.

Workstations obtain both their IPv6 global unicast address and all network configuration parameters exclusively from a stateful DHCPv6 server.

Workstations use SLAAC to generate their IPv6 address via EUI-64 but are prevented from receiving any domain name resolution settings because DHCPv6 is disabled by the

M=0

flag.

Workstations derive their link-local address via SLAAC, but must compress their global unicast address using double-colon notation before registering with the local router.

Show answer & explanation

Answer: Workstations generate their IPv6 global unicast address autonomously using SLAAC from the advertised prefix, and query a stateless DHCPv6 server for additional parameters such as DNS server addresses.

Answer

In IPv6 deployment, ICMPv6 Router Advertisement (RA) flags direct host configuration behavior. When the Managed Address flag is off (

M=0

) and the Other flag is on (

O=1

), clients utilize Stateless Address Autoconfiguration (SLAAC) to build their own global unicast IP address using the prefix provided in the RA message. Concurrently, clients send information requests to a stateless DHCPv6 server to acquire non-address parameters such as DNS server addresses and domain names.

Step-by-Step Solution

Analyze the Router Advertisement (RA) flag settings provided in the scenario.

The Managed Address Configuration flag is set to 0 (

M=0

), and the Other Stateful Configuration flag is set to 1 (

O=1

RA flags dictate whether clients use SLAAC, stateful DHCPv6, or stateless DHCPv6.

Evaluate the impact of

M=0

on address autoconfiguration.

Hosts do not use a stateful DHCPv6 server for IPv6 address assignment; instead, they use Stateless Address Autoconfiguration (SLAAC) to derive their IP address from the network prefix.

M=0

indicates that addresses are not managed by a stateful DHCPv6 server.

Evaluate the impact of

O=1

on additional network options.

Hosts query a stateless DHCPv6 server for non-address configuration information such as DNS server IP addresses and domain search lists.

O=1

specifies that 'other' configuration details are available via DHCPv6.

Key Concept

IPv6 Router Advertisement (RA) Flags and Autoconfiguration Modes (SLAAC vs. Stateless/Stateful DHCPv6)

Question 1625Question →

A network engineer is deploying IPv6 across an enterprise environment and configuring dynamic address assignment and routing mechanisms. Match each IPv6 deployment component or address feature on the left to its corresponding technical operation or scope on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Stateless DHCPv6 (M=0, O=1) matches hosts auto-generating IP addresses via SLAAC while querying DHCPv6 for non-address settings. EUI-64 Interface Identifier matches inserting 0xFFFE into a MAC address and flipping the 7th bit. Unique Local Address (fc00::/7) matches globally unique non-routable addressing for internal sites. All-DHCPv6-Relay-Agents/Servers Multicast (ff02::1:2) matches the link-local destination address used for relaying client Solicit messages.

Stateless DHCPv6 (M=0, O=1) allows hosts to use SLAAC for IP generation while fetching extra options like DNS via DHCPv6. Modified EUI-64 expands a 48-bit MAC address into a 64-bit interface ID by inserting 0xFFFE and inverting the 7th bit. Unique Local Addresses (fc00::/7) provide internal private IP routing within enterprise boundaries without internet accessibility. The multicast address ff02::1:2 specifically targets local DHCPv6 relay agents and servers.

Step-by-Step Solution

Analyze Router Advertisement flags for dynamic IPv6 address deployment modes.

M=0 indicates that stateful address assignment is disabled, and O=1 indicates that other configuration parameters (such as DNS) are available via DHCPv6. This corresponds to Stateless DHCPv6 combined with SLAAC.

Understanding ICMPv6 RA flag combinations is essential for deploying stateless versus stateful IPv6 autoconfiguration.

Evaluate host interface identifier generation mechanisms using EUI-64.

Modified EUI-64 splits the MAC address into two 24-bit halves, inserts 16 bits (0xFFFE), and flips the Universal/Local (U/L) bit (the 7th bit of the first octet).

Verifying correct EUI-64 translation rules ensures proper identification of automatically derived IPv6 host addresses.

Classify IPv6 address scopes and prefixes for enterprise design.

The fc00::/7 prefix defines Unique Local Addresses (ULA), which are internal, non-globally-routable IPv6 addresses.

Distinguishing between Global Unicast, Link-Local, and Unique Local prefixes is critical for secure network architecture.

Identify specialized IPv6 multicast addresses used during address assignment.

ff02::1:2 represents the All-DHCPv6-Relay-Agents/Servers link-local multicast group.

DHCPv6 clients communicate with local servers or relay agents using designated multicast groups rather than broadcast mechanisms.

Key Concept

IPv6 Address Scopes, EUI-64 Derivation, and Autoconfiguration Mechanisms (SLAAC / DHCPv6)

Question 1626Question →

A network technician is configuring an 802.1Q trunk link between core switch `SW-CORE-01` and access switch `SW-ACCESS-02` to allow traffic for multiple VLANs across the site. Interface `GigabitEthernet 1/0/24` on `SW-CORE-01` has been configured with the following command sequence:

text
interface GigabitEthernet 1/0/24
switchport mode trunk
switchport trunk native vlan 99
switchport trunk allowed vlan 10,20,30,40

On `SW-ACCESS-02`, interface `GigabitEthernet 0/1` connects to `SW-CORE-01` and is configured as follows:

text
interface GigabitEthernet 0/1
switchport mode trunk
switchport trunk native vlan 99
switchport trunk allowed vlan 10,20,40

End devices connected to `SW-ACCESS-02` on VLAN 30 are unable to reach the default gateway located on `SW-CORE-01`, while hosts on VLAN 10 and VLAN 20 experience no connectivity issues. Which configuration change on `SW-ACCESS-02` will resolve the connectivity issue for VLAN 30 hosts?

Add VLAN 30 to the allowed VLAN list on interface GigabitEthernet 0/1 of SW-ACCESS-02.

Change the native VLAN on interface GigabitEthernet 0/1 of SW-ACCESS-02 to VLAN 30.

Reconfigure interface GigabitEthernet 0/1 on SW-ACCESS-02 into access mode assigned to VLAN 30.

Modify the native VLAN on SW-CORE-01 to match the default untagged VLAN 1.

Show answer & explanation

Answer: Add VLAN 30 to the allowed VLAN list on interface GigabitEthernet 0/1 of SW-ACCESS-02.

Answer

The connectivity issue for VLAN 30 hosts is resolved by adding VLAN 30 to the allowed VLAN list on interface GigabitEthernet 0/1 of SW-ACCESS-02.

The trunk configuration on the access switch explicitly filters out VLAN 30 because its allowed list is limited to VLANs 10, 20, and 40. Updating the allowed VLAN list on interface GigabitEthernet 0/1 of the access switch to include VLAN 30 permits 802.1Q tagged frames for VLAN 30 to pass across the trunk interface to the core switch default gateway.

Step-by-Step Solution

Analyze the configuration on both ends of the 802.1Q trunk link.

SW-CORE-01 allows VLANs 10, 20, 30, and 40, while SW-ACCESS-02 only allows VLANs 10, 20, and 40.

When a VLAN is omitted from the trunk's allowed list (`switchport trunk allowed vlan`), the switch drops all ingress and egress frames tagged for that specific VLAN ID on that port.

Correlate the symptom with the missing configuration.

Hosts on VLAN 30 attached to SW-ACCESS-02 cannot transmit frames across the trunk to SW-CORE-01 because SW-ACCESS-02 blocks VLAN 30 on interface GigabitEthernet 0/1.

The allowed VLAN list acts as a traffic filter for 802.1Q tagged frames across trunk interfaces.

Determine the necessary remediation command.

Modify the trunk configuration on SW-ACCESS-02 to append VLAN 30 to the allowed list (`switchport trunk allowed vlan add 30` or updating the list to include 30).

This allows 802.1Q frames tagged with VLAN 30 to pass bidirectionally across the link.

Key Concept

802.1Q Trunk Allowed VLAN Pruning and Filtering

Question 1627Question →

A client host on VLAN 10 ( $10.10.10.0/24$ ) boots up and requires an IPv4 address lease from a centralized DHCP server located at $192.168.100.50$ . A Layer 3 switch serves as the DHCP relay agent with an IP helper address configured on SVI 10 ( $10.10.10.1$ ). What is the correct sequence of network operations during the initial DHCP discover and offer phase?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order of network operations is: 1) Client host broadcasts a DHCPDISCOVER packet on UDP port 67 within VLAN 10; 2) Layer 3 switch populates the `giaddr` field with

10.10.10.1

and sends a unicast DHCPDISCOVER packet to

192.168.100.50

on UDP port 67; 3) DHCP server matches `giaddr` to the

10.10.10.0/24

address pool, selects an available IP, and sends a unicast DHCPOFFER message back to

10.10.10.1

; 4) Layer 3 switch receives the unicast DHCPOFFER and forwards it onto VLAN 10 to the requesting client on UDP port 68.

The sequence begins when the host client transmits a broadcast DHCPDISCOVER frame on UDP port 67 inside VLAN 10. The Layer 3 switch (relay agent) intercepts this broadcast on SVI 10, sets the `giaddr` field to its local gateway IP (

10.10.10.1

), and unicasts the packet to the remote DHCP server at

192.168.100.50

on UDP port 67. Upon receiving the unicast packet, the DHCP server evaluates `giaddr` to pick an IP address from the matching

10.10.10.0/24

scope and sends a unicast DHCPOFFER reply back to the relay agent (

10.10.10.1

). Finally, the relay agent forwards the DHCPOFFER onto VLAN 10 where the client receives it on UDP port 68.

Step-by-Step Solution

Client broadcast initialization

DHCPDISCOVER broadcast generated on local VLAN 10 on UDP port 67.

Unconfigured hosts cannot unicast without an assigned IP address.

Relay agent interception and `giaddr` insertion

Layer 3 switch intercepts broadcast, inserts its SVI IP (

10.10.10.1

) into the `giaddr` header field, and converts the packet into a unicast payload directed to

192.168.100.50

on UDP port 67.

Broadcast packets do not cross router boundaries; unicast conversion and `giaddr` tagging allow cross-subnet DHCP routing.

DHCP server pool matching and reply generation

DHCP server receives unicast request, examines `giaddr` (

10.10.10.1

), queries the corresponding

10.10.10.0/24

scope, reserves an available IPv4 address, and sends a unicast DHCPOFFER back to

10.10.10.1

The `giaddr` field tells the central server exactly which IP range to allocate from.

Relay agent local delivery

Layer 3 switch receives the unicast DHCPOFFER and transmits it onto VLAN 10 addressed to the client on UDP port 68.

The client software listens on UDP port 68 to receive incoming DHCP offer and acknowledgment responses.

Key Concept

DHCP Relay Agent Operational Packet Flow and `giaddr` Header Processing

Question 1628Question →

A network engineer is configuring subnets within the $10.50.0.0/16$ address space for a new office building. Each subnet must support a minimum of $150$ usable host IP addresses. What is the maximum number of equal-sized subnets that can be created from this block while meeting the host requirement?

Show answer & explanation

Answer: 256

Answer

The maximum number of equal-sized subnets that can be created is 256.

To support at least 150 host devices, 8 host bits are necessary (

2^8 - 2 = 254

usable hosts). Subtracting 8 host bits from 32 total address bits yields a /24 subnet mask. Borrowing 8 bits from the original /16 network prefix produces

2^8 = 256

subnets.

Step-by-Step Solution

Determine the number of host bits (

h

) required for each subnet to accommodate at least 150 usable host IP addresses.

8 host bits are required because

2^8 - 2 = 254

usable host addresses (7 host bits only yield

2^7 - 2 = 126

usable host addresses).

Two IP addresses per subnet are reserved for the network ID and broadcast address.

Calculate the prefix length required for the new subnets.

Prefix length

/24

(

32 \text{ total bits} - 8 \text{ host bits} = 24

IPv4 addresses consist of 32 bits divided between network/subnet and host portions.

Determine the number of subnet bits borrowed from the original /16 network mask.

8 bits borrowed (

24 - 16 = 8

Subtracting the original network prefix length from the new subnet prefix length gives the borrowed bits.

Calculate the total number of equal-sized subnets that can be created.

256 subnets (

2^8 = 256

The number of subnets generated is

2^n

, where

n

is the number of borrowed subnet bits.

Key Concept

IPv4 Subnet Masking and Host Capacity Calculations

Question 1629Question →

An administrator connects two switches, SW-ALPHA and SW-BETA, via an 802.1Q trunk link. Switch SW-ALPHA is configured with an 802.1Q trunking interface where VLAN 50 is explicitly defined as the native VLAN. Switch SW-BETA has a freshly unboxed configuration with default 802.1Q trunking parameters enabled (using VLAN 1 as the native VLAN). Trunking is operational on both sides. Which of the following technical outcomes will occur on this link? (Select TWO.)

Select all that apply

Untagged frames transmitted by SW-ALPHA from VLAN 50 will be placed into VLAN 1 upon arrival at SW-BETA.

Frames tagged for VLAN 20 sent by SW-ALPHA will be dropped by SW-BETA due to the native VLAN configuration mismatch.

Switches will generate CDP or syslog warning messages indicating a native VLAN mismatch condition.

Traffic between devices on VLAN 50 and VLAN 1 will automatically be routed by the switches without requiring a Layer 3 device.

Show answer & explanation

Answer: Untagged frames transmitted by SW-ALPHA from VLAN 50 will be placed into VLAN 1 upon arrival at SW-BETA.; Switches will generate CDP or syslog warning messages indicating a native VLAN mismatch condition.

Answer

Untagged frames transmitted by the first switch from VLAN 50 will be placed into VLAN 1 upon arrival at the second switch, and the switches will generate CDP/syslog warnings indicating a native VLAN mismatch condition.

In 802.1Q trunking, native VLAN traffic is sent untagged across the link. When one switch transmits VLAN 50 traffic untagged and the receiving switch expects VLAN 1 traffic to be untagged, the receiving switch assigns those incoming untagged frames to VLAN 1. Additionally, management protocols such as CDP detect this parameter mismatch across neighboring ports and generate explicit syslog error alerts.

Step-by-Step Solution

Analyze how 802.1Q handles native VLAN traffic.

Frames belonging to the native VLAN are transmitted across the trunk link untagged (without an 802.1Q header).

802.1Q standard dictates that frames belonging to the port's native VLAN carry no 4-byte VLAN tag.

Evaluate ingress frame processing on the receiving switch.

When untagged frames arrive at the second switch, it assigns them to its own configured native VLAN (VLAN 1).

Switches implicitly map all incoming untagged trunk frames to their locally configured native VLAN ID.

Determine impact on tagged traffic and protocol logging.

Tagged traffic (VLAN 20) remains properly tagged and processed, while protocol daemons (like CDP) log native VLAN mismatch errors.

Native VLAN mismatches affect untagged frame mapping and trigger control-plane diagnostics, but do not stop processing of explicitly tagged frames.

Key Concept

802.1Q Native VLAN Operation and Mismatch Behavior

Question 1630Question →

A network engineer is configuring a static host entry on a datacenter firewall for a server deployed on the IPv6 subnet $2001:\text{db8}:\text{cafe}:10::/64$ . The server interface uses the MAC address $00:50:56:\text{A1}:\text{B2}:\text{C3}$ and generates its host portion automatically via the IEEE EUI-64 mechanism. Which of the following represents the fully compliant RFC 5952 IPv6 global unicast address for this server?

2001:db8:cafe:10:250:56ff:fea1:b2c3

2001:db8:cafe:10:50:56ff:fea1:b2c3

2001:db8:cafe:10::250:56ff:fea1:b2c3

2001:db8:cafe:10:0250:56ff:ffff:a1b2:c3

Show answer & explanation

Answer: 2001:db8:cafe:10:250:56ff:fea1:b2c3

Answer

The correct IPv6 address is 2001:db8:cafe:10:250:56ff:fea1:b2c3.

To construct an EUI-64 IPv6 interface identifier from a MAC address (00:50:56:A1:B2:C3), insert FF:FE into the middle (00:50:56:FF:FE:A1:B2:C3) and invert the 7th bit of the first byte (0x00 becomes 0x02). Grouping into 16-bit hextets yields 0250:56ff:fea1:b2c3. Appending this to the subnet prefix 2001:db8:cafe:10::/64 gives 2001:0db8:cafe:0010:0250:56ff:fea1:b2c3. Applying RFC 5952 rules suppresses leading zeros in each hextet (0db8 -> db8, 0010 -> 10, 0250 -> 250), producing 2001:db8:cafe:10:250:56ff:fea1:b2c3.

Step-by-Step Solution

Split the MAC address into two 24-bit halves and insert the hex value FF:FE in the middle.

The MAC address 00:50:56:A1:B2:C3 becomes 00:50:56:FF:FE:A1:B2:C3.

EUI-64 requires expanding a 48-bit MAC address into a 64-bit interface identifier by inserting 16 specific bits (FF:FE).

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

The first byte 0x00 (binary 00000000) becomes 0x02 (binary 00000010). The expanded address becomes 0250:56ff:fea1:b2c3.

IEEE EUI-64 specifies that the Universal/Local bit must be inverted (0 for universally administered MAC addresses changed to 1 for local scope representation).

Combine the network prefix 2001:db8:cafe:10::/64 with the calculated 64-bit interface ID.

Uncompressed full address: 2001:0db8:cafe:0010:0250:56ff:fea1:b2c3.

The first 64 bits represent the network prefix and subnet, and the last 64 bits represent the host interface identifier.

Apply RFC 5952 address compression rules.

2001:db8:cafe:10:250:56ff:fea1:b2c3.

Leading zeros in each hextet must be suppressed (0db8 -> db8, 0010 -> 10, 0250 -> 250). The double colon (::) is not used when no hextet consists entirely of zeros.

Key Concept

EUI-64 Interface Identifier Generation and RFC 5952 Address Compression

Estimated Time:2m 0s

Question 1631Question →

A network administrator is publishing DNS records for a newly deployed corporate service. The domain requires web clients to resolve `www.example.com` to the IPv4 address `192.0.2.50`, and incoming domain email must be routed to `mail.example.com`.

Which TWO of the following DNS resource record types must be configured to achieve this functionality?

Select all that apply

An A record pointing www.example.com to 192.0.2.50

An MX record specifying mail.example.com as the mail exchanger

An AAAA record pointing www.example.com to 192.0.2.50

A PTR record mapping 192.0.2.50 to www.example.com

A CNAME record aliasing www.example.com directly to 192.0.2.50

Show answer & explanation

Answer: An A record pointing www.example.com to 192.0.2.50; An MX record specifying mail.example.com as the mail exchanger

Answer

The configuration requires an A record to map the website hostname to its IPv4 address and an MX record to specify the mail gateway handling domain email.

An Address (A) record is required to map the web server hostname to its 32-bit IPv4 address. A Mail Exchanger (MX) record is required to specify the mail server responsible for accepting incoming email on behalf of the domain.

Step-by-Step Solution

Identify the IPv4 address mapping requirement for the web server hostname.

Recognize that mapping a hostname (`www.example.com`) to a 32-bit IPv4 address (`192.0.2.50`) requires an Address (A) record.

Standard forward IPv4 resolution in DNS uses A records.

Identify the email routing requirement for the domain.

Recognize that designating a mail server (`mail.example.com`) to receive email for the domain requires a Mail Exchanger (MX) record.

Simple Mail Transfer Protocol (SMTP) servers query MX records to determine destination mail handlers.

Key Concept

DNS Host (A) and Mail Exchanger (MX) Record Functions

Estimated Time:1m 0s

Question 1632Question →

A network administrator is configuring a firewall rule to allow secure remote graphical management from an internal administrative jump host to a pool of Windows application servers. To enforce the principle of least privilege, the firewall policy should strictly limit access to the standard Remote Desktop Protocol (RDP). Which transport protocol and destination port combination must be permitted in the firewall policy?

TCP port 3389

TCP port 22

UDP port 3389

TCP port 23

Show answer & explanation

Answer: TCP port 3389

Answer

TCP port 3389 is the standard transport protocol and destination port used for Remote Desktop Protocol (RDP) traffic.

Remote Desktop Protocol (RDP) uses TCP port 3389 by default to establish reliable, connection-oriented virtual graphical desktop sessions between a client and a remote server.

Step-by-Step Solution

Identify the service and management protocol requested in the scenario.

The requirement specifies standard Remote Desktop Protocol (RDP) for graphical server administration.

Different remote administration services use distinct transport layer protocols and default well-known ports.

Determine the transport layer protocol and default well-known port number for RDP.

RDP natively uses TCP as its primary transport protocol and listends on default port 3389.

Connection-oriented delivery guaranteed by TCP is required for control and screen rendering reliability.

Match the identified protocol and port to the corresponding firewall rule option.

Configuring a firewall rule permitting TCP port 3389 satisfies the least privilege requirement for RDP traffic.

Permitting TCP 3389 opens only the strictly required path for RDP management sessions.

Key Concept

Common Ports and Protocols: Remote Desktop Protocol (RDP)

Estimated Time:1m 0s

Question 1633Question →

A network administrator is evaluating dynamic routing protocol behaviors for a campus network infrastructure. Which of the following operational features specifically characterize Open Shortest Path First (OSPF) as a link-state routing protocol? (Select TWO.)

Select all that apply

It constructs a complete topological database of the network area by flooding Link-State Advertisements (LSAs).

It executes Dijkstra's Shortest Path First (SPF) algorithm to calculate the shortest path tree to destination networks.

It encapsulates routing protocol updates directly into Layer 2 Ethernet frames to bypass Layer 3 routing lookup overhead.

It maps multiple private internal host IP addresses to a single external public IP address using transport-layer port translation.

Show answer & explanation

Answer: It constructs a complete topological database of the network area by flooding Link-State Advertisements (LSAs).; It executes Dijkstra's Shortest Path First (SPF) algorithm to calculate the shortest path tree to destination networks.

Answer

The correct characteristics of OSPF as a link-state routing protocol are constructing a complete topological database via Link-State Advertisements (LSAs) and executing Dijkstra's Shortest Path First (SPF) algorithm to determine optimal paths.

The correct options accurately describe how link-state routing protocols function. OSPF routers flood Link-State Advertisements (LSAs) to build a synchronized topological map (LSDB) of the area, and each router independently runs Dijkstra's Shortest Path First (SPF) algorithm on this map to identify the shortest path to each destination network.

Step-by-Step Solution

Identify the fundamental mechanics of link-state routing protocols.

Link-state protocols require routers to advertise the state of their directly connected links to all routers within an area.

Flooding LSAs enables every router in the area to maintain an identical Link-State Database (LSDB).

Determine how link-state protocols compute optimal routes from the topology map.

Each router independently runs Dijkstra's SPF algorithm using the LSDB as the topology map.

This builds a shortest-path tree rooted at the local router to compute loop-free routes to every network prefix.

Evaluate and discard distractor statements that misattribute OSI layer operations or NAT mechanisms.

Discard the statements regarding Layer 2 frame encapsulation and IP address/port translation.

OSPF relies on IP Layer 3 encapsulation (protocol 89), and IP/port translation is an edge NAT feature rather than a dynamic routing mechanism.

Key Concept

Link-State Routing Protocol Mechanics (OSPF LSDB and SPF Algorithm)

Question 1634Question →

A network technician is troubleshooting an inter-switch connectivity issue. Host PC-A (IP address 192.168.10.15/24) connected to switch SW-EDGE-01 in VLAN 10 cannot communicate with Host PC-B (IP address 192.168.10.25/24) connected to switch SW-EDGE-02 in VLAN 10. The two switches are interconnected via an 802.1Q trunk link on interface GigabitEthernet0/24.

Upon inspecting interface GigabitEthernet0/24 on SW-EDGE-01, the technician reviews the following command output:

SW-EDGE-01# show interfaces gigabitethernet0/24 switchport
Name: Gi0/24
Administrative Mode: trunk
Operational Mode: trunk
Administrative Trunking Encapsulation: dot1q
Operational Trunking Encapsulation: dot1q
Negotiation of Trunking: On
Access Mode VLAN: 1 (default)
Trunking Native Mode VLAN: 1 (default)
Administrative Native VLAN tagging: disabled
Voice VLAN: none
Administrative Mode VLANs Allowed: 1-9,11-4094

Which of the following configuration commands on interface GigabitEthernet0/24 of SW-EDGE-01 will resolve the communication failure between PC-A and PC-B?

`switchport trunk allowed vlan add 10`

`switchport trunk native vlan 10`

`switchport mode access`

`switchport access vlan 10`

Show answer & explanation

Answer: `switchport trunk allowed vlan add 10`

Answer

Executing `switchport trunk allowed vlan add 10` on interface GigabitEthernet0/24 appends VLAN 10 to the allowed list, enabling VLAN 10 traffic to traverse the trunk link.

The output demonstrates that interface GigabitEthernet0/24 has an administrative allowed VLAN list of `1-9,11-4094`. Because VLAN 10 is excluded from this list, switch SW-EDGE-01 prunes all ingress and egress frames belonging to VLAN 10 on this trunk interface. Executing the command `switchport trunk allowed vlan add 10` explicitly appends VLAN 10 to the permitted range, allowing traffic between PC-A and PC-B to pass across the switch link.

Step-by-Step Solution

Analyze the output of `show interfaces gigabitethernet0/24 switchport`.

The parameter `Administrative Mode VLANs Allowed` is set to `1-9,11-4094`.

VLAN 10 is omitted from the allowed range, causing the switch to drop all tagged frames for VLAN 10 arriving at or departing from interface GigabitEthernet0/24.

Determine the corrective Cisco IOS switchport configuration command.

Using `switchport trunk allowed vlan add 10` modifies the trunk filter to permit VLAN 10 frames.

Using the `add` keyword ensures that VLAN 10 is appended to the current list of allowed VLANs without replacing the existing list.

Key Concept

802.1Q Trunk Allowed VLAN List Pruning and Configuration

Estimated Time:1m 15s

Question 1635Question →

An enterprise network contains multiple subnets connected through a core Layer 3 switch. Hosts on VLAN 50 ( $10.50.0.0/23$ ) fail to acquire IPv4 address leases from the centralized DHCP server located on VLAN 10 ( $192.168.1.100$ ). A network technician verifies that the DHCP server is operational and reachable via ICMP from the core switch. Where must the `ip helper-address 192.168.1.100` command be configured on the Layer 3 switch to enable DHCP relay functionality for the client hosts?

On the Switch Virtual Interface (SVI) for VLAN 50

On the Switch Virtual Interface (SVI) for VLAN 10

In global configuration mode to enable TCP port 67 relaying across all switch interfaces

On the trunk port interface connecting the switch directly to the remote DHCP server

Show answer & explanation

Answer: On the Switch Virtual Interface (SVI) for VLAN 50

Answer

The helper address command must be configured on the Switch Virtual Interface (SVI) for VLAN 50.

DHCP relay agents must be configured on the ingress Layer 3 interface (default gateway) of the subnet where client hosts reside. In this scenario, hosts on VLAN 50 send broadcast DHCP requests to their gateway (VLAN 50 SVI). Placing `ip helper-address 192.168.1.100` on the VLAN 50 SVI instructs the switch to convert those broadcast UDP port 67 packets into unicast traffic directed to the remote DHCP server at

192.168.1.100

Step-by-Step Solution

Identify the ingress layer 3 boundary for client broadcast traffic.

Client hosts on VLAN 50 transmit DHCPDISCOVER packets as layer 2 broadcasts (

255.255.255.255

), which are restricted to VLAN 50.

Layer 3 boundaries block broadcast traffic by default.

Determine the interface requiring DHCP relay agent capabilities.

The gateway interface for VLAN 50 (SVI for VLAN 50) must receive the broadcast and convert it to a unicast DHCPDISCOVER packet sent to

192.168.1.100

The IP helper address must be enabled on the specific interface receiving the client's local broadcast.

Verify protocol and port operation.

The relay agent listens for UDP port 67 client requests on VLAN 50 SVI and populates the `giaddr` field with

10.50.0.1

before unicasting the packet to the server.

The server uses `giaddr` to assign an IP address from the correct pool (

10.50.0.0/23

Key Concept

DHCP Relay Agent Interface Placement

Estimated Time:1m 30s

Question 1636Question →

An organization is configuring a core router that has multiple active routing protocols running simultaneously. The router receives two competing routes for the destination network 192.168.50.0/24. The first route is learned via OSPF with a metric of 45. The second route is learned via EIGRP (Internal) with a metric of 156800. Assuming default administrative distance values are used, which path will the router select for traffic destined to 192.168.50.0/24?

The route learned via EIGRP, because internal EIGRP has a lower administrative distance than OSPF.

The route learned via OSPF, because OSPF has a significantly lower route metric cost than EIGRP.

Both routes equally using Equal-Cost Multi-Path (ECMP) load balancing across both protocols.

Neither route, because competing routing protocols destination prefixes create an invalid loop condition.

Show answer & explanation

Answer: The route learned via EIGRP, because internal EIGRP has a lower administrative distance than OSPF.

Answer

The router will select the route learned via EIGRP because its default Administrative Distance (90) is lower than OSPF's default Administrative Distance (110).

When a router receives identical network destination prefixes from two different dynamic routing protocols, it compares their Administrative Distance (AD) values. Default AD for internal EIGRP is 90, and for OSPF it is 110. Because lower AD indicates higher trustworthiness, the router selects the EIGRP path regardless of raw metric values.

Step-by-Step Solution

Identify the destination prefixes and compare their subnet mask lengths.

Both protocols report the identical network prefix length (192.168.50.0/24), so longest prefix match does not eliminate either route.

Longest prefix match is always the first evaluation criteria when choosing between routes.

Compare the Administrative Distance (AD) of the competing dynamic routing protocols.

Internal EIGRP has a default AD of 90, while OSPF has a default AD of 110.

When different routing protocols advertise the exact same network prefix, the router uses Administrative Distance to determine source trustworthiness before comparing metrics.

Select the preferred route based on the lower Administrative Distance.

The EIGRP route (AD 90) is selected and placed into the IP routing table.

A lower Administrative Distance indicates a more trustworthy routing source.

Key Concept

Administrative Distance vs. Routing Metric in Route Selection

Question 1637Question →

A network technician is provisioning a new subnet for an office department that requires static IPv4 addresses for $50$ workstation computers and $4$ network printers. Which of the following CIDR prefix lengths represents the smallest subnet mask that will accommodate all required devices while conserving address space?

/27

/26

/25

/28

Show answer & explanation

Answer: $/26$

Answer

The prefix length

/26

is the smallest subnet mask that provides enough usable host addresses (

62

) for the

54

required devices.

The correct answer specifying

/26

allocates

6

bits for the host portion of the address space (

32 - 26 = 6

). This creates

2^6 = 64

total IP addresses. Subtracting

2

addresses for the network ID and broadcast address results in

62

usable IP addresses, which efficiently accommodates the required

54

devices (

50

workstations +

4

printers) with minimal waste.

Step-by-Step Solution

Calculate total required host addresses

50 \text{ workstations} + 4 \text{ printers} = 54 \text{ host addresses}

Both workstations and printers require individual usable host IP addresses on the network.

Determine the required number of host bits (

h

)

For

h = 6

host bits, total addresses =

2^6 = 64

. Usable addresses =

64 - 2 = 62

. Since

62 \ge 54

6

host bits are required.

The formula for usable hosts is

2^h - 2

, reserving one address for the network ID and one for the broadcast address.

Calculate the CIDR prefix length

Prefix length =

32 - 6 = 26

, denoted as

/26

IPv4 addresses consist of 32 total bits. Subtracting the host bits from 32 gives the network prefix length.

Key Concept

Calculating host requirements and CIDR prefix masks for IPv4 subnets

Estimated Time:1m 30s

Question 1638Question →

A network administrator is deploying a Voice over IP (VoIP) system across an enterprise network. The administrator needs to configure a DNS record that allows client IP phones to dynamically discover the hostname and specific port number of the Session Initiation Protocol (SIP) server. Which DNS resource record type should the administrator create to fulfill this requirement?

An SRV record

A PTR record

A CNAME record

An MX record

Show answer & explanation

Answer: An SRV record

Answer

An SRV record is the correct DNS record type for specifying service hostnames and port numbers.

SRV (Service Location) records enable clients to discover hostnames and designated port numbers for specific services and protocols (such as SIP or LDAP).

Step-by-Step Solution

Identify the technical requirements of the scenario.

The requirement is to map a network service (SIP for VoIP) to its hostname, priority, weight, and port number in DNS.

Client applications need both server identification and destination port information for service auto-discovery.

Evaluate standard DNS record capabilities.

SRV records are designed specifically to provide service location data including port numbers and protocols.

Standard host records (like A or CNAME) only provide IP addresses or aliases without port details.

Key Concept

DNS Service Location (SRV) Records

Estimated Time:45s

Question 1639Question →

A network administrator is analyzing packet captures on a core router interface to troubleshoot routing updates. The administrator observes routing protocol packets being encapsulated directly into Layer 3 IP payloads using IP protocol number 89, without using Layer 4 Transport Layer headers such as TCP or UDP. Which of the following dynamic routing protocols is generating these packet captures?

Open Shortest Path First (OSPF)

Border Gateway Protocol (BGP)

Routing Information Protocol (RIPv2)

Enhanced Interior Gateway Routing Protocol (EIGRP) over TCP

Show answer & explanation

Answer: Open Shortest Path First (OSPF)

Answer

Open Shortest Path First (OSPF) is the dynamic routing protocol that encapsulates its protocol messages directly into Layer 3 IP payloads using IP protocol number 89.

Open Shortest Path First (OSPF) formats its protocol messages directly inside IP packets using IP protocol number 89. It does not use Layer 4 protocols like TCP or UDP, managing packet delivery reliability internally through link-state acknowledgments.

Step-by-Step Solution

Identify the OSI layer encapsulation characteristics described in the scenario.

The routing protocol operates directly above Layer 3 (IP) with IP protocol number 89 and does not use a Layer 4 transport protocol (TCP or UDP).

Different dynamic routing protocols operate at different layers of the network protocol stack.

Evaluate the protocol encapsulation mechanisms of the listed dynamic routing protocols.

OSPF uses IP protocol number 89 directly over IP, BGP uses TCP port 179, and RIPv2 uses UDP port 520.

Knowing protocol numbers and transport mechanisms helps distinguish protocol behaviors during packet analysis.

Select the matching routing protocol.

Open Shortest Path First (OSPF) matches IP protocol number 89 direct IP encapsulation.

OSPF implements its own reliable transport mechanism via link-state acknowledgments rather than relying on TCP or UDP.

Key Concept

Dynamic Routing Protocol Encapsulation and Layer 3 Operations

Question 1640Question →

A network administrator is allocating IP addresses for a newly created VoIP segment assigned the IPv4 subnet block $172.31.80.0/21$ . What is the broadcast address of this subnet?

Show answer & explanation

Answer: 172.31.87.255

Answer

172.31.87.255

For the subnet

172.31.80.0/21

, the subnet mask is

255.255.248.0

. The block size in the third octet is 8. The network range extends from

172.31.80.0

172.31.87.255

, making

172.31.87.255

the broadcast address.

Step-by-Step Solution

Determine the subnet mask and block size for a /21 prefix.

A /21 prefix corresponds to a subnet mask of

255.255.248.0

. The block size in the third octet is

256 - 248 = 8

Prefix length /21 leaves 3 subnet bits in the third octet (

2^3 = 8

block size).

Calculate the next subnet network address.

Adding the block size of 8 to the third octet of

172.31.80.0

yields the next network address:

172.31.88.0

The next subnet boundary defines the end of the current subnet range.

Subtract 1 from the next network address to find the broadcast address.

172.31.88.0 - 1 = 172.31.87.255

The broadcast address is the highest address within the subnet range where all host bits are set to 1.

Key Concept

Subnet Boundary and Broadcast Address Calculation

Estimated Time:1m 0s

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