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Network Fundamentals

398 questions

All types398Multiple choice205Multiple select92Numeric18Matching68Ordering7Short answer8
Question 361Question →

An administrator is verifying IPv6 stateless address autoconfiguration (SLAAC) on an Ethernet interface with a burned-in MAC address of 0011.2233.4455. Which TWO statements accurately describe the standard EUI-64 interface identifier generation process for this interface? (Select two.)

Select all that apply

Show answer & explanation

Answer: The hexadecimal value fffe is inserted into the middle of the 48-bit MAC address between the third and fourth bytes.; The resulting 64-bit interface identifier derived from the MAC address is 0211:22ff:fe33:4455.

Answer

The correct statements are that the hexadecimal value fffe is inserted into the middle of the MAC address between the third and fourth bytes, and the resulting derived 64-bit interface identifier is 0211:22ff:fe33:4455.
Generating an IPv6 EUI-64 interface identifier from a 48-bit MAC address requires two key steps: inserting the 16-bit hex value fffe into the middle of the MAC address between the 3rd and 4th bytes, and inverting the 7th bit (Universal/Local bit) of the first byte. For MAC 0011.2233.4455, the first byte 0x00 (0000 0000) becomes 0x02 (0000 0010), and inserting fffe produces 0211:22ff:fe33:4455.

Step-by-Step Solution

1
Split the MAC address into two 24-bit halves and insert fffe into the midpoint.
Splitting 0011.2233.4455 into 0011.22 and 33.4455 and inserting fffe yields 0011:22ff:fe33:4455.
IEEE EUI-64 expansion converts a 48-bit MAC address into a 64-bit interface identifier by inserting the 16-bit pattern fffe in the middle.
2
Invert the 7th bit (Universal/Local bit) of the first byte.
First byte 0x00 binary is 0000 0000. Inverting the 7th bit gives 0000 0010, which is 0x02.
IPv6 modifies the 7th bit of the OUI to represent universal scope when derived via EUI-64.
3
Combine the inverted first byte with the rest of the expanded interface identifier.
The final 64-bit interface identifier is 0211:22ff:fe33:4455.
Replacing 00 with 02 in the initial 16-bit block forms the complete, valid EUI-64 host portion.

Key Concept

EUI-64 Interface Identifier Derivation
Question 362Question →

A network engineer is allocated the IPv4 address block 172.16.64.0/21172.16.64.0/21 to provision point-to-point WAN links between branch offices. Each link is configured with a /30/30 subnet mask to conserve address space. What is the maximum number of /30/30 subnets that can be created from this /21/21 block?

Show answer & explanation

Answer: 512

Answer

The maximum number of /30/30 subnets that can be created is 512.
To calculate the total number of subnets created, subtract the starting prefix length from the target prefix length: 30−21=930 - 21 = 9 borrowed subnet bits. Applying the formula 2n2^n yields 29=5122^9 = 512 subnets.

Step-by-Step Solution

1
Calculate the number of borrowed subnet bits.
9 subnet bits borrowed (30−21=930 - 21 = 9).
Subnetting a /21/21 network block into /30/30 subnets requires extending the prefix by 9 bits.
2
Calculate the total number of available subnets.
512 subnets (29=5122^9 = 512).
Each borrowed bit doubles the number of created subnets, yielding 29=5122^9 = 512 total subnets.

Key Concept

Calculating the total number of subnets created by borrowing host bits from an IPv4 CIDR prefix.
Question 363Question →

A network administrator executes the `show ipv6 interface gigabitethernet0/0/0` command on a Cisco IOS router. The interface has a burned-in MAC address of `3c4a.9211.2233` and is configured to generate its link-local address dynamically using the IEEE EUI-64 process. Which IPv6 link-local address will be assigned to this interface?

Show answer & explanation

Answer: fe80::3e4a:92ff:fe11:2233

Answer

fe80::3e4a:92ff:fe11:2233
IPv6 link-local addresses use the FE80::/10 prefix. When generating a 64-bit interface identifier using EUI-64 from a 48-bit MAC address (3c4a.9211.2233), FFFE is inserted between the 3rd and 4th bytes (after 92), and the 7th bit (Universal/Local bit) of the first byte (0x3C = 0011 1100) is inverted to 1 (0x3E = 0011 1110). Combining FE80:: with 3e4a:92ff:fe11:2233 produces fe80::3e4a:92ff:fe11:2233.

Step-by-Step Solution

1
Identify the IPv6 address scope prefix required for interface link-local configuration.
Link-local addresses always begin with the prefix fe80::/10.
IPv6 interfaces automatically assign a link-local address in the fe80::/10 range when IPv6 is enabled.
2
Insert the 16-bit hex value FFFE into the middle of the 48-bit MAC address (3c4a.9211.2233).
Splitting 3c4a.92 and 11.2233 yields 3c4a:92ff:fe11:2233.
EUI-64 expands a 48-bit MAC address into a 64-bit interface identifier by inserting FFFE (1111 1111 1111 1110) in the middle.
3
Invert the 7th bit (Universal/Local bit) of the first byte of the MAC address.
First byte 0x3C in binary is 0011 1100. Inverting the 7th bit (from left) gives 0011 1110, which is 0x3E in hexadecimal.
IEEE EUI-64 rules dictate that the 7th bit of the first byte must be inverted (0 for globally unique/burned-in MACs becomes 1 to indicate local scope in the EUI-64 format).
4
Combine the link-local prefix with the resulting 64-bit EUI-64 interface identifier.
fe80::3e4a:92ff:fe11:2233
Concatenating the prefix fe80:: and the interface identifier 3e4a:92ff:fe11:2233 forms the final IPv6 link-local address.

Key Concept

EUI-64 Interface Identifier Derivation and Link-Local Address Formation
Question 364Question →

Network physical interfaces use specific transceiver standards designed for different cabling media, connector types, and maximum distances. Match each Ethernet transceiver standard to its corresponding physical medium and operational distance specification.

Click a left item, then click its matching right item

Items

10GBASE-LR
1000BASE-SX
100GBASE-SR4
10GBASE-T

Matches

Show answer & explanation

Answer

10GBASE-LR matches Single-Mode Fiber (SMF) using LC connectors up to 10 km at 1310 nm; 1000BASE-SX matches Multimode Fiber (MMF) using LC connectors up to 550 m over OM3; 100GBASE-SR4 matches Multimode Fiber (MMF) using MPO/MTP multi-fiber connectors up to 100 m; 10GBASE-T matches Category 6A (Cat6a) balanced twisted-pair copper cabling up to 100 m.
Each Ethernet optical and copper transceiver designation specifies its exact physical media type, wavelength, connector, and maximum supported distance according to IEEE 802.3 standards.

Step-by-Step Solution

1
Identify the optical fiber classification and range for 10GBASE-LR
The 'LR' stands for Long Range, which utilizes 1310 nm lasers over single-mode fiber (SMF) up to 10 km.
Single-mode fiber is required for long-range high-bandwidth transmission due to minimal modal dispersion.
2
Identify the optical fiber classification and range for 1000BASE-SX
The 'SX' suffix denotes short-wavelength 850 nm optics designed for multimode fiber (MMF) reaching up to 550 m over OM3 grade fiber.
Multimode fiber utilizes LED or VCSEL light sources for shorter intra-building runs.
3
Identify the multi-lane 100G fiber specification for 100GBASE-SR4
The 'SR4' suffix indicates short-range 4-lane optics requiring parallel MPO/MTP multi-fiber push-on connectors up to 100 m.
Parallel optics aggregate four 25 Gbps lanes across separate fiber strands inside a ribbon cable.
4
Identify the copper Ethernet standard for 10GBASE-T
The 'T' suffix indicates twisted-pair copper media using standard RJ-45 connectors, specifically requiring Cat6a for full 100-meter 10 Gbps runs.
Category 6A cabling mitigates alien crosstalk required for 10 Gbps operation up to 100 meters.

Key Concept

Ethernet Physical Layer Standards and Media Specifications
Question 365Question →

A network host sends a unicast Ethernet frame to a Layer 2 switch. The source MAC address of the frame is not currently present in the switch's MAC address table, but the destination MAC address is already mapped to an active interface. In what sequence does the switch perform operations to process and forward this frame? Arrange the switch operations in the correct chronological order from first to last.

Drag items to arrange them in the correct order

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Answer

The correct sequence of switch operations is: receiving the frame on the ingress interface, examining and recording the source MAC address in the MAC address table, looking up the destination MAC address in the table, identifying the corresponding egress interface, and forwarding the unicast frame directly out that egress interface.
When an Ethernet frame arrives on a switch port, the switch first learns or updates the mapping of the frame's source MAC address to the ingress interface in its MAC address table. Next, it performs a lookup for the destination MAC address. Because the destination MAC address is already known, the switch identifies the single mapped egress interface and forwards the frame directly out that port without flooding.

Step-by-Step Solution

1
Frame Ingress
The Ethernet frame enters the switch port buffer.
Layer 2 processing begins as soon as the physical frame is received on an interface.
2
Source MAC Learning
The unknown source MAC address is mapped to the ingress port in the MAC address table.
Switches continuously build their MAC address table by examining the source MAC address of every incoming frame.
3
Destination MAC Lookup
The switch searches its MAC address table for the destination MAC address.
Forwarding decisions are determined by comparing the destination MAC address against existing table entries.
4
Egress Port Selection
The switch identifies the target egress port linked to the destination MAC address.
A valid table match maps the destination MAC address directly to a specific physical or logical interface.
5
Unicast Forwarding
The frame is transmitted out the target egress port.
Known unicast frames are forwarded strictly out the designated interface rather than being flooded out all ports.

Key Concept

Layer 2 MAC Address Table Learning and Known Unicast Forwarding Logic
Question 366Question →

A network technician is troubleshooting a newly configured workstation that is unable to communicate with its local gateway or external networks. The workstation interface has been manually assigned the IPv4 address 172.31.25.127172.31.25.127 with a prefix length of /26/26. The default gateway interface is configured as 172.31.25.65/26172.31.25.65/26. What is the primary reason for the connectivity failure on this workstation?

Show answer & explanation

Answer: The assigned address 172.31.25.127 is the subnet broadcast address for the 172.31.25.64/26 subnetwork and cannot be assigned to an end-device host interface.

Answer

The assigned address 172.31.25.127 is the subnet broadcast address for the 172.31.25.64/26 subnetwork and cannot be assigned to an end-device host interface.
For a /26 prefix, the subnet block size is 64. The subnetwork containing the gateway 172.31.25.65 starts at network ID 172.31.25.64 and ends at broadcast address 172.31.25.127. Usable host IPs span 172.31.25.65 to 172.31.25.126. Assigning 172.31.25.127 to a host attempts to use the subnetwork broadcast address, which is invalid for host interfaces.

Step-by-Step Solution

1
Determine the subnet mask block size for a /26 prefix.
A /26 prefix corresponds to a subnet mask of 255.255.255.192. The block size in the 4th octet is 256−192=64256 - 192 = 64.
Calculating the block size identifies the subnetwork boundaries.
2
Identify the subnetwork boundaries containing IP address 172.31.25.65.
Subnet multiples of 64 are 0, 64, 128, 192. The IP address 172.31.25.65 falls into the subnetwork starting at 172.31.25.64.
Determines the network ID for the subnetwork.
3
Determine the usable host range and broadcast address for 172.31.25.64/26.
Network ID: 172.31.25.64. Broadcast address: 172.31.25.64+63=172.31.25.127172.31.25.64 + 63 = 172.31.25.127. Usable host range: 172.31.25.65 to 172.31.25.126.
Verifies whether 172.31.25.127 is a valid host address.
4
Evaluate the validity of assigning 172.31.25.127 to a workstation.
Because 172.31.25.127 is the broadcast address, host network stacks will reject or fail to operate correctly with this IP assignment.
Broadcast addresses are reserved for sending packets to all hosts on the subnetwork.

Key Concept

Subnet Boundary and Broadcast Address Identification
Question 367Question →

A network technician needs to connect a workstation's Ethernet network interface card directly to a Cisco router's GigabitEthernet interface for initial device staging without using an intermediate switch. Auto-MDIX has been manually disabled on the router interface. Which cable type is required to successfully establish Layer 1 link integrity between these two devices?

Show answer & explanation

Answer: A crossover cable with all four wire pairs transposed between opposing connectors

Answer

A crossover cable with all four wire pairs transposed between opposing connectors
The correct option is the choice describing a crossover cable with all four wire pairs transposed. Ethernet interfaces on end-user PCs and routers both use standard MDI pinouts. When connecting two MDI devices directly without an intervening switch (MDI-X) and with Auto-MDIX disabled, the physical transmit conductors of one device must connect to the receive conductors of the other. Furthermore, 1000BASE-T GigabitEthernet utilizes all four wire pairs for data, requiring all four pairs to be crossed over.

Step-by-Step Solution

1
Identify the pinout role (MDI vs MDI-X) of each connected device
Both the host workstation NIC and the Cisco router interface are classified as MDI (Media Dependent Interface) devices.
MDI devices transmit data on specific pins and receive on others. Like devices (MDI-to-MDI or MDIX-to-MDIX) transmit on the same pins.
2
Determine the required cable configuration based on Auto-MDIX state
Because Auto-MDIX is disabled, a crossover cable must be used to swap the transmit and receive line pairs.
Without Auto-MDIX automatically remapping signals in hardware, a straight-through cable results in transmit signals connecting directly to transmit signals on the opposite end.
3
Account for the interface speed specification (GigabitEthernet / 1000BASE-T)
All four wire pairs (1/2, 3/6, 4/5, 7/8) must be transposed.
Unlike 10/100 Mbps Ethernet which uses two pairs, 1000BASE-T operates over all four copper pairs simultaneously.

Key Concept

MDI/MDI-X pinouts and 1000BASE-T crossover requirements
Estimated Time:1m 0s
Question 368Question →

An administrator enables IPv6 autoconfiguration on a network host interface with a burned-in hardware MAC address of 70-69-5A-12-34-B5. Which two statements regarding the generated EUI-64 interface identifier and link-local address are correct?

Select all that apply

Show answer & explanation

Answer: The generated interface identifier is 7269:5aff:fe12:34b5.; The automatically assigned link-local address uses the prefix fe80::/10.

Answer

The correct statements are that the generated interface identifier is 7269:5aff:fe12:34b5 and the link-local address uses the prefix fe80::/10.
Generating an IPv6 EUI-64 interface identifier from a 48-bit MAC address requires splitting the MAC address into two 24-bit halves, inserting fffe into the middle, and inverting the 7th bit (Universal/Local bit) of the first byte. For the MAC 70-69-5A-12-34-B5, the byte 70 (0111 0000) becomes 72 (0111 0010), producing the interface ID 7269:5aff:fe12:34b5. Furthermore, all IPv6 link-local autoconfigured addresses use the standardized prefix fe80::/10.

Step-by-Step Solution

1
Split the 48-bit MAC address into two 24-bit halves
Left half: 70-69-5A, Right half: 12-34-B5
EUI-64 creation requires inserting a fixed 16-bit value between the Organizationally Unique Identifier (OUI) and the NIC-specific identifier.
2
Insert the hexadecimal string fffe between the two halves
7069:5AFF:FE12:34B5
The IEEE EUI-64 standard specifies the insertion of FFFE into the midpoint of a 48-bit MAC address.
3
Invert the 7th bit (Universal/Local bit) of the first byte
Hex 70 (binary 0111 0000) becomes Hex 72 (binary 0111 0010), yielding the interface ID 7269:5aff:fe12:34b5
In EUI-64 formatting, the 7th bit of the first octet is inverted to indicate local scope/modification.
4
Combine the interface identifier with the IPv6 link-local prefix
fe80::7269:5aff:fe12:34b5 (prefix range fe80::/10)
Link-local addresses in IPv6 use the reserved prefix range fe80::/10.

Key Concept

IPv6 EUI-64 Interface Identifier Derivation and Link-Local Addressing
Estimated Time:1m 30s
Question 369Question →

A network switch interface has a burned-in MAC address of `3464.a912.3456`. When calculating an IPv6 address using the standard IEEE EUI-64 process, what is the generated 64-bit interface identifier expressed in standard colon-separated IPv6 hexadecimal notation?

Show answer & explanation

Answer: 3664:a9ff:fe12:3456; 3664:A9FF:FE12:3456

Answer

3664:a9ff:fe12:3456
Converting the MAC address 34-64-A9-12-34-56 to EUI-64 format requires inserting FFFE into the middle (34-64-A9-FF-FE-12-34-56) and inverting the 7th bit of the first byte (0x34 becomes 0x36). Grouping into 16-bit hexadecimal blocks separated by colons yields 3664:a9ff:fe12:3456.

Step-by-Step Solution

1
Split the 48-bit MAC address into two 24-bit halves.
Left half: 34:64:A9, Right half: 12:34:56.
EUI-64 requires inserting a specific 16-bit hexadecimal value into the midpoint of the MAC address.
2
Insert the hexadecimal value FFFE between the two halves.
34-64-A9-FF-FE-12-34-56
IEEE EUI-64 standard specifies the insertion of FFFE to expand a 48-bit MAC address to 64 bits.
3
Invert the 7th bit (Universal/Local bit) of the first byte (0x34).
Binary 0x34 = 0011 0100; inverted 7th bit = 0011 0110 = 0x36. The updated byte sequence is 36-64-A9-FF-FE-12-34-56.
The Universal/Local (U/L) bit is flipped to indicate a globally unique, modified EUI-64 address.
4
Format the resulting 64-bit value into IPv6 colon-separated hexadecimal quads.
3664:a9ff:fe12:3456
IPv6 interface identifiers are written as four 16-bit hexadecimal blocks separated by colons.

Key Concept

EUI-64 Interface Identifier Derivation
Question 370Question →

A network administrator is planning a high-density switch-to-server interconnect within a data center that requires a 10GBASE-T copper Ethernet link over an end-to-end cable run of 70 meters. To guarantee full 10 Gbps performance across this distance without exceeding the standard cable specification, which minimum copper twisted-pair cable category must be specified?

Show answer & explanation

Answer: Category 6A UTP

Answer

Category 6A UTP is the correct answer, as it is the minimum specification required to operate 10GBASE-T up to the full standard distance of 100 meters.
The option specifying Category 6A UTP is correct because Category 6A is rated for 500 MHz bandwidth and explicitly guarantees 10GBASE-T performance up to 100 meters (328 feet). At 70 meters, Category 6A is the minimum category that meets the requirement.

Step-by-Step Solution

1
Identify the performance requirement and link distance.
The scenario requires 10GBASE-T (10 Gbps Ethernet over twisted-pair copper) across a distance of 70 meters.
Cable categories have specific bandwidth ratings and maximum distance thresholds for higher-speed Ethernet standards.
2
Evaluate standard Category 6 limitations for 10GBASE-T.
Category 6 cable can run 10GBASE-T, but only for distances between 37 and 55 meters max due to alien crosstalk issues at higher frequencies (up to 250 MHz).
A 70-meter run exceeds the maximum 55-meter operating envelope for Category 6 at 10 Gbps.
3
Select the minimum category that satisfies the full 70-meter distance.
Category 6A (Augmented Category 6) operates at frequencies up to 500 MHz and features stricter alien crosstalk specifications, supporting 10GBASE-T for up to 100 meters.
Category 6A fulfills the minimum technical requirement for a 70-meter 10GBASE-T connection.

Key Concept

Copper Ethernet Cabling Standards and 10GBASE-T Distance Limits
Question 371Question →

An enterprise network team is allocated the IPv4 address block 10.240.16.0/2010.240.16.0/20 to provision subnets for remote branch offices. Each branch subnet must support at least 60 usable host addresses. What is the maximum number of equal-sized subnets that can be created from this address block?

Show answer & explanation

Answer: 64

Answer

The maximum number of equal-sized subnets that can be created is 64.
To support 60 usable host addresses, a subnet requires 6 host bits because 26−2=622^6 - 2 = 62 usable addresses (25−2=302^5 - 2 = 30 is insufficient). A subnet with 6 host bits corresponds to a prefix length of /26 (32−6=2632 - 6 = 26). Subtracting the original block prefix (/20) from the subnet prefix (/26) yields 6 borrowed subnet bits. Calculating 262^6 gives 64 maximum equal-sized subnets.

Step-by-Step Solution

1
Calculate required host bits for 60 usable hosts
6 host bits (26−2=62≥602^6 - 2 = 62 \ge 60)
Network and broadcast addresses are reserved and cannot be assigned to hosts, requiring subtraction of 2 from total addresses.
2
Determine target prefix length
Prefix length of /26
Subtracting 6 host bits from the 32 bits of an IPv4 address results in a 26-bit network prefix.
3
Calculate borrowed subnet bits
6 subnet bits
The difference between the new prefix (/26) and the allocated prefix (/20) represents the borrowed bits (26−20=626 - 20 = 6).
4
Calculate total usable subnets
64 subnets
The formula 2s2^s, where ss is the number of borrowed bits, gives 26=642^6 = 64 available subnets.

Key Concept

IPv4 Subnetting and Host Requirement Sizing
Question 372Question →

Match each physical network cabling or interface standard on the left with its corresponding physical characteristic or specification on the right.

Click a left item, then click its matching right item

Items

Single-Mode Fiber (SMF) OS2
Multimode Fiber (MMF) OM3
Category 6 UTP (10GBASE-T)
Rollover Cable

Matches

Show answer & explanation

Answer

Single-Mode Fiber (SMF) OS2 matches the 9 µm core diameter specification for long-haul links; Multimode Fiber (MMF) OM3 matches the 50 µm core supporting 10 Gbps up to 300 m; Category 6 UTP (10GBASE-T) matches 10 Gbps support up to 55 m; Rollover Cable matches reversing pin connections for console access.
Each physical media type has distinct physical properties: Single-Mode Fiber (OS2) uses a 9 µm core to eliminate modal dispersion over long distances; Multimode Fiber (OM3) uses a laser-optimized 50 µm core to achieve 10 Gbps over 300 m; Category 6 UTP supports 10 Gbps up to 55 m; and Rollover Cable completely flips the pin connections (pin 1 to pin 8) for out-of-band console access.

Step-by-Step Solution

1
Analyze Single-Mode Fiber (SMF) OS2 characteristics.
Identify that SMF features a small 9 µm core for long-range links with zero modal dispersion.
SMF core size allows only one mode of light propagation, enabling distances of 10 km or more.
2
Analyze Multimode Fiber (MMF) OM3 characteristics.
Identify that OM3 fiber uses a laser-optimized 50 µm core rated for 10 Gbps up to 300 meters.
OM3 fiber is engineered for 850 nm VCSEL lasers to extend 10G reach beyond older OM1/OM2 fibers.
3
Evaluate Category 6 UTP copper constraints for 10GBASE-T.
Match Category 6 UTP to 10 Gbps transmission capped at 55 meters.
Category 6 UTP can support 10 Gbps up to 55 meters, whereas Category 6A extends this reach to 100 meters.
4
Identify the purpose and pinout structure of a Rollover Cable.
Match Rollover Cable to reversing pin connections for switch console management.
Rollover cables flip all 8 pins (pin 1 to 8, 2 to 7) to map DB-9/USB serial adapters to RJ-45 Cisco console ports.

Key Concept

Physical interface cabling specifications, core diameters, distance limits, and pinout roles in Cisco enterprise networks.
Question 373Question →

A network engineer configures SLAAC on a dual-stack Cisco IOS router interface with a burned-in MAC address of `a036.9f12.3456`. Based on standard IEEE EUI-64 calculations, which IPv6 link-local address is automatically assigned to this interface?

Show answer & explanation

Answer: fe80::a236:9fff:fe12:3456

Answer

The automatically assigned link-local address is fe80::a236:9fff:fe12:3456.
The correct answer derives the link-local address by taking the MAC address a036.9f12.3456, inserting FFFE in the middle (a036.9fff.fe12.3456), and inverting the 7th bit of the first byte from 0 to 1 (0xA0 binary 10100000 -> 0xA2 binary 10100010). Combining this 64-bit interface identifier with the link-local prefix FE80:: results in fe80::a236:9fff:fe12:3456.

Step-by-Step Solution

1
Split the MAC address into two 24-bit halves.
High-order OUI: a0:36:9f; Low-order NIC ID: 12:34:56.
EUI-64 formatting requires inserting a 16-bit payload into the middle of a standard 48-bit MAC address.
2
Insert the hexadecimal value FFFE between the two halves.
a0:36:9f:ff:fe:12:34:56.
IEEE EUI-64 expansion rule inserts FFFE (11111111 11111110 binary) at the midpoint.
3
Invert the 7th bit (Universal/Local bit) of the first byte.
First byte 0xA0 (10100000 binary) becomes 0xA2 (10100010 binary), yielding interface ID a236:9fff:fe12:3456.
In EUI-64 interface identifiers derived from IEEE MAC addresses, the 7th bit is flipped to indicate universal scope when set to 1.
4
Combine the calculated 64-bit interface ID with the link-local prefix fe80::/10.
fe80::a236:9fff:fe12:3456.
IPv6 link-local autoconfiguration appends the EUI-64 interface ID to fe80::/10.

Key Concept

Generating an IPv6 EUI-64 interface identifier from a 48-bit MAC address requires inserting FFFE into the middle of the MAC address, flipping the 7th bit (Universal/Local bit) of the first octet, and prepending the fe80::/10 prefix for link-local addresses.
Question 374Question →

A network technician is connecting two legacy Cisco switches together using their 100BASE-TX FastEthernet ports. Auto-MDIX is manually disabled on both switch interfaces. Which unshielded twisted-pair (UTP) cable pinout specification is required to establish successful Layer 1 connectivity between these two switches?

Show answer & explanation

Answer: A crossover cable with transmit pins 1 and 2 on one connector wired to receive pins 3 and 6 on the opposite connector

Answer

A crossover cable with transmit pins 1 and 2 on one connector wired to receive pins 3 and 6 on the opposite connector is required.
Switches default to MDI-X pinouts, transmitting on pins 3 and 6 and receiving on pins 1 and 2. For 100BASE-TX FastEthernet connections between two switches where Auto-MDIX is disabled, pins 1 and 2 on one end must connect to pins 3 and 6 on the other end to link transmit pairs directly to receive pairs.

Step-by-Step Solution

1
Identify the device types being connected and their default interface pinout roles.
Switches use MDI-X pinouts by default, transmitting data on pins 3 and 6 and receiving data on pins 1 and 2.
Connecting identical device types (MDI-X to MDI-X) means both sides transmit on the same pair of pins by default.
2
Determine the impact of disabling Auto-MDIX on both switch interfaces.
Without Auto-MDIX, the interface hardware cannot dynamically swap its internal transmit/receive pin assignments.
Physical pin crossover must be provided by the cabling infrastructure when hardware auto-detection is inactive.
3
Select the proper UTP pinout for 100BASE-TX FastEthernet crossover cabling.
Pin 1 connects to Pin 3, and Pin 2 connects to Pin 6 across the cable connectors.
This bridges the transmit pair of one switch to the receive pair of the opposing switch and vice versa.

Key Concept

Ethernet UTP Cable Pinouts and Auto-MDIX Operations
Estimated Time:1m 0s
Question 375Question →

A network administrator is assigning static IP addresses to devices on a LAN segment allocated the network subnet block 10.88.160.0/2010.88.160.0/20. Which of the following IPv4 addresses are valid, usable host addresses within this subnet? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: 10.88.160.110.88.160.1; 10.88.175.25010.88.175.250

Answer

The IPv4 addresses 10.88.160.110.88.160.1 and 10.88.175.25010.88.175.250 are valid usable host addresses within the 10.88.160.0/2010.88.160.0/20 subnet.
For the subnet 10.88.160.0/2010.88.160.0/20, the usable host address range is 10.88.160.110.88.160.1 through 10.88.175.25410.88.175.254. The IPv4 addresses 10.88.160.110.88.160.1 and 10.88.175.25010.88.175.250 fall directly within this valid host range.

Step-by-Step Solution

1
Determine the subnet mask and block size for the /20 prefix.
A /20 prefix translates to the dotted-decimal subnet mask 255.255.240.0255.255.240.0. In the third octet, the block size is calculated as 256−240=16256 - 240 = 16.
Calculating the block size defines the boundary of consecutive subnets.
2
Calculate the network address, broadcast address, and host range for 10.88.160.0/2010.88.160.0/20.
The subnet starts at Network ID 10.88.160.010.88.160.0. The third octet ranges from 160 up to 160+16−1=175160 + 16 - 1 = 175. The Broadcast address is 10.88.175.25510.88.175.255. The usable host range spans from 10.88.160.110.88.160.1 through 10.88.175.25410.88.175.254.
Usable host addresses consist of all IP addresses strictly between the Network ID and Broadcast address.
3
Evaluate the given choices against the calculated host range.
The addresses 10.88.160.110.88.160.1 and 10.88.175.25010.88.175.250 lie within 10.88.160.1−10.88.175.25410.88.160.1 - 10.88.175.254. The address 10.88.175.25510.88.175.255 is the broadcast address, while 10.88.176.110.88.176.1 belongs to the next network segment (10.88.176.0/2010.88.176.0/20).
Only addresses within the usable range can be assigned to host devices.

Key Concept

Determining usable host address ranges in IPv4 subnetting with CIDR notation.
Estimated Time:1m 30s
Question 376Question →

An interface on a Cisco IOS router is dynamically generating its IPv6 address using the EUI-64 standard. Output from the command `show ipv6 interface GigabitEthernet0/0` reveals that the link-local address assigned to the interface is `fe80::1a2b:3cff:fe4d:5e6f`. Based on this address, what is the burned-in MAC address of the Ethernet interface?

Show answer & explanation

Answer: 182b.3c4d.5e6f

Answer

The MAC address of the interface is 182b.3c4d.5e6f.
To reverse the EUI-64 generation process and recover the original 48-bit MAC address from the interface ID `1a2b:3cff:fe4d:5e6f`: first, remove the inserted `FFFE` bytes from the midpoint to get `1a2b:3c4d:5e6f`. Next, convert the first byte (`1A` hex) to binary (`0001 1010`) and flip the 7th bit (the Universal/Local bit), resulting in `0001 1000` (`18` hex). Assembling the modified octets gives the hardware MAC address `182b.3c4d.5e6f`.

Step-by-Step Solution

1
Extract the 64-bit Interface Identifier from the IPv6 address.
The Interface ID portion of `fe80::1a2b:3cff:fe4d:5e6f` is `1a2b:3cff:fe4d:5e6f`.
In IPv6 link-local addresses, the last 64 bits represent the Interface Identifier.
2
Remove the inserted 16-bit hex sequence `ff:fe` (`FFFE`) from the middle of the Interface ID.
Removing `ff:fe` from `1a:2b:3c:ff:fe:4d:5e:6f` yields 6 bytes: `1a:2b:3c:4d:5e:6f`.
The IEEE EUI-64 process inserts `FFFE` into the middle of a 48-bit MAC address to expand it to 64 bits.
3
Invert the 7th bit (Universal/Local bit) of the first byte.
The first byte `1A` in binary is `0001 1010`. Inverting the 7th bit (2nd bit from the left) changes it to `0001 1000`, which is `18` in hexadecimal.
EUI-64 inverts the Universal/Local bit (7th bit of the first byte) when creating an interface identifier from a MAC address.
4
Format the resulting 48 bits into standard Cisco MAC address notation.
Combining `18`, `2b`, `3c`, `4d`, `5e`, and `6f` produces `182b.3c4d.5e6f`.
Cisco CLI format uses three sets of four hexadecimal digits separated by dots.

Key Concept

Reversing the EUI-64 interface identifier process to determine the burned-in MAC address
Question 377Question →

A network administrator is evaluating physical cabling and optical transceiver specifications for a new campus network deployment. Which two statements accurately describe Ethernet physical interface and cabling requirements? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: 10GBASE-SR transceivers utilize multimode fiber (MMF) with LC connectors for short-distance transmission up to several hundred meters.; 1000BASE-T operation requires all four wire pairs of Category 5e or higher unshielded twisted-pair (UTP) cabling.

Answer

The correct statements are that 10GBASE-SR transceivers utilize multimode fiber (MMF) with LC connectors for short distances, and 1000BASE-T requires all four pairs of UTP cabling.
10GBASE-SR is designed for short-reach multimode fiber up to 300 meters using LC connectors. Additionally, 1000BASE-T standard mandates the use of all four pairs (8 wires) in Cat5e or higher twisted-pair cabling to transmit 1 Gbps data rates.

Step-by-Step Solution

1
Analyze optical transceiver standards and media compatibility
10GBASE-SR specifies Short Range operation over 850 nm multimode fiber using LC connectors, making the 10GBASE-SR MMF statement correct. 1000BASE-SX also uses MMF (not SMF), making the 1000BASE-SX SMF statement incorrect.
Optical optics are differentiated by wavelength, distance capability (SR vs LR/LX), and fiber modal type (MMF vs SMF).
2
Analyze copper twisted-pair cabling specifications and pinouts
1000BASE-T requires all four pairs of Category 5e/6 UTP for full Gigabit transmission. Category 6a supports 10GBASE-T up to 100 meters (not restricted to 37 meters).
Gigabit Ethernet utilizes five-level pulse amplitude modulation (PAM-5) across all four pairs simultaneously.
3
Verify MDI/MDI-X cable requirements
Connecting identical layer devices (switch to switch) when Auto-MDIX is turned off requires crossover wiring so transmit pins connect to receive pins.
Straight-through cables connect MDI to MDI-X devices (e.g., switch to host); like-to-like connections require crossover pinouts.

Key Concept

Physical Ethernet standards specify distinct transceiver optics, fiber types, cable categories, and pinout pairings based on bandwidth and distance requirements.
Question 378Question →

A network technician needs to connect a router FastEthernet interface directly to a server FastEthernet interface without using an intermediate switch. Both device interfaces have Auto-MDIX disabled. The technician terminates the first end of an Ethernet cable using the T568B standard (Pin 1: White/Orange, Pin 2: Orange, Pin 3: White/Green, Pin 6: Green). Which color wiring pinout must be used on the opposite RJ-45 connector to ensure a functional connection?

Show answer & explanation

Answer: Pin 1: White/Green, Pin 2: Green, Pin 3: White/Orange, Pin 6: Orange (T568A standard)

Answer

Terminating the opposite end using the T568A standard (Pin 1: White/Green, Pin 2: Green, Pin 3: White/Orange, Pin 6: Orange) creates a functional crossover cable by crossing transmit pins 1 and 2 with receive pins 3 and 6.
Because both the router and server act as DTE devices, transmitting on pins 1 and 2 and receiving on pins 3 and 6, connecting them directly without Auto-MDIX requires a crossover cable. Terminating one end with T568B and the other with T568A ensures that the transmit pins on one end connect directly to the receive pins on the other.

Step-by-Step Solution

1
Identify the roles of the interconnected devices and Auto-MDIX status
Both the router interface and server interface act as DTE devices that transmit on pins 1 and 2 and receive on pins 3 and 6.
Direct connection between like DTE devices requires a crossover cable when Auto-MDIX is unavailable to automatically swap pin assignments.
2
Determine the standard pinout requirement for an Ethernet crossover cable
One end must be terminated with T568B and the opposing end terminated with T568A.
This configuration swaps pair 2 (orange, pins 1 and 2) with pair 3 (green, pins 3 and 6).
3
Select the correct T568A wire sequence for the second connector
Pin 1: White/Green, Pin 2: Green, Pin 3: White/Orange, Pin 6: Orange.
Matching T568A on the second end aligns the first end's transmit pins with the second end's receive pins.

Key Concept

Ethernet Crossover Cable Pinout (T568A to T568B)
Estimated Time:1m 0s
Question 379Question →

A network administrator is verifying the IPv6 dynamic address configuration on a host interface. The host interface has a burned-in MAC address of `0012.7F34.5678`. Which two statements correctly describe the process and result of generating the interface identifier and IPv6 address using standard EUI-64?

Select all that apply

Show answer & explanation

Answer: The hexadecimal string `FFFE` is inserted into the middle of the 48-bit MAC address to expand it to 64 bits.; The resulting 64-bit interface identifier for the host is `0212:7fff:fe34:5678`.

Answer

The correct statements are that FFFE is inserted into the midpoint of the 48-bit MAC address and that the resulting 64-bit interface identifier is 0212:7fff:fe34:5678.
Generating an IPv6 EUI-64 interface identifier requires two key transformations on a 48-bit MAC address: inserting the hexadecimal value FFFE between the OUI and the vendor-assigned payload (the middle 16 bits) and inverting the 7th bit (Universal/Local bit) of the first byte. For MAC address 0012.7F34.5678, inserting FFFE produces 0012:7FFF:FE34:5678, and flipping the 7th bit converts the first byte from 00 to 02, producing 0212:7fff:fe34:5678.

Step-by-Step Solution

1
Split MAC address and insert FFFE
Splitting `00-12-7F` and `34-56-78` and placing `FF-FE` in the middle yields `00-12-7F-FF-FE-34-56-78`.
EUI-64 standard requires expanding a 48-bit MAC address to 64 bits by inserting FFFE into the middle.
2
Invert the 7th bit (Universal/Local bit)
The first byte `00` in binary is `0000 0000`. Inverting the 7th bit (bit index 1 from right in 0-indexed byte) yields `0000 0010`, which equals `02` in hex.
EUI-64 specifies that the 7th bit of the first byte must be inverted to indicate local scope within the IPv6 address framework.
3
Format as IPv6 Interface Identifier
Combining the modified first byte with the rest gives `0212:7fff:fe34:5678`.
IPv6 interface identifiers are formatted in four 16-bit hexadecimal quads separated by colons.

Key Concept

IPv6 EUI-64 Interface Identifier Derivation
Question 380Question →

A network engineer is allocated the IPv4 address block 10.200.0.0/1910.200.0.0/19 to provision dedicated subnets for IP surveillance cameras across multiple industrial facilities. Each facility subnet must be configured using a 255.255.255.192255.255.255.192 subnet mask. How many distinct subnets can be created from this block?

Show answer & explanation

Answer: 128

Answer

128 subnets can be created.
To determine the number of subnets created, convert the subnet mask 255.255.255.192255.255.255.192 to /26/26 CIDR notation. Borrowing bits from the original /19/19 prefix yields 26−19=726 - 19 = 7 subnet bits. Calculating 272^7 gives exactly 128 subnets.

Step-by-Step Solution

1
Determine the prefix length corresponding to the given subnet mask.
The dotted-decimal subnet mask 255.255.255.192255.255.255.192 corresponds to a /26/26 prefix (2424 network bits + 22 subnet bits in the fourth octet).
Converting the dotted-decimal mask to CIDR notation establishes the target prefix length.
2
Calculate the number of borrowed subnet bits.
26 bits (new prefix)−19 bits (original prefix)=7 borrowed subnet bits26 \text{ bits (new prefix)} - 19 \text{ bits (original prefix)} = 7 \text{ borrowed subnet bits}.
Subnetting extends the network prefix length from /19/19 to /26/26.
3
Compute the total number of subnets.
27=1282^7 = 128 total subnets.
The formula to calculate the number of subnets generated by borrowing ss bits is 2s2^s.

Key Concept

Subnet Count Calculation
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