Network Fundamentals

398 questions

All types398 Multiple choice205 Multiple select92 Numeric18 Matching68 Ordering7 Short answer8

A network engineer deploys a Layer 2 switch to connect 24 host workstations within a single IP subnet. The switch connects upstream directly to a single Ethernet interface on a enterprise router. Assuming standard full-duplex switch port configurations and no VLAN segmentation, how do the Layer 2 switch and the router function regarding collision domains and broadcast domains for this segment?

The Layer 2 switch provides an independent collision domain on each port, creating 25 total collision domains, while the router interface bounds the segment into a single broadcast domain.

The Layer 2 switch divides the connected hosts into 25 distinct broadcast domains, while the router interface consolidates all traffic into a single collision domain.

The Layer 2 switch maintains a single collision domain across all ports, while the router interface creates 25 individual broadcast domains.

Both the Layer 2 switch and the router maintain a single shared collision domain and a single shared broadcast domain across the entire subnet.

Show answer & explanation

Answer: The Layer 2 switch provides an independent collision domain on each port, creating 25 total collision domains, while the router interface bounds the segment into a single broadcast domain.

Answer

The Layer 2 switch isolates collision domains per port (25 total), while the router interface defines a single broadcast domain boundary.

Layer 2 switches process frames at the Data Link layer, providing dedicated bandwidth and isolating collision domains on each individual port. Because there are 24 host links plus 1 router connection, 25 collision domains exist. Routers operate at Layer 3 and stop Layer 2 broadcasts from traversing to other interface segments, creating a single broadcast domain boundary for the subnet.

Step-by-Step Solution

Analyze the operational role of the Layer 2 switch regarding collision domains.

Each micro-segmented port on a Layer 2 switch acts as its own independent collision domain. With 24 hosts and 1 router uplink connected across 25 active ports, there are 25 distinct collision domains.

Layer 2 switches buffer frames and operate at the Data Link layer to prevent packet collisions across separate physical ports.

Analyze the operational role of the Layer 2 switch and Router regarding broadcast domains.

Layer 2 switches forward broadcast frames (FF:FF:FF:FF:FF:FF) out all switch ports within the default VLAN (VLAN 1), keeping all 24 hosts and the router interface in one broadcast domain. The router interface drops Layer 2 broadcasts by default and does not forward them to other network segments.

Routers operate at Layer 3 and establish the boundary of a broadcast domain.

Key Concept

Collision and Broadcast Domain Separation across Layer 2 Switches and Routers

Question 122Question →

Match each transport layer characteristic or mechanism on the left to its corresponding protocol behavior on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Sliding Windowing pairs with dynamically regulating data transmission volume based on receiver buffer capacity; 8-Byte Header Overhead pairs with minimizing protocol processing delay and encapsulation size for low-latency applications; Three-Way Handshake pairs with synchronizing initial sequence numbers and establishing session state prior to data exchange; Best-Effort Stateless Delivery pairs with transmitting individual datagrams independently without session management or delivery confirmation.

TCP relies on connection-oriented mechanisms including sequence number synchronization via a three-way handshake (SYN, SYN-ACK, ACK) and dynamic sliding window flow control to manage receiver buffers. Conversely, UDP is connectionless and lightweight, utilizing a minimal fixed 8-byte header and best-effort delivery without state tracking or retransmission overhead.

Step-by-Step Solution

Identify the connection-oriented reliability and flow control features associated with TCP.

Sliding Windowing regulates buffer data rate (TCP Flow Control), and the Three-Way Handshake synchronizes sequence numbers during session setup (TCP Connection Establishment).

TCP requires explicit state management, sequence verification, and flow negotiation between endpoints.

Identify the connectionless, low-overhead characteristics associated with UDP.

An 8-Byte Header Overhead minimizes processing latency, and Best-Effort Stateless Delivery transmits data without acknowledgments or session tracking.

UDP trades error recovery and session control for speed and reduced encapsulation header space.

Map each left transport layer term to its matching operational behavior on the right.

left_1 matches right_1, left_2 matches right_2, left_3 matches right_3, and left_4 matches right_4.

Matches align directly with TCP stateful reliability features versus UDP lightweight connectionless characteristics.

Key Concept

Transport Layer TCP Connection-Oriented Reliability vs UDP Connectionless Overhead

Question 123Question →

An enterprise network architect is designing a Variable Length Subnet Mask (VLSM) address allocation plan starting at the base private IPv4 block $10.200.16.0/20$ . Three internal subnets must be allocated sequentially in contiguous order from the lowest available boundary:

- Subnet A (WLAN Clients): Requires at least $1,000$ usable host IP addresses
- Subnet B (VoIP Phones): Requires at least $450$ usable host IP addresses
- Subnet C (Management): Requires at least $60$ usable host IP addresses

If Subnet A is allocated first, followed immediately by Subnet B, and then Subnet C, what is the valid usable host IPv4 address range for Subnet C?

10.200.22.1 to 10.200.22.62

10.200.22.0 to 10.200.22.63

10.200.20.1 to 10.200.20.62

10.200.21.1 to 10.200.21.62

Show answer & explanation

Answer: 10.200.22.1 to 10.200.22.62

Answer

The valid usable host IPv4 address range for Subnet C is 10.200.22.1 to 10.200.22.62.

To calculate VLSM subnets sequentially from the base block 10.200.16.0/20:
1. Subnet A requires 1,000 usable host IPs. The smallest power of 2 minus 2 that satisfies this is 2^10 - 2 = 1,022 (10 host bits, /22 prefix). Allocated block: 10.200.16.0 to 10.200.19.255.
2. Subnet B starts at 10.200.20.0 and requires 450 usable host IPs. 2^9 - 2 = 510 (9 host bits, /23 prefix). Allocated block: 10.200.20.0 to 10.200.21.255.
3. Subnet C starts at 10.200.22.0 and requires 60 usable host IPs. 2^6 - 2 = 62 (6 host bits, /26 prefix). Network ID is 10.200.22.0 and broadcast is 10.200.22.63.
Thus, the usable host range for Subnet C is 10.200.22.1 through 10.200.22.62.

Step-by-Step Solution

Calculate host bits and prefix length for Subnet A (1,000 hosts).

Host bits needed:

h = 10

(

2^{10} - 2 = 1,022 \ge 1,000

). Prefix length:

/22

(

32 - 10

). Address block:

10.200.16.0/22

, spanning

10.200.16.0

through

10.200.19.255

Subnetting formula

2^h - 2

requires 10 host bits to fit 1,000 usable host addresses.

Determine starting address and size for Subnet B (450 hosts).

Start address:

10.200.20.0

. Host bits needed:

h = 9

(

2^9 - 2 = 510 \ge 450

). Prefix length:

/23

(

32 - 9

). Address block:

10.200.20.0/23

, spanning

10.200.20.0

through

10.200.21.255

Sequential allocation places Subnet B immediately after Subnet A's broadcast boundary.

Determine starting address, size, and usable range for Subnet C (60 hosts).

Start address:

10.200.22.0

. Host bits needed:

h = 6

(

2^6 - 2 = 62 \ge 60

). Prefix length:

/26

(

32 - 6

). Network ID:

10.200.22.0/26

. Broadcast address:

10.200.22.63

. Usable range:

10.200.22.1

10.200.22.62

Subnet C starts at

10.200.22.0/26

. Usable hosts range from Network ID

+ 1

to Broadcast

- 1

Key Concept

Variable Length Subnet Masking (VLSM) host requirement sizing and sequential subnet boundary calculation.

Question 124Question →

A network engineer is configuring a point-to-point connection and internal VLANs within an enterprise infrastructure. An interface is assigned the IPv4 address $10.144.67.100/21$ . What is the decimal value of the third octet of the broadcast address for the subnet to which this host belongs?

Show answer & explanation

Answer: 71

Answer

The third octet of the broadcast address is 71 (broadcast address is 10.144.71.255).

For the prefix /21 (subnet mask 255.255.248.0), the third octet operates with a block size of 8 (256 - 248 = 8). The third octet value 67 falls into the subnet starting at 64 (10.144.64.0). The next subnet starts at 72 (10.144.72.0), making the broadcast address for the current subnet 10.144.71.255. The third octet of this broadcast address is 71.

Step-by-Step Solution

Convert CIDR prefix /21 to dotted-decimal mask.

Mask is 255.255.248.0 (21 network bits: 8 + 8 + 5).

The third octet contains 5 subnet bits and 3 host bits.

Calculate the block size (increment) for the third octet.

256 - 248 = 8.

Subnets increment in steps of 8 in the third octet.

Find the third octet network boundary for IP 10.144.67.100.

Network address is 10.144.64.0.

The greatest multiple of 8 less than or equal to 67 is 64.

Determine the broadcast address of the subnet.

Broadcast address is 10.144.71.255.

The subnet spans from 10.144.64.0 to 10.144.71.255, one value prior to the next subnet 10.144.72.0.

Key Concept

Subnet Boundary and Broadcast Address Calculation

Question 125Question →

A network engineer is interconnecting two legacy network switches via their FastEthernet interfaces. Neither switch interface supports IEEE 802.3ab Auto-MDIX. Which physical cable type and pin mapping must be used on this link to establish Layer 1 connectivity?

A crossover cable with pins 1 and 2 on one end connected to pins 3 and 6 on the opposite end

A straight-through cable with pins 1 and 2 on one end connected to pins 1 and 2 on the opposite end

A rollover cable with pin 1 connected to pin 8 and pin 2 connected to pin 7

A crossover cable with pins 1 and 2 on one end connected to pins 4 and 5 on the opposite end

Show answer & explanation

Answer: A crossover cable with pins 1 and 2 on one end connected to pins 3 and 6 on the opposite end

Answer

A crossover cable with pins 1 and 2 on one end connected to pins 3 and 6 on the opposite end must be used to interconnect the two switches.

When connecting two switch interfaces (both MDI-X) without Auto-MDIX support, an Ethernet crossover cable is mandatory. In FastEthernet (100BASE-TX), pins 1 and 2 transmit data while pins 3 and 6 receive data. The crossover cable swaps pins 1 and 2 on one connector to pins 3 and 6 on the opposite connector so transmit signals reach the receiver on the other side.

Step-by-Step Solution

Identify the device types and port pinouts.

Both switches use MDI-X (Medium Dependent Interface Crossover) port pinouts on their FastEthernet interfaces.

Like network devices (such as switch-to-switch or router-to-PC) operate with identical pin configurations.

Evaluate the requirement for crossover wiring when Auto-MDIX is unavailable.

The transmit pair (pins 1 and 2) of one switch must connect to the receive pair (pins 3 and 6) of the opposing switch.

Without Auto-MDIX, the interface cannot automatically reassign transmit and receive functions dynamically.

Key Concept

Ethernet Crossover Cable Pinout Requirements

Estimated Time:1m 0s

Question 126Question →

A network engineer is configuring the default gateway interface for a enterprise VLAN assigned the IPv4 network block $172.24.128.0/21$ . According to network policy, the default gateway must be assigned the highest usable IPv4 address within the subnet. Which IPv4 address must be assigned to the gateway interface?

172.24.135.254

172.24.135.255

172.24.131.254

172.32.135.254

Show answer & explanation

Answer: $172.24.135.254$

Answer

The IPv4 address

172.24.135.254

must be assigned to the gateway interface.

For a

/21

network (

255.255.248.0

), the third octet increments by 8. Given the network ID

172.24.128.0

, the subnet range encompasses

172.24.128.0

through

172.24.135.255

. The broadcast address is

172.24.135.255

, making

172.24.135.254

the highest usable host IPv4 address.

Step-by-Step Solution

Determine the subnet mask and block size for a

/21

prefix

/21

prefix corresponds to a subnet mask of

255.255.248.0

. The block size in the third octet is

256 - 248 = 8

Calculating the block size identifies the boundary for the network increment.

Calculate the network ID and broadcast address

Starting at network ID

172.24.128.0

, adding the block size of

8

gives the next subnet starting at

172.24.136.0

. Therefore, the broadcast address for this subnet is

172.24.135.255

The broadcast address is the last address in the subnet before the start of the next subnet.

Identify the highest usable host IP address

Subtracting 1 from the broadcast address

172.24.135.255

yields

172.24.135.254

as the highest usable host address.

The network ID (

172.24.128.0

) and broadcast address (

172.24.135.255

) are reserved and cannot be assigned to network interfaces.

Key Concept

IPv4 Subnet Boundaries and Usable Host Range Calculation

Question 127Question →

A network host generates its IPv6 link-local address using stateless address autoconfiguration (SLAAC) and the standard IEEE EUI-64 process. If the Ethernet interface has a burned-in MAC address of `0800.27a4.12b3`, what is the resulting IPv6 link-local address assigned to the interface?

fe80::a00:27ff:fea4:12b3

fe80::800:27ff:fea4:12b3

fe80::a00:27ff:ffa4:12b3

fe80::200:27ff:fea4:12b3

Show answer & explanation

Answer: fe80::a00:27ff:fea4:12b3

Answer

The resulting link-local address is fe80::a00:27ff:fea4:12b3.

The correct option correctly converts the MAC address `0800.27a4.12b3` into an EUI-64 interface identifier by inserting `FFFE` into the middle (`0800:27FF:FEa4:12b3`) and toggling the 7th bit of the first byte (`0x08` to `0x0A`). Prepending the `fe80::` link-local prefix results in `fe80::a00:27ff:fea4:12b3`.

Step-by-Step Solution

Split the 48-bit MAC address into two 24-bit halves and insert FFFE into the midpoint.

The MAC address `08-00-27-a4-12-b3` becomes `0800:27FF:FEA4:12B3`.

EUI-64 requires expanding a 48-bit MAC address into a 64-bit interface identifier by placing FFFE between the OUI and the vendor-assigned payload.

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

First byte `0x08` (`0000 1000` in binary) has its 7th bit inverted to become `0000 1010` (`0x0A`).

IEEE EUI-64 rules dictate that the 7th bit of the first byte must be inverted (0 for global uniqueness becomes 1 for local scope in IPv6).

Prepend the standard link-local network prefix fe80::/10.

The complete IPv6 link-local address is `fe80::a00:27ff:fea4:12b3` (leading zeroes in hex blocks are omitted per standard IPv6 notation).

Link-local addresses generated via SLAAC automatically use the prefix fe80::/10 followed by 54 zero bits and the 64-bit EUI-64 interface identifier.

Key Concept

EUI-64 Interface Identifier Generation and IPv6 Link-Local Addressing

Question 128Question →

An infrastructure team is provisioning a new subnet from the IPv4 address block $172.20.96.0/23$ for a server farm that requires support for up to $250$ usable host interfaces. If the team applies the most efficient prefix length that satisfies this requirement, which IPv4 address is the last usable host address in the first subnet created?

172.20.96.254

172.20.96.255

172.20.97.254

172.32.96.254

Show answer & explanation

Answer: 172.20.96.254

Answer

172.20.96.254

To support

250

usable host interfaces efficiently,

8

host bits are required (

2^8 - 2 = 254

usable hosts), corresponding to a

/24

subnet prefix. The first

/24

subnet carved from the parent block

172.20.96.0/23

has a network address of

172.20.96.0

and a broadcast address of

172.20.96.255

. The last usable host IP address is one position below the broadcast address, which is

172.20.96.254

Step-by-Step Solution

Determine the required host bits and prefix length for 250 hosts.

Using

2^n - 2 \ge 250

n = 8

host bits are needed (

2^8 - 2 = 254

usable addresses). The corresponding prefix length is

32 - 8 = /24

Selecting 8 host bits fulfills the requirement of at least 250 usable hosts with minimum wasted IP space.

Calculate the boundaries for the first /24 subnet from 172.20.96.0/23.

The first subnet is 172.20.96.0/24, spanning network address 172.20.96.0 to broadcast address 172.20.96.255.

A /24 subnet increments in the third octet by 1, starting at the base network address 172.20.96.0.

Identify the last usable host address of the 172.20.96.0/24 subnet.

Subtracting 1 from the broadcast address (172.20.96.255 - 1) gives 172.20.96.254.

Usable host IP addresses lie strictly between the network address and the broadcast address.

Key Concept

IPv4 Subnetting, Prefix Derivation, and Host Range Boundaries

Question 129Question →

A network administrator is configuring an internal host interface and must assign an IP address within the RFC 1918 Class B private address space. Which of the following IP addresses is a valid RFC 1918 private IPv4 address?

172.20.50.1

172.32.1.10

192.169.100.1

11.10.0.1

Show answer & explanation

Answer: 172.20.50.1

Answer

172.20.50.1

The address 172.20.50.1 is a valid RFC 1918 private IPv4 address because it falls within the 172.16.0.0/12 block, which ranges from 172.16.0.0 to 172.31.255.255.

Step-by-Step Solution

Identify the RFC 1918 Class B private address boundaries

RFC 1918 defines Class B private space as 172.16.0.0 through 172.31.255.255 (/12 prefix).

Understanding RFC 1918 ranges is essential for properly distinguishing private non-routable internal addresses from globally routable public addresses.

Evaluate the candidate addresses against the RFC 1918 boundaries

The address 172.20.50.1 falls between 172.16.0.0 and 172.31.255.255. The other addresses (172.32.1.10, 192.169.100.1, 11.10.0.1) fall outside RFC 1918 ranges.

Only addresses within the designated RFC 1918 ranges are non-routable on the public Internet and reserved for private network deployment.

Key Concept

Private IPv4 Address Ranges (RFC 1918)

Estimated Time:45s

Question 130Question →

An enterprise network engineering team is upgrading a data center fabric to support high-density East-West microsegmentation traffic using a Spine-Leaf architecture. Which of the following statements accurately characterize the topology rules and traffic handling of a standard two-tier Spine-Leaf design? (Select TWO.)

Select all that apply

Every leaf switch connects to every spine switch, and spine switches do not connect directly to each other.

Traffic traveling between endpoints connected to different leaf switches experiences a consistent, predictable hop count across the fabric.

End-user servers and virtual machine hypervisors are attached directly to spine switches to maximize uplink bandwidth.

Inter-switch trunk connections between spine switches require PortFast to accelerate spanning tree convergence across the core.

Show answer & explanation

Answer: Every leaf switch connects to every spine switch, and spine switches do not connect directly to each other.; Traffic traveling between endpoints connected to different leaf switches experiences a consistent, predictable hop count across the fabric.

Answer

The correct statements are: 1) Every leaf switch connects to every spine switch, and spine switches do not connect directly to each other; and 2) Traffic traveling between endpoints connected to different leaf switches experiences a consistent, predictable hop count across the fabric.

In a standard two-tier Spine-Leaf architecture, every leaf switch connects to every spine switch, and spine switches do not connect directly to each other. This creates a uniform bipartite topology where any endpoint on a leaf switch reaches an endpoint on another leaf switch in exactly three hops (Leaf -> Spine -> Leaf), yielding consistent and predictable latency for East-West traffic.

Step-by-Step Solution

Analyze Spine-Leaf physical connectivity rules

Confirm that Spine-Leaf fabrics require a bipartite connection pattern where leaf switches connect to all spine switches, while spine-to-spine and leaf-to-leaf direct links are omitted.

This structural pattern allows Equal-Cost Multi-Pathing (ECMP) to load-balance traffic dynamically across all available spine paths.

Evaluate traffic path consistency for East-West data flows

Determine that any leaf-to-leaf communication crosses exactly one spine switch.

Uniform path lengths prevent latency variance for inter-server communication.

Evaluate attachment points for endpoints and hypervisors

Identify that servers, hypervisors, and security appliances attach strictly to leaf switches (access layer of the fabric).

Attaching hosts to spine switches breaks the bipartite Clos architecture and compromises ECMP forwarding.

Key Concept

Spine-Leaf Topology Connectivity & Traffic Dynamics

Estimated Time:1m 30s

Question 131Question →

When an interface automatically constructs an IPv6 EUI-64 interface identifier using the MAC address 00:1A:2B:3C:4D:5E, which string represents the correctly generated 64-bit interface ID?

021a:2bff:fe3c:4d5e

001a:2bff:fe3c:4d5e

021a:2bff:ff3c:4d5e

021a:2b3c:4d5e:fffe

Show answer & explanation

Answer: 021a:2bff:fe3c:4d5e

Answer

The correct interface identifier is 021a:2bff:fe3c:4d5e.

To create a 64-bit EUI-64 interface ID from a 48-bit MAC address, the 16-bit value FFFE is inserted between the third and fourth bytes of the MAC address. In addition, the 7th bit (universal/local bit) of the first byte must be inverted. Inverting the 7th bit of 0x00 results in 0x02, producing the complete interface ID 021a:2bff:fe3c:4d5e.

Step-by-Step Solution

Split the 48-bit MAC address into two 24-bit halves.

The left half is 00:1A:2B and the right half is 3C:4D:5E.

EUI-64 generation requires inserting a specific 16-bit hex value between the OUI and NIC specific bytes.

Insert the 16-bit hexadecimal value FFFE between the two halves.

The combined value becomes 00:1A:2B:FF:FE:3C:4D:5E.

FFFE expands the 48-bit MAC address into a 64-bit structure.

Invert the 7th bit (Universal/Local bit) of the first byte (00).

Hexadecimal 00 (0000 0000 in binary) becomes hexadecimal 02 (0000 0010 in binary).

IEEE EUI-64 rules require modifying the 7th bit to represent universal scope.

Format the resulting 64 bits into standard IPv6 colon-separated hextets.

021a:2bff:fe3c:4d5e

IPv6 interface identifiers are presented as four 16-bit hexadecimal hextets.

Key Concept

EUI-64 Interface ID Generation

Estimated Time:45s

Question 132Question →

A network engineer is configuring internal host interfaces on a private enterprise network. According to RFC 1918, which of the following IP addresses falls within a reserved private IPv4 address block and can be assigned to internal network devices without public Internet routing?

172.20.45.10

172.32.10.1

192.169.1.25

11.10.0.1

Show answer & explanation

Answer: 172.20.45.10

Answer

172.20.45.10 is a valid private IPv4 address as defined by RFC 1918.

The address 172.20.45.10 is part of the RFC 1918 Class B private address range (172.16.0.0/12), which covers all IP addresses from 172.16.0.0 through 172.31.255.255. Since the second octet is 20, it falls within this reserved range and can be used on internal private networks without public routing.

Step-by-Step Solution

Recall the three designated RFC 1918 private IPv4 address ranges

The reserved ranges are: 10.0.0.0/8 (10.0.0.0 – 10.255.255.255), 172.16.0.0/12 (172.16.0.0 – 172.31.255.255), and 192.168.0.0/16 (192.168.0.0 – 192.168.255.255).

RFC 1918 specifies exact address ranges reserved strictly for internal private enterprise use.

Evaluate the second octet of the Class B candidate address 172.20.45.10

The second octet is 20, which lies inclusively between 16 and 31.

Addresses starting with 172 are only private if the second octet is within the 16 to 31 range inclusive.

Verify candidates against public address boundaries

The address 172.20.45.10 is private, while 172.32.10.1, 192.169.1.25, and 11.10.0.1 fall outside the RFC 1918 designated ranges and are public addresses.

Only addresses inside the designated RFC 1918 prefixes are valid private addresses.

Key Concept

RFC 1918 Private IPv4 Address Ranges

Question 133Question →

A network architect is designing an enterprise infrastructure integrating physical networking devices, wireless control elements, and virtualized compute resources. Match each network component on the left to its primary operational function on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Next-Generation Firewall matches application visibility and DPI inspection; Centralized WLC matches CAPWAP, RF management, and centralized authentication; Multilayer Switch matches ASIC-based Layer 2 switching and Layer 3 CEF routing; Type 1 Hypervisor matches running directly on host hardware to virtualize guest compute resources.

Each component is correctly paired with its defining architectural role: NGFW provides Layer 7 deep packet inspection and threat security; WLC centralizes CAPWAP and wireless control plane logic; Multilayer Switches process L2 switching and L3 IP routing at line rate using hardware ASICs; Type 1 hypervisors virtualize physical compute directly on bare metal.

Step-by-Step Solution

Identify the primary role of the Next-Generation Firewall (NGFW).

Map NGFW to stateful security policy enforcement, deep packet inspection (DPI), and Layer 7 application control.

NGFWs extend traditional Layer 3/4 firewall capabilities into full application-layer inspection and threat prevention.

Identify the primary role of the Centralized Wireless LAN Controller (WLC).

Map Centralized WLC to control-plane operations like CAPWAP tunneling, RF management, and client authentication.

Centralized wireless deployment offloads control plane tasks from Lightweight APs to the centralized controller.

Identify the primary role of the Multilayer Switch.

Map Multilayer Switch to wire-speed Layer 2 switching and Layer 3 CEF routing via ASICs.

Multilayer switches handle both broadcast domain switching and inter-VLAN IP routing in hardware.

Identify the primary role of the Type 1 Hypervisor.

Map Type 1 Hypervisor to bare-metal hardware virtualization for guest virtual machines.

Type 1 hypervisors execute directly on host physical hardware rather than on top of an existing host OS.

Key Concept

Operational Roles and Control/Data Plane Functions of Enterprise Network Infrastructure Components

Question 134Question →

A network administrator is diagnosing physical layer and data link layer interface issues on Cisco Catalyst switches. Match each Cisco IOS interface state and CLI error counter signature on the left to its primary physical or configuration root cause on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair: (1) Runts, giants, and alignment errors with physical cable corruption or NEXT; (2) Late collisions and FCS errors with a local half-duplex mismatch; (3) Fiber link down despite active lasers with Tx/Rx polarity reversal; (4) Deferred frames without late collisions or FCS errors with standard half-duplex CSMA/CD backoff.

Each physical symptom maps to its exact root cause: distorted frame dimensions (runts/giants) result from signal corruption or bad cabling; late collisions stem from full/half duplex mismatch timing conflicts; down fiber links with active lasers indicate swapped Tx/Rx strands; and deferred frames without errors represent normal half-duplex CSMA/CD deferral behavior.

Step-by-Step Solution

Analyze symptom 1 (runts, giants, and frame alignment errors).

Corrupted frame sizes and framing errors reflect Layer 1 electrical noise, bad terminations, or cable length violations.

Damaged copper media distorts bits in flight, causing the receiving MAC layer to receive malformed frame lengths.

Analyze symptom 2 (late collisions and FCS errors).

Late collisions indicate a collision occurring after transmitting the first 64 bytes of a frame.

This symptom is indicative of a duplex mismatch where the opposite end is configured for full-duplex and transmits asynchronously.

Analyze symptom 3 (fiber link down/notconnect with active Tx lasers).

Optical transceivers require Tx-to-Rx cross-wiring.

If light is emitting from both ends but link is not established, the optical strands are connected Tx-to-Tx and Rx-to-Rx.

Analyze symptom 4 (high deferred transmission counter without late collisions).

Deferred frames count the number of times an interface delays sending a frame due to carrier detect on a half-duplex link.

This is normal CSMA/CD behavior on a busy segment and does not indicate physical cable or hardware failure.

Key Concept

Cisco IOS Interface Counters and Physical Layer Diagnostics

Question 135Question →

A network engineer is configuring a Cisco router interface with a MAC address of `0050.568A.B1C2`. The interface is enabled for IPv6 using SLAAC with EUI-64 interface identifier generation under the prefix `2001:db8:1:1::/64`. Additionally, the engineer plans to configure a Unique Local Address (ULA) on the same interface. Which TWO statements regarding the IPv6 addressing for this interface are correct? (Select TWO.)

Select all that apply

The dynamically generated link-local IPv6 address on the interface is `fe80::250:56ff:fe8a:b1c2`.

Any valid Unique Local Address configured on this interface must fall within the prefix range `fc00::/7`.

The automatically generated EUI-64 interface identifier derived from the MAC address is `0050:56ff:fe8a:b1c2`.

The `2001:db8:1:1::/64` subnet contains exactly

2^{64} - 2

usable host addresses because the all-zeros address is reserved for the subnet ID and the all-ones address is reserved for the IPv6 broadcast domain.

Show answer & explanation

Answer: The dynamically generated link-local IPv6 address on the interface is `fe80::250:56ff:fe8a:b1c2`.; Any valid Unique Local Address configured on this interface must fall within the prefix range `fc00::/7`.

Answer

The correct statements are that the dynamically generated link-local IPv6 address is `fe80::250:56ff:fe8a:b1c2` and that any Unique Local Address must fall within the `fc00::/7` prefix block.

Generating an EUI-64 address from MAC `0050.568A.B1C2` involves inserting `FFFE` into the middle (`0050:56FF:FE8A:B1C2`) and toggling the 7th bit from 0 to 1 (`00` hex becomes `02` hex), producing interface ID `0250:56ff:fe8a:b1c2`. Prepended with `fe80::`, this forms the valid link-local address `fe80::250:56ff:fe8a:b1c2`. Furthermore, RFC 4193 specifies `fc00::/7` as the designated address range for IPv6 Unique Local Addresses.

Step-by-Step Solution

Calculate the EUI-64 interface identifier from MAC address `0050.568A.B1C2`.

Insert `FFFE` in the middle of the 48-bit MAC address: `0050:56FF:FE8A:B1C2`. Invert the 7th bit (Universal/Local bit) of the first byte: `00` (`0000 0000`) becomes `02` (`0000 0010`). The resulting EUI-64 interface ID is `0250:56ff:fe8a:b1c2`.

IEEE EUI-64 conversion rules mandate inserting `FFFE` between the OUI and NIC specific identifier and setting the 7th bit to 1 for globally unique MAC addresses.

Form the link-local address using the EUI-64 interface ID.

Combine `fe80::/10` with interface ID `0250:56ff:fe8a:b1c2` to get `fe80::250:56ff:fe8a:b1c2`.

Link-local addresses start with the `fe80::/10` prefix followed by 54 zero bits and the 64-bit interface identifier.

Identify the reserved prefix scope for IPv6 Unique Local Addresses (ULA).

ULAs are defined under `fc00::/7`.

RFC 4193 designates `fc00::/7` for Unique Local Unicast addresses, serving a role similar to RFC 1918 private IPv4 addresses.

Key Concept

EUI-64 Interface Identifier Generation & IPv6 Address Scopes

Question 136Question →

Network engineers must evaluate transport layer characteristics to optimize application performance and troubleshoot communication issues across enterprise networks. Match each transport layer protocol feature or header characteristic on the left with its corresponding operational behavior on the right. Which pairs correctly match each feature to its correct transport protocol mechanism?

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings match TCP dynamic windowing and SACK to TCP Reliability and Flow Control Mechanics; the 4-field 8-byte header to UDP Header Structure and Overhead; SYN handshake and ISN synchronization to TCP Connection Establishment and State Tracking; and the connectionless IPv6 checksum requirement to UDP Stateless Operation and IPv6 Checksum Rule.

TCP uses a minimum 20-byte header to carry sequence numbers, acknowledgment numbers, and dynamic window sizes for flow control. UDP uses an 8-byte header containing only source port, destination port, length, and checksum. TCP establishes state using the 3-way handshake with SYN/ACK flags, whereas UDP operates statelessly and requires a checksum in IPv6.

Step-by-Step Solution

Analyze transport layer header size and flow control features.

Identified that a 20-byte base header with dynamic windowing and SACK corresponds to TCP flow control and reliability.

TCP requires a minimum of 20 bytes of header space to store sequence numbers, acknowledgment numbers, flags, and window size.

Analyze UDP header composition.

Identified that the 8-byte header with Source Port, Destination Port, Length, and Checksum defines UDP header structure.

UDP avoids connection states and sequence tracking, allowing its header to remain fixed at 8 bytes (64 bits).

Evaluate connection setup mechanisms.

Matched SYN/SYN-ACK control flags and Initial Sequence Number (ISN) synchronization to TCP session establishment.

TCP must establish bidirectionally synchronized sequence numbers before data transfer can begin.

Examine protocol behavior differences across IP versions.

Matched connectionless delivery with mandatory IPv6 checksum calculation to UDP operation.

Because IPv6 removes the Layer 3 header checksum field, it forces Layer 4 protocols like UDP to perform mandatory checksum calculations for integrity.

Key Concept

TCP vs UDP Header Fields, Connection Management, and IPv4/IPv6 Operational Rules

Question 137Question →

A network engineer is troubleshooting performance issues across several switch interfaces using Cisco IOS CLI outputs. Match each interface status or error counter symptom on the left to its corresponding Layer 1 or Layer 2 root cause on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Each interface symptom matches its specific root cause based on Cisco IOS interface operation: Late collisions match a duplex mismatch; physical down/down with straight-through and disabled Auto-MDIX matches pinout misconfiguration requiring a crossover cable; up/down (disabled) matches Layer 2 framing or keepalive failure; incrementing giants with CRC errors matches oversized frame transmission exceeding MTU.

Each symptom accurately maps to its fundamental physical or data-link root cause: late collisions indicate duplex mismatch; down/down with straight-through cabling on switch-to-switch links without Auto-MDIX indicates incorrect pinout; up/down indicates Layer 2 framing/keepalive failure; and giants indicate frames received that exceed the allowable maximum frame size.

Step-by-Step Solution

Analyze the late collisions symptom

Identified duplex mismatch as the root cause

Late collisions happen when one side sends frames mid-transmission because it is configured as full-duplex while the local side is half-duplex.

Analyze the line status down / line protocol down scenario with Auto-MDIX disabled

Identified pinout misconfiguration requiring a crossover cable

Switch-to-switch links connect MDI-X to MDI-X. Without Auto-MDIX, a straight-through cable causes Tx-to-Tx mapping, resulting in Layer 1 link failure.

Analyze the up / line protocol down (disabled) state

Identified Layer 2 framing or encapsulation mismatch

Layer 1 is active (up), but Layer 2 fails to establish framing or keepalive communication, causing line protocol to be down.

Analyze incrementing 'giants' and CRC counters

Identified oversized frame transmission exceeding MTU limits

'Giants' explicitly refer to frames received over 1518 bytes (or configured MTU), typically caused by jabbering NICs or MTU mismatch.

Key Concept

Cisco IOS Interface Counter Analysis and Physical/Data-Link Troubleshooting

Question 138Question →

A network engineer executes the following configuration commands on a Cisco router interface with a MAC address of $54\text{e}1.\text{ad}12.3456$ :

text
interface GigabitEthernet0/0/0
ipv6 address 2001:db8:abc:10::/64 eui-64
ipv6 address fe80::1 link-local

Which two statements accurately describe the resulting IPv6 addressing and prefix properties for this interface? (Select two.)

Select all that apply

The dynamically generated Global Unicast Address on the interface is

2001:\text{db8}:\text{abc}:10:56\text{e}1:\text{adff}:\text{fe}12:3456/64

The explicitly configured link-local address

\text{fe80}::1

replaces the auto-generated EUI-64 link-local address while staying within the reserved

\text{fe80}::/10

scope.

The automatically generated Global Unicast Address on the interface is

2001:\text{db8}:\text{abc}:10:54\text{e}1:\text{adff}:\text{fe}12:3456/64

because EUI-64 preserves the original MAC address byte values.

The interface prefix length of

/64

allows for

2^{64} - 2

usable host addresses after deducting reserved subnet network and broadcast identifiers.

Show answer & explanation

Answer: The dynamically generated Global Unicast Address on the interface is $2001:\text{db8}:\text{abc}:10:56\text{e}1:\text{adff}:\text{fe}12:3456/64$ .; The explicitly configured link-local address $\text{fe80}::1$ replaces the auto-generated EUI-64 link-local address while staying within the reserved $\text{fe80}::/10$ scope.

Answer

The correct statements are that the dynamically generated Global Unicast Address is 2001:db8:abc:10:56e1:adff:fe12:3456/64, and the explicitly configured link-local address fe80::1 replaces the auto-generated EUI-64 link-local address within the fe80::/10 scope.

The EUI-64 process takes the 48-bit MAC address 54e1.ad12.3456, inserts FFFE in the middle to make it 64 bits, and flips the 7th bit of the first byte from 54 hex to 56 hex, yielding the interface ID 56e1:adff:fe12:3456. Combining this with the configured prefix 2001:db8:abc:10::/64 produces 2001:db8:abc:10:56e1:adff:fe12:3456/64. Additionally, manual link-local configuration overrides the default EUI-64 link-local address and uses the fe80::/10 scope.

Step-by-Step Solution

Calculate the EUI-64 Interface Identifier from the MAC address

Interface ID is 56e1:adff:fe12:3456

Split MAC address 54e1.ad12.3456 into two 24-bit halves (54-E1-AD and 12-34-56). Insert FFFE in the center (54E1:ADFF:FE12:3456). Invert the 7th bit of the first byte: 54 in hex is 01010100 in binary; flipping bit 7 yields 01010110 binary, which is 56 in hex.

Combine prefix with EUI-64 Interface Identifier

Global Unicast Address is 2001:db8:abc:10:56e1:adff:fe12:3456/64

The prefix specified is 2001:db8:abc:10::/64. Appending the 64-bit interface identifier derived in Step 1 forms the full address.

Evaluate link-local address configuration and reserved scope

fe80::1 is assigned as link-local within fe80::/10

Explicit link-local configuration replaces the automatically derived EUI-64 link-local address. The prefix fe80::/10 covers fe80:: through febf::, so fe80::1 is valid.

Key Concept

EUI-64 Interface ID Generation and IPv6 Address Scopes

Estimated Time:2m 0s

Question 139Question →

A Cisco Catalyst switch maintains the following dynamic MAC address table entries across multiple VLANs:

VLAN	MAC Address	Type	Port
10	0050.56a1.1a01	DYNAMIC	GigabitEthernet0/1
10	0050.56a2.2b02	DYNAMIC	GigabitEthernet0/2
20	0050.56a1.1a01	DYNAMIC	GigabitEthernet0/3

A host connected to port GigabitEthernet0/1 sends an ingress Ethernet frame tagged for VLAN 20 with a Source MAC address of $0050.56a1.1a01$ and a Destination MAC address of $0050.56a3.3c03$ .

How does the switch update its MAC address table and handle the forwarding of this ingress frame?

The switch updates the VLAN 20 MAC address table entry for 0050.56a1.1a01 to port GigabitEthernet0/1 while preserving the VLAN 10 entry, and floods the frame out all active VLAN 20 ports except GigabitEthernet0/1.

The switch flushes the VLAN 10 entry for MAC address 0050.56a1.1a01 and drops the frame because the destination MAC address does not exist in VLAN 20.

The switch forwards the frame directly out GigabitEthernet0/3 based on the pre-existing VLAN 20 entry and drops the frame on GigabitEthernet0/1 due to a port mismatch.

The switch updates the VLAN 20 entry to GigabitEthernet0/1 and floods the frame out all active interfaces across all VLANs, including the ingress port GigabitEthernet0/1.

Show answer & explanation

Answer: The switch updates the VLAN 20 MAC address table entry for 0050.56a1.1a01 to port GigabitEthernet0/1 while preserving the VLAN 10 entry, and floods the frame out all active VLAN 20 ports except GigabitEthernet0/1.

Answer

The switch updates the VLAN 20 MAC table entry for source MAC 0050.56a1.1a01 to GigabitEthernet0/1 while preserving the VLAN 10 entry, and floods the unknown unicast frame out all active interfaces belonging to VLAN 20 except the ingress port GigabitEthernet0/1.

The correct response reflects two core switching mechanics: (1) Per-VLAN MAC learning updates the MAC address table for VLAN 20 on port GigabitEthernet0/1 without impacting entries in other VLANs (such as VLAN 10). (2) Destination MAC address lookup for an unlisted MAC address causes the switch to treat the frame as unknown unicast, flooding it out all active ports in VLAN 20 except the port on which the frame was received.

Step-by-Step Solution

Examine the ingress frame's VLAN tag, Source MAC address, and ingress interface.

The ingress frame is tagged for VLAN 20, has Source MAC 0050.56a1.1a01, and arrives on port GigabitEthernet0/1.

Source MAC learning occurs on every incoming frame within the context of the frame's assigned VLAN.

Perform MAC address learning/station move update for VLAN 20.

The MAC table entry for VLAN 20 and MAC 0050.56a1.1a01 is updated from GigabitEthernet0/3 to GigabitEthernet0/1. The VLAN 10 table entry remains intact.

Cisco Catalyst switches maintain independent per-VLAN MAC address tables (PVST / per-VLAN learning context). Updating an entry in VLAN 20 does not purge entries in VLAN 10.

Perform Destination MAC lookup for 0050.56a3.3c03 in VLAN 20.

Destination MAC 0050.56a3.3c03 is not found in the VLAN 20 MAC address table (Unknown Unicast).

Layer 2 switches check the destination MAC address against the forwarding table for the matching VLAN.

Determine the Layer 2 forwarding decision for an unknown unicast frame.

The switch floods the frame out all operational access/trunk interfaces forwarding for VLAN 20, excluding the ingress port GigabitEthernet0/1.

Unknown unicast traffic must be flooded within the broadast domain (VLAN boundary) to reach the intended destination, but ingress interface split-horizon rules prevent returning the frame out its receiving port.

Key Concept

Per-VLAN MAC Address Table Operation and Unknown Unicast Flooding

Question 140Question →

Match each network topology architecture to its corresponding structural design trait and traffic flow optimization behavior.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Routed Access 3-Tier Campus Architecture pairs with terminating VLANs at the access switch using IGP routing. Spine-Leaf Data Center Architecture pairs with predictable single-hop latency via a bipartite non-interconnected tier design. Collapsed Core 2-Tier Architecture pairs with consolidating core routing and aggregation into a unified switch pair. Traditional Layer 2 Access 3-Tier Architecture pairs with extending VLANs across access switches requiring STP loop blocking.

Each topology pattern is accurately matched based on its architectural boundaries and operational characteristics: Routed Access eliminates STP on uplinks via Layer 3 edge routing; Spine-Leaf uses a bipartite design for consistent East-West latency; Collapsed Core combines core and aggregation layers to save cost; and Traditional Layer 2 Access relies on STP to block redundant Layer 2 uplinks.

Step-by-Step Solution

Analyze the Layer 2/Layer 3 boundary for Routed Access.

In a Routed Access model, routing occurs at the access switch, so uplinks are routed point-to-point links. This eliminates Layer 2 loops and Spanning Tree Protocol blocking on uplinks.

Moving the routing boundary to the access switch limits broadcast domains to a single switch.

Evaluate the connectivity rules of Spine-Leaf (Clos) topologies.

Spine-Leaf uses a bipartite graph (leaf switches connect to all spines; spine switches connect to all leaves; no intra-tier links).

This guarantees equal distance (two hops) and equal-cost multi-pathing (ECMP) for East-West server traffic.

Examine the physical consolidation in a Collapsed Core design.

A Collapsed Core merges Distribution (policy/aggregation) and Core (high-speed transport) into a single layer.

This reduces equipment count, cabling complexity, and cost for smaller deployment environments.

Identify the primary operational requirement of Traditional Layer 2 Access designs.

Extending Layer 2 VLANs across multiple access switches to distribution switches creates Layer 2 loops across redundant uplinks.

Spanning Tree Protocol must block redundant paths to maintain a loop-free topology.

Key Concept

Comparison of Enterprise Campus and Data Center Network Topology Architectures

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