Examkin

Network Fundamentals

398 questions

All types398Multiple choice205Multiple select92Numeric18Matching68Ordering7Short answer8
Question 281Question →

A network administrator is inspecting intra-host traffic flows within a Type 1 hypervisor environment. Two virtual machines, VM-A and VM-B, reside on the same physical host and are attached to the same VLAN configured on a virtual switch (vSwitch). When VM-A transmits Ethernet frames destined for VM-B, how does the hypervisor handle the frame forwarding?

Show answer & explanation

Answer: The vSwitch performs Layer 2 forwarding entirely in software within host memory, delivering the frames directly to VM-B without sending them out the physical Network Interface Card (pNIC).

Answer

The vSwitch performs Layer 2 forwarding entirely in software within host memory, delivering the frames directly to VM-B without sending them out the physical Network Interface Card (pNIC).
When virtual machines reside on the same physical host and are attached to the same VLAN on a vSwitch, the vSwitch performs Layer 2 frame switching entirely within host memory. The frames move directly between virtual ports (vNICs) via hypervisor software, so traffic does not exit out to physical network interface cards (pNICs) or external switches.

Step-by-Step Solution

1
Identify the location of the source VM, destination VM, and virtual network configuration.
Both VM-A and VM-B reside on the same physical host and belong to the same VLAN on the internal vSwitch.
Traffic flow destination analysis determines whether physical media must be traversed.
2
Analyze how a virtual switch (vSwitch) operates inside a bare-metal hypervisor.
The vSwitch maintains an internal software MAC address table mapping virtual ports to VM vNIC MAC addresses.
A vSwitch functions like a physical Layer 2 switch, but forwarding occurs via system RAM and bus operations inside the hypervisor.
3
Determine physical network interface card (pNIC) involvement.
Since both virtual interfaces are internal to the host on the same vSwitch and VLAN, frame switching occurs completely within host memory without utilizing uplink pNICs.
Uplink pNICs are only required when traffic is destined for devices outside the physical host.

Key Concept

Virtual switch (vSwitch) intra-host frame forwarding mechanics
Estimated Time:1m 0s
Question 282Question →

An organization is deploying a dedicated network segment for industrial control systems. The network team needs to enforce deep packet inspection, application-level policy enforcement, and threat prevention between this industrial segment and the corporate intranet. Which network component is best suited to perform these specific security functions at the boundary?

Show answer & explanation

Answer: Next-Generation Firewall (NGFW)

Answer

Next-Generation Firewall (NGFW)
A Next-Generation Firewall (NGFW) integrates traditional stateful firewall capabilities with Layer 7 application inspection, intrusion prevention systems (IPS), and threat protection, making it the appropriate choice to protect boundaries between different functional network zones.

Step-by-Step Solution

1
Analyze the operational requirements specified in the scenario
The requirement identifies deep packet inspection, application-level security enforcement, and threat mitigation between two network zones.
Security functions above Layer 3 require specialized appliances capable of inspecting traffic payloads beyond basic header routing.
2
Evaluate the capabilities of network components at boundary points
Next-Generation Firewalls combine stateful inspection with advanced services like application control and intrusion prevention.
Deploying an NGFW at the network boundary ensures policy control and threat protection for incoming and outgoing segment traffic.

Key Concept

Network Security Appliance Roles (NGFW vs. L2 Devices & Hypervisors)
Question 283Question →

A network administrator enters the command `ipv6 address 2001:db8:1111:2222::/64 eui-64` on a Cisco router interface. If the MAC address of the interface is `1422.db88.99aa`, which IPv6 address will be automatically configured on the interface?

Show answer & explanation

Answer: 2001:db8:1111:2222:1622:dbff:fe88:99aa

Answer

The IPv6 address automatically configured on the interface is 2001:db8:1111:2222:1622:dbff:fe88:99aa.
The correct option demonstrates the precise three-step EUI-64 calculation process: splitting MAC address 1422.db88.99aa into two 24-bit halves, inserting FFFE in the center to form 64 bits, and flipping the seventh bit of the first byte (converting 0x14 to 0x16), resulting in 2001:db8:1111:2222:1622:dbff:fe88:99aa.

Step-by-Step Solution

1
Split the 48-bit MAC address into two 24-bit halves.
First half: 14-22-DB, Second half: 88-99-AA.
EUI-64 formatting requires inserting a 16-bit delimiter into the center of a standard 48-bit MAC address.
2
Insert the hex value FFFE between the two halves.
1422:DBFF:FE88:99AA.
FFFE is the standard 16-bit field inserted to expand a 48-bit MAC to a 64-bit interface identifier.
3
Invert the 7th bit (Universal/Local bit) of the first byte.
Hex 14 (binary 00010100) becomes Hex 16 (binary 00010110), yielding the interface ID 1622:dbff:fe88:99aa.
The IEEE EUI-64 standard flips the 7th bit of the OUI to represent universal scope when derived from a MAC address.
4
Combine the /64 IPv6 prefix with the calculated 64-bit interface ID.
2001:db8:1111:2222:1622:dbff:fe88:99aa.
The full 128-bit IPv6 address is constructed by appending the EUI-64 interface ID to the assigned prefix.

Key Concept

Modified EUI-64 Interface Identifier Generation
Question 284Question →

A network technician is installing a new backbone link between two network switches located in separate campus buildings that are 2 kilometers apart. The project specification requires a Gigabit Ethernet connection over fiber-optic cabling. Which combination of optical transceiver standard and fiber cabling type must be selected to support this link?

Show answer & explanation

Answer: 1000BASE-LX transceiver with Single-Mode Fiber (SMF)

Answer

1000BASE-LX transceiver with Single-Mode Fiber (SMF)
The combination of a 1000BASE-LX transceiver and Single-Mode Fiber (SMF) is designed for long-wavelength laser transmission over long distances, supporting Gigabit Ethernet reaches up to 5 km. This fully covers the required 2-kilometer span.

Step-by-Step Solution

1
Identify the required bandwidth and distance specs.
The requirement specifies Gigabit Ethernet (1000 Mbps) over a distance of 2 kilometers.
Choosing physical layer optics requires matching speed rating and distance limitations.
2
Evaluate candidate Ethernet transceiver standards against the 2 km requirement.
1000BASE-LX over Single-Mode Fiber supports Gigabit speeds up to 5 km, whereas 1000BASE-SX maxes out at 550m and copper 1000BASE-T maxes out at 100m.
Single-mode fiber features a tiny core (9μm9\mu m) that minimizes modal dispersion over long distances.

Key Concept

Gigabit Ethernet Fiber Transceiver Specifications (1000BASE-LX vs 1000BASE-SX)
Question 285Question →

During the dynamic creation of an IPv6 interface identifier using the EUI-64 method, a 48-bit MAC address is split into two equal 24-bit halves. Which 16-bit hexadecimal value is inserted directly between these two halves?

Show answer & explanation

Answer: FFFE

Answer

FFFE
The standard IEEE EUI-64 process expands a 48-bit MAC address into a 64-bit interface identifier by inserting the specific 16-bit hexadecimal value FFFE directly into the center between the first 24 bits (OUI) and the last 24 bits (NIC specific extension), followed by inverting the 7th bit (Universal/Local bit).

Step-by-Step Solution

1
Identify the MAC address splitting point
The 48-bit MAC address is divided into two 24-bit (3-byte) halves.
EUI-64 requires extending a 48-bit MAC address to a 64-bit length by inserting 16 bits in the middle.
2
Determine the inserted hexadecimal value
The 16-bit hexadecimal value FFFE is inserted between the two 24-bit halves.
By standard definition (RFC 4291), FFFE expands the 48-bit MAC address to 64 bits.

Key Concept

EUI-64 Interface Identifier Derivation
Question 286Question →

An enterprise network infrastructure relies on dedicated hardware devices to perform specialized functions across the network topology. Match each network component on the left to its primary operational role on the right.

Click a left item, then click its matching right item

Items

Router
Layer 2 Switch
Next-Generation Firewall (NGFW)
Wireless LAN Controller (WLC)

Matches

Show answer & explanation

Answer

Router matches with evaluating packet headers against a routing table to forward traffic between Layer 3 networks and isolate broadcast domains; Layer 2 Switch matches with making frame forwarding decisions based on destination MAC addresses while isolating collision domains; Next-Generation Firewall matches with providing stateful packet inspection and application-level threat prevention; Wireless LAN Controller matches with centralizing management and RF control across lightweight access points.
Each network component is paired according to its core architectural role in an enterprise network: Routers perform Layer 3 forwarding and broadcast isolation; Layer 2 Switches perform MAC-based switching within a local broadcast domain; NGFWs enforce boundary security and application awareness; WLCs manage lightweight access points and radio frequencies centrally.

Step-by-Step Solution

1
Analyze the primary operation of a Router
Identified as a Layer 3 device that routes between IP networks and terminates broadcast domains.
Routers forward packets using destination IP addresses and routing table lookups.
2
Analyze the primary operation of a Layer 2 Switch
Identified as a Layer 2 device that switches Ethernet frames using MAC address tables.
Switches separate collision domains on each physical interface within a single broadcast domain.
3
Analyze the primary operation of a Next-Generation Firewall (NGFW)
Identified as a security component providing stateful inspection and deep packet/application monitoring.
NGFWs go beyond basic Layer 3/4 ACLs by inspecting Layer 7 application payloads and threats.
4
Analyze the primary operation of a Wireless LAN Controller (WLC)
Identified as a centralized management device for lightweight access points (LAPs).
WLCs handle control plane activities, radio frequency (RF) management, and LAP provisioning via CAPWAP.

Key Concept

Enterprise Network Components and Functional Roles
Question 287Question →

Match each Ethernet physical interface transceiver standard to its corresponding physical cabling medium, optical wavelength, and maximum distance specification.

Click a left item, then click its matching right item

Items

10GBASE-SR
1000BASE-LX
1000BASE-T
10GBASE-LR

Matches

Show answer & explanation

Answer

10GBASE-SR matches 850 nm OM3 MMF up to 300 m; 1000BASE-LX matches 1310 nm SMF up to 5 km; 1000BASE-T matches 4-pair Cat 5e/6 UTP up to 100 m; 10GBASE-LR matches 1310 nm SMF up to 10 km.
Each physical interface type is uniquely defined by its transmission speed (1G vs 10G), medium (Copper UTP, MMF, or SMF), optical wavelength (850 nm vs 1310 nm), and distance limit (100 m, 300 m, 5 km, 10 km). Matching these characteristics correctly ensures physical layer compatibility and prevents signal degradation over specified link lengths.

Step-by-Step Solution

1
Identify the transmission speed and medium type for each standard.
Standards starting with '1000BASE' denote 1 Gbps, while '10GBASE' standards denote 10 Gbps. 'T' indicates twisted-pair copper, 'SR' indicates short-range multimode fiber, and 'LX'/'LR' denote long-reach single-mode fiber.
Ethernet standard nomenclature directly describes operational speed and physical media characteristics.
2
Map optical wavelength and maximum distance limits to each optic spec.
Short-range multimode optics (SR) utilize 850 nm VCSEL lasers (300 m reach on OM3). Long-range single-mode optics (LX/LR) utilize 1310 nm lasers (5 km for 1G LX, 10 km for 10G LR). Copper Gigabit (1000BASE-T) uses RJ-45 over Cat5e/6 for 100 m.
Wavelength and fiber core design (SMF vs MMF) dictate distance limitations due to modal dispersion and attenuation.

Key Concept

Physical interface transceiver optics and copper cabling distance/speed standards
Question 288Question →

An administrator is inspecting SLAAC autoconfiguration logs on a host workstation connected to an enterprise switch port. The host has dynamically derived the link-local IPv6 address fe80::a8bb:ccff:fe11:2233fe80::a8bb:ccff:fe11:2233 using the standard IEEE EUI-64 process. Which 48-bit MAC address belongs to the physical network interface card of this host?

Show answer & explanation

Answer: aabb.cc11.2233

Answer

The physical MAC address of the host network interface is aabb.cc11.2233.
The MAC address aabb.cc11.2233 is correct because reversing the EUI-64 process involves removing the inserted midpoint sequence FFFE and toggling the 7th bit (Universal/Local bit) of the first byte. The first byte of the interface ID is hex A8 (binary 1010 1000). Toggling the 7th bit (the second least-significant bit of the byte) changes it to binary 1010 1010, which corresponds to hex AA. Combining AA with the remaining bytes after removing FFFE yields aabb.cc11.2233.

Step-by-Step Solution

1
Extract the 64-bit Interface Identifier from the link-local address.
Interface Identifier is a8bb:ccff:fe11:2233a8bb:ccff:fe11:2233.
The prefix fe80::/64fe80::/64 forms the network portion, while the last 64 bits represent the EUI-64 interface ID.
2
Remove the 16-bit ff:feff:fe sequence inserted into the midpoint of the EUI-64 address.
Splitting at ff:feff:fe yields a8:bb:cca8:bb:cc and 11:22:3311:22:33, forming a8bb.cc11.2233a8bb.cc11.2233.
During EUI-64 generation, 0xfffe0xfffe (1616 bits) is inserted between the OUI (first 24 bits) and the NIC-assigned portion (last 24 bits).
3
Invert the 7th bit (Universal/Local bit) of the first MAC octet.
First octet a8a8 (1010 100021010\ 1000_2) with 7th bit inverted becomes aaaa (1010 101021010\ 1010_2). Combining all octets gives aabb.cc11.2233aabb.cc11.2233.
EUI-64 creation toggles the 7th bit (U/L bit) of the first MAC byte. Reversing the operation requires toggling bit 7 back to its original state.

Key Concept

Deriving original MAC address from IPv6 EUI-64 interface identifier by removing FFFE midpoint insertion and toggling the Universal/Local (7th) bit.
Estimated Time:2m 0s
Question 289Question →

Match each network virtualization component or abstraction on the left with its corresponding operational role in an enterprise host environment on the right.

Click a left item, then click its matching right item

Items

Virtual Network Interface Card (vNIC)
Physical Network Interface Card (pNIC)
Virtual Switch (vSwitch)
Type 1 Hypervisor

Matches

Show answer & explanation

Answer

Virtual Network Interface Card (vNIC) matches with the virtual network adapter presented to a guest OS; Physical Network Interface Card (pNIC) matches with the physical hardware adapter connecting to external switches; Virtual Switch (vSwitch) matches with the software-based switching entity in host memory; Type 1 Hypervisor matches with the bare-metal software layer operating directly on server hardware.
Each virtualization element is accurately mapped to its functional role: vNIC provides virtual adapter presentation to guest VMs, pNIC acts as physical network uplink hardware, vSwitch performs host memory frame switching, and Type 1 hypervisor executes directly on bare-metal hardware.

Step-by-Step Solution

1
Analyze guest network adapter abstractions.
Identify that a vNIC is software exposed to the virtual machine operating system to provide virtual MAC and network connectivity.
Virtual machines require software-emulated interface cards to send and receive frames.
2
Analyze host hardware uplinks.
Associate pNIC with the physical host server hardware interface card.
The pNIC serves as the physical bridge carrying vSwitch traffic onto the physical infrastructure.
3
Evaluate internal frame forwarding constructs.
Match Virtual Switch (vSwitch) to software-based switching inside hypervisor memory.
vSwitches manage Layer 2 forwarding between VM interfaces and physical uplink ports within the hypervisor.
4
Determine hypervisor architecture classification.
Match Type 1 Hypervisor to bare-metal execution operating directly on server hardware.
Type 1 hypervisors control host hardware directly rather than running on top of an OS.

Key Concept

Virtualization Fundamentals and Host Components
Question 290Question →

A network engineer is implementing a Variable Length Subnet Mask (VLSM) addressing scheme for a regional facility using the summary block 10.200.16.0/2210.200.16.0/22. Subnet A must support up to 120120 usable host addresses for a departmental LAN segment. Subnet B must support a point-to-point WAN link between two routers. Subnet A is allocated the lowest numerical range within the summary block, starting at 10.200.16.010.200.16.0. If Subnet B is assigned the next immediately adjacent subnet block with the tightest prefix length that satisfies its requirement, which IPv4 address is a valid host address for Subnet B?

Show answer & explanation

Answer: 10.200.16.13010.200.16.130

Answer

The IPv4 address 10.200.16.13010.200.16.130 is a valid usable host address for Subnet B.
The address 10.200.16.13010.200.16.130 is a valid usable host address on Subnet B. Subnet A requires 120120 usable host addresses, which requires 77 host bits (27−2=1262^7 - 2 = 126), resulting in a /25/25 subnet mask. Starting at 10.200.16.010.200.16.0, Subnet A occupies 10.200.16.010.200.16.0 through 10.200.16.12710.200.16.127. Subnet B starts immediately at 10.200.16.12810.200.16.128. Because Subnet B is a point-to-point link requiring 22 host IPs, a /30/30 mask (22 host bits) is assigned. The /30/30 block 10.200.16.128/3010.200.16.128/30 has usable host addresses 10.200.16.12910.200.16.129 and 10.200.16.13010.200.16.130.

Step-by-Step Solution

1
Determine the prefix length and address boundary for Subnet A
Subnet A uses a /25/25 mask, spanning 10.200.16.010.200.16.0 to 10.200.16.12710.200.16.127.
Subnet A requires 120120 usable host addresses. Using the formula 2h−2≥1202^h - 2 \ge 120, h=7h = 7 host bits are needed (27−2=1262^7 - 2 = 126). The prefix length is 32−7=/2532 - 7 = /25. Starting at 10.200.16.010.200.16.0, the subnet range runs from network ID 10.200.16.010.200.16.0 to broadcast address 10.200.16.12710.200.16.127.
2
Identify the starting network address and required mask for Subnet B
Subnet B network address is 10.200.16.12810.200.16.128 with a /30/30 prefix length.
The next immediately adjacent unallocated IP address is 10.200.16.12810.200.16.128. Subnet B needs 22 usable host addresses for a point-to-point link. Using 2h−2≥22^h - 2 \ge 2, h=2h = 2 host bits are needed (22−2=22^2 - 2 = 2). The prefix length is 32−2=/3032 - 2 = /30.
3
Calculate the valid host address range for Subnet B
Valid host addresses are 10.200.16.12910.200.16.129 and 10.200.16.13010.200.16.130.
For the network 10.200.16.128/3010.200.16.128/30, 10.200.16.12810.200.16.128 is the network address, 10.200.16.12910.200.16.129 is the first usable host, 10.200.16.13010.200.16.130 is the last usable host, and 10.200.16.13110.200.16.131 is the broadcast address.

Key Concept

VLSM block allocation and usable host range calculation for sequential subnets
Estimated Time:2m 0s
Question 291Question →

An engineer is deploying a high-performance database server in a virtualized data center. To minimize latency and resource overhead, the virtualization software is installed directly onto the physical server hardware without an intervening host operating system. Which virtualization component is responsible for managing physical hardware resources and allocating them directly to the guest virtual machines in this architecture?

Show answer & explanation

Answer: Type 1 hypervisor

Answer

The correct architecture is a Type 1 hypervisor, which operates directly on server bare-metal hardware to allocate compute, memory, and network resources to virtual machines.
A Type 1 hypervisor (also known as a bare-metal hypervisor) installs directly onto the physical server hardware. It manages system resources—including CPU, RAM, and storage—and schedules guest virtual machines with minimal abstraction overhead, making it ideal for enterprise deployments.

Step-by-Step Solution

1
Analyze the deployment requirement from the scenario.
The requirement specifies installing virtualization software directly onto physical server hardware without an underlying host operating system.
Bare-metal deployments eliminate host operating system overhead to maximize hardware performance.
2
Distinguish between hypervisor execution models.
Type 1 hypervisors run directly on bare metal, while Type 2 hypervisors rely on an existing host OS (such as Windows or Linux).
Understanding the direct hardware interaction helps identify the correct component for low-latency enterprise server environments.

Key Concept

Type 1 vs Type 2 Hypervisor Architecture
Question 292Question →

A network administrator is interconnecting two legacy Cisco routers directly using their GigabitEthernet0/0 interfaces (both configured as MDI ports) in a high-security environment where Auto-MDIX is manually disabled. The administrator needs to build a custom RJ-45 copper patch cable to achieve a functional 1000BASE-T connection between the devices. If the first end of the cable is terminated using the standard T568B pinout, which pin wiring scheme must be applied to the second end to guarantee full Gigabit Ethernet Layer 1 operation?

Show answer & explanation

Answer: Pins 1 and 2 crossed with pins 3 and 6, and pins 4 and 5 crossed with pins 7 and 8

Answer

Pins 1 and 2 crossed with pins 3 and 6, and pins 4 and 5 crossed with pins 7 and 8
For Gigabit Ethernet (1000BASE-T) links between two MDI devices (such as routers) where Auto-MDIX is turned off, a full 4-pair crossover cable is required. This configuration crosses Pair 1 (pins 1 and 2) with Pair 2 (pins 3 and 6), as well as Pair 3 (pins 4 and 5) with Pair 4 (pins 7 and 8).

Step-by-Step Solution

1
Identify the device roles and interface pinout mode (MDI vs MDI-X).
Both router GigabitEthernet interfaces act as MDI (Medium Dependent Interface) ports.
Like devices (router to router, switch to switch) require pin crossover on the cable when connected directly without Auto-MDIX.
2
Determine the physical medium transmission standard requirements (1000BASE-T).
1000BASE-T operates over Category 5e/6 UTP using all four pairs (8 wires) simultaneously for full-duplex transmission (BI_DA, BI_DB, BI_DC, BI_DD).
Unlike 10BASE-T/100BASE-TX which only utilize pairs 1-2 and 3-6, 1000BASE-T requires all four pairs to be properly paired and crossed.
3
Map the crossover pinout required for 1000BASE-T operation.
Pair 1 (pins 1, 2) swaps with Pair 2 (pins 3, 6), and Pair 3 (pins 4, 5) swaps with Pair 4 (pins 7, 8).
This guarantees that transmit and receive signals across all four bi-directional channels align correctly between the two MDI interfaces.

Key Concept

1000BASE-T Crossover Cabling Pinouts and MDI/MDI-X Interconnection
Question 293Question →

A network administrator is allocated the IPv6 prefix 2001:db8:acad:4000::/582001:\text{db8}:\text{acad}:4000::/58 for a new branch location. The administrator needs to divide this block into standard /64/64 subnets for local LAN segments. How many /64/64 subnets can be created from this block?

Show answer & explanation

Answer: 64

Answer

64 subnets can be created from the given /58/58 prefix block.
Subtracting the assigned prefix length (/58) from the standard interface subnet prefix length (/64) results in 6 subnet bits (64−58=664 - 58 = 6). Computing 262^6 yields 64 total subnets.

Step-by-Step Solution

1
Calculate the number of subnetting bits.
64−58=664 - 58 = 6 bits.
Subtracting the starting prefix length (/58) from the target prefix length (/64) determines the number of bits available for subnetting.
2
Calculate the total number of subnets.
26=642^6 = 64 subnets.
Raising 2 to the power of the subnet bit count (262^6) provides the total number of valid /64/64 subnets.

Key Concept

IPv6 Subnetting and Prefix Length Calculation
Estimated Time:1m 0s
Question 294Question →

A network engineer is inspecting an IPv6 neighbor discovery cache entry on a Cisco switch and identifies a connected host with the SLAAC-generated link-local address fe80::1e1b:0dff:fe44:5566fe80::1e1b:0dff:fe44:5566. Based on the EUI-64 address generation standard, what is the original burned-in MAC address of the host's network interface card?

Show answer & explanation

Answer: 1c1b.0d44.5566

Answer

The original burned-in MAC address of the host is 1c1b.0d44.5566.
To reverse the EUI-64 process for the interface ID 1e1b:0dff:fe44:55661e1b:0dff:fe44:5566, first remove the 16-bit FFFEFFFE inserted in the middle to restore 48 bits: 1E−1B−0D−44−55−661E-1B-0D-44-55-66. Next, convert the first byte 1E1E to binary (0001 111020001\ 1110_2) and invert the 7th bit (counted from left to right), changing it from 1 to 0 to get 0001 110020001\ 1100_2 (1C1C). Reassembling the MAC address yields 1c1b.0d44.5566.

Step-by-Step Solution

1
Identify the 64-bit interface identifier portion of the IPv6 link-local address.
The interface identifier is 1e1b:0dff:fe44:5566.
In a link-local address (fe80::/64fe80::/64), the last 64 bits represent the interface identifier.
2
Remove the 16-bit FFFEFFFE hexadecimal value inserted into the midpoint during EUI-64 derivation.
Removing FFFE from 1e1b:0dff:fe44:5566 yields the 48-bit pattern 1E-1B-0D-44-55-66.
EUI-64 generation inserts FFFEFFFE between the OUI (first 24 bits) and the vendor-assigned extension identifier (last 24 bits).
3
Invert the 7th bit (Universal/Local bit) of the first byte.
The first byte 1E1E (0001 111020001\ 1110_2) with its 7th bit inverted becomes 1C1C (0001 110020001\ 1100_2).
During EUI-64 formation, the 7th bit of the first MAC octet is inverted (0 for universal/burned-in becomes 1 for EUI-64). Reversing the process converts the 1 back to 0.
4
Reconstruct the 48-bit MAC address in standard Cisco dotted-quad hexadecimal notation.
Combining 1C−1B−0D−44−55−661C-1B-0D-44-55-66 into Cisco MAC format gives 1c1b.0d44.5566.
Cisco IOS formats MAC addresses as three groups of four hexadecimal digits separated by dots.

Key Concept

Reversing EUI-64 Interface Identifier Derivation to Extract Original MAC Address
Question 295Question →

A network engineer is configuring a point-to-point WAN link between two branch routers using the 192.168.10.4/30192.168.10.4/30 network segment. Which IPv4 address represents a valid usable host address for one of the router interfaces?

Show answer & explanation

Answer: 192.168.10.5192.168.10.5

Answer

The address 192.168.10.5192.168.10.5 is a valid usable host IPv4 address in the 192.168.10.4/30192.168.10.4/30 subnet.
In a /30/30 subnet starting at 192.168.10.4192.168.10.4, the total block size is 44 IP addresses (192.168.10.4192.168.10.4 through 192.168.10.7192.168.10.7). The network address is 192.168.10.4192.168.10.4 and the broadcast address is 192.168.10.7192.168.10.7. The valid usable host IP addresses for router interfaces are 192.168.10.5192.168.10.5 and 192.168.10.6192.168.10.6. Therefore, 192.168.10.5192.168.10.5 is a correct host address.

Step-by-Step Solution

1
Determine the block size for a /30/30 prefix length.
A /30/30 mask uses 3030 network bits and 22 host bits (32−30=232 - 30 = 2). The block size is 22=42^2 = 4.
Calculating the block size defines the size and boundaries of the subnet segment.
2
Identify the network address and broadcast address for the segment.
For the network starting at 192.168.10.4192.168.10.4, the block ends before 192.168.10.8192.168.10.8. The network address is 192.168.10.4192.168.10.4 and the broadcast address is 192.168.10.7192.168.10.7.
The first address in the block is reserved for the network ID and the last address is reserved for broadcast.
3
Determine the range of valid usable host IP addresses.
Usable host addresses span from 192.168.10.5192.168.10.5 through 192.168.10.6192.168.10.6.
Usable host addresses are the IP addresses strictly between the network ID and the broadcast address.

Key Concept

IPv4 Subnet Boundaries and Usable Host Address Ranges
Question 296Question →

A network technician is configuring a new subnet with an IPv4 prefix length of /28/28. How many usable host IPv4 addresses are available on this subnet?

Show answer & explanation

Answer: 14

Answer

14 usable IPv4 host addresses are available on a /28 subnet.
For a /28/28 prefix, the host portion contains 32−28=432 - 28 = 4 bits. The formula to calculate usable hosts is 2h−22^h - 2, where hh is the number of host bits. Therefore, 24−2=16−2=142^4 - 2 = 16 - 2 = 14 usable host IPv4 addresses.

Step-by-Step Solution

1
Determine the number of host bits available.
Host bits = 32−28=432 - 28 = 4 bits.
An IPv4 address consists of 32 bits. Subtracting the prefix length (/28) leaves the host portion of the address.
2
Calculate total IP address space for the subnet.
Total addresses = 24=162^4 = 16 addresses.
The formula 2h2^h determines total IP address combinations, where hh is the number of host bits.
3
Subtract reserved addresses to get usable host addresses.
Usable addresses = 16−2=1416 - 2 = 14 addresses.
The first address (subnet ID) and last address (broadcast) cannot be assigned to individual hosts.

Key Concept

IPv4 Subnet Usable Host Calculation
Question 297Question →

A network engineer analyzes a Wireshark capture of a custom network monitoring stream. The Layer 4 protocol header contains only four fields: Source Port (16 bits), Destination Port (16 bits), Length (16 bits), and Checksum (16 bits). Which statement accurately describes the operational behavior of this transport-layer protocol?

Show answer & explanation

Answer: It provides connectionless, best-effort delivery with a fixed 8-byte header overhead and no inherent flow control or retransmission mechanisms.

Answer

The protocol in the packet capture is UDP, which provides connectionless, best-effort delivery with a minimal 8-byte header structure and no native retransmission or flow control mechanisms.
The described header consists of four 16-bit fields totaling 64 bits (8 bytes): Source Port, Destination Port, Length, and Checksum. This structure uniquely identifies UDP. Because UDP has no flags, sequence numbers, acknowledgment fields, or windowing fields, it operates connectionlessly without guaranteeing delivery or providing flow control.

Step-by-Step Solution

1
Calculate the total header size from the fields described in the prompt.
Source Port (16 bits) + Destination Port (16 bits) + Length (16 bits) + Checksum (16 bits) = 64 bits total, which equals 8 bytes.
Determining the total header length helps identify whether the protocol is TCP (minimum 20 bytes) or UDP (fixed 8 bytes).
2
Identify the protocol associated with this specific header structure.
The 8-byte header consisting solely of Source Port, Destination Port, Length, and Checksum corresponds to User Datagram Protocol (UDP).
UDP is designed for low overhead and fast delivery by omitting connection establishment and state tracking fields.
3
Evaluate the delivery properties of UDP compared to TCP.
UDP operates in a connectionless mode without acknowledgments, sequence numbering, or flow control (window size).
Upper-layer applications requiring low latency choose UDP and rely on upper layers if error handling or sequencing is needed.

Key Concept

UDP Header Structure and Protocol Characteristics
Estimated Time:1m 0s
Question 298Question →

An enterprise network administrator is evaluating the operational roles of different hardware and virtual components deployed across a branch office. Which two statements correctly describe the boundary handling and forwarding behaviors of these network components?

Select all that apply

Show answer & explanation

Answer: Layer 3 switches establish distinct broadcast domains per VLAN and utilize internal routing hardware to forward IP packets between VLANs.; Routers separate broadcast domains at each routed interface and make packet forwarding decisions using destination Layer 3 IP addresses.

Answer

The statements describing Layer 3 switches establishing distinct broadcast domains per VLAN with internal IP routing, and routers separating broadcast domains per routed interface using destination IP addresses, are both correct.
Layer 3 switches combine switching speed with Layer 3 IP routing functions, creating logical Layer 2 broadcast domains per configured VLAN and routing packets between them using internal Switch Virtual Interfaces (SVIs) or routed ports. Routers function at Layer 3 of the OSI model, naturally preventing broadcast frames from leaking across interfaces, thus isolating broadcast domains and forwarding traffic using network-layer destination IP addresses.

Step-by-Step Solution

1
Analyze the role of Layer 2 switches, Layer 3 switches, and routers regarding collision and broadcast domains.
Layer 2 switches divide collision domains per port but maintain a single broadcast domain. Layer 3 switches use VLANs and internal routing to create and interconnect separate broadcast domains. Routers inherently break broadcast domains at every routed interface.
Understanding domain boundaries is fundamental to identifying the correct operational capabilities of network devices.
2
Evaluate hypervisor virtual switching capabilities.
Type 1 hypervisors implement internal virtual switches (vSwitches) to switch local traffic between VM virtual NICs internally.
Traffic between VMs on the same hypervisor host does not need to exit to an external physical switch unless configured for external monitoring or security policies.
3
Identify the two accurate statements based on the analysis.
The statement regarding Layer 3 switch VLAN routing and the statement regarding router Layer 3 interface forwarding are correct.
Both statements accurately reflect standard Cisco networking principles and hardware component roles.

Key Concept

Network Component Operational Roles and Broadcast Domain Boundaries
Estimated Time:1m 30s
Question 299Question →

A network engineer is assigned the IPv4 network block 10.140.0.0/1810.140.0.0/18 to provision subnets for standard branch offices. Each branch office requires a subnet capable of accommodating at least 110110 usable host IP addresses. What is the maximum number of equal-sized subnets supporting this host requirement that can be created from the 10.140.0.0/1810.140.0.0/18 block?

Show answer & explanation

Answer: 128

Answer

The maximum number of valid branch subnets that can be created is 128.
To support at least 110 usable hosts per subnet, 7 host bits are required (27−2=1262^7 - 2 = 126 usable hosts). Subtracting 7 host bits from 32 total IPv4 bits results in a /25/25 prefix length. Borrowing bits from the allocated /18/18 summary block gives 25−18=725 - 18 = 7 subnet bits. Therefore, 27=1282^7 = 128 valid subnets can be created.

Step-by-Step Solution

1
Determine the required host bits for at least 110 usable host IP addresses.
7 host bits are required because 27−2=126≥1102^7 - 2 = 126 \ge 110, whereas 6 host bits (26−2=622^6 - 2 = 62) are insufficient.
Every IPv4 subnet reserves two IP addresses for the network ID and the broadcast ID, making the usable host formula 2h−22^h - 2.
2
Determine the required CIDR prefix length for each subnet.
The prefix length is /25/25 (32−7=2532 - 7 = 25).
An IPv4 address consists of 32 bits; subtracting the 7 host bits leaves 25 network bits.
3
Calculate the number of borrowed subnet bits relative to the parent block.
7 subnet bits are borrowed (25−18=725 - 18 = 7).
Extending the parent network prefix from /18/18 to /25/25 borrows 7 bits for subnetting.
4
Calculate the total number of /25/25 subnets within the /18/18 parent block.
The total number of subnets is 27=1282^7 = 128.
Borrowing nn subnet bits yields 2n2^n distinct subnets.

Key Concept

Calculating IPv4 subnet capacity and prefix derivation based on host requirements
Estimated Time:2m 0s
Question 300Question →

A network administrator is evaluating virtualization architecture requirements for an enterprise branch deployment. The design calls for a bare-metal Type 1 hypervisor installation on physical host hardware. Which two operational characteristics are associated with this virtualization approach? (Select two.)

Select all that apply

Show answer & explanation

Answer: The hypervisor installs directly on host hardware without requiring an underlying host operating system.; Virtual machines interact with hardware resources through a management software layer operating directly on bare metal.

Answer

The correct characteristics of a Type 1 hypervisor architecture are that the hypervisor installs directly on host hardware without requiring an underlying host operating system, and virtual machines access physical hardware resources through a management software layer operating directly on bare metal.
Type 1 hypervisors (bare-metal) install and run directly on physical host hardware without a general-purpose host operating system underneath. As a result, the hypervisor directly controls physical CPU, memory, storage, and network interfaces, delivering higher performance and lower latency for tenant virtual machines.

Step-by-Step Solution

1
Analyze the virtualization deployment requirement.
Identify that the scenario requires a bare-metal Type 1 hypervisor deployment.
Type 1 hypervisors interact directly with underlying physical hardware rather than relying on a host OS layer.
2
Compare Type 1 (bare-metal) and Type 2 (hosted) hypervisor operational traits.
Determine that Type 1 hypervisors execute on bare metal directly and manage hardware resources without an intermediate host operating system, while Type 2 hypervisors run as applications within a host OS.
Selecting the bare-metal characteristics correctly distinguishes Type 1 architecture from hosted Type 2 solutions.

Key Concept

Type 1 Bare-Metal Hypervisor Architecture vs Type 2 Hosted Hypervisor
Estimated Time:1m 30s
PreviousPage 15 / 20Next