Examkin

Network Fundamentals

398 questions

All types398Multiple choice205Multiple select92Numeric18Matching68Ordering7Short answer8
Question 341Question →

A network administrator is inspecting a host connected to a Cisco switch. The host network interface has a hardware MAC address of `5254.0012.3456`. Assuming IPv6 stateless autoconfiguration uses the standard IEEE EUI-64 format, what is the resulting 64-bit interface identifier?

Show answer & explanation

Answer: 5054:00ff:fe12:3456; 5054:00FF:FE12:3456; 5054:ff:fe12:3456; 5054:FF:FE12:3456

Answer

5054:00ff:fe12:3456
To derive the EUI-64 interface identifier from MAC address `5254.0012.3456`:
1. Split the MAC address: `52:54:00` and `12:34:56`.
2. Insert `FFFE` in the middle: `5254:00FF:FE12:3456`.
3. Invert the 7th bit of the first byte: `0x52` (`0101 0010` in binary) becomes `0x50` (`0101 0000` in binary).

Combining these operations yields the final 64-bit interface identifier `5054:00ff:fe12:3456`.

Step-by-Step Solution

1
Split the 48-bit MAC address into two 24-bit halves.
Left half: `52:54:00`, Right half: `12:34:56`
EUI-64 insertion occurs directly between the OUI and the vendor-assigned NIC portion.
2
Insert the 16-bit hex value `FFFE` into the middle.
`5254:00FF:FE12:3456`
Inserting `FFFE` expands the 48-bit MAC address into a 64-bit structure.
3
Invert the 7th bit (Universal/Local bit) of the first byte.
First byte `0x52` in binary is `0101 0010`. Inverting the 7th bit yields `0101 0000`, which is `0x50` in hex.
IEEE EUI-64 standards dictate that the 7th bit is flipped (0 becomes 1 for local uniqueness, 1 becomes 0 for universal).
4
Assemble the final 64-bit interface identifier in standard IPv6 hexadecimal notation.
`5054:00ff:fe12:3456`
Combines the modified first byte, remaining MAC components, and inserted `FFFE` into 16-bit colon-separated fields.

Key Concept

EUI-64 Interface Identifier Derivation
Question 342Question →

A network engineering team is evaluating topological designs for a expanding enterprise facility. The current infrastructure uses a Collapsed Core topology where distribution and core functions are combined on a single redundant switch pair. As the facility adds multiple new buildings with high port densities and complex policy enforcement requirements, the team considers migrating to a traditional 3-Tier Campus design. Which technical requirement or operational constraint best justifies transitioning to a full 3-Tier architecture?

Show answer & explanation

Answer: The quantity of access-layer switches and distribution-layer interconnects exceeds the interface density and processing capacity of a single consolidated switch pair.

Answer

The transition to a 3-Tier architecture is justified when the volume of access switches and interconnections exceeds the port density, backplane performance, and policy processing capacity of a collapsed distribution/core pair.
In expanding campus networks, a Collapsed Core topology can become a bottleneck due to limited physical uplink ports, buffer sizes, and control-plane processing under high policy loads. Transitioning to a traditional 3-Tier design adds a dedicated Core layer that interconnects multiple distribution blocks, providing scalable throughput and switch port availability.

Step-by-Step Solution

1
Analyze the functional limits of a Collapsed Core topology.
Collapsed Core combines the Core and Distribution layers into a single switch pair, which works well for small to medium environments but encounters scaling limits as port counts and routing policies grow.
Understanding architectural boundaries is essential for network design scaling.
2
Evaluate the benefits of introducing a dedicated Core tier in a 3-Tier design.
A 3-Tier design separates the Distribution layer (which handles routing, ACL policies, and VLAN aggregation) from the Core layer (which focuses strictly on ultra-fast, unthrottled packet forwarding).
Decoupling roles prevents a single tier from suffering hardware port exhaustion or CPU overload.

Key Concept

Cisco 3-Tier Campus vs Collapsed Core Architectural Scaling
Estimated Time:1m 30s
Question 343Question →

A network engineer is provisioning interface IP addresses for a server cluster assigned to the 172.16.140.0/22172.16.140.0/22 IPv4 subnet. Which two IP addresses are valid usable host addresses within this subnet?

Select all that apply

Show answer & explanation

Answer: 172.16.141.255; 172.16.143.254

Answer

The addresses 172.16.141.255 and 172.16.143.254 are valid usable host addresses for the 172.16.140.0/22 subnet.
For the network block 172.16.140.0/22, the subnet spans 172.16.140.0 to 172.16.143.255. The unusable boundary addresses are 172.16.140.0 (network ID) and 172.16.143.255 (broadcast ID). Any address strictly between these boundaries (172.16.140.1 to 172.16.143.254) is usable. Consequently, 172.16.141.255 (where host bits are binary 01 11111111) and 172.16.143.254 are both valid usable host addresses.

Step-by-Step Solution

1
Determine the subnet mask and block size from CIDR prefix /22
A /22 prefix corresponds to mask 255.255.252.0. The third octet block size is 256 - 252 = 4.
Prefix length of 22 leaves 10 host bits (32 - 22 = 10). 2 bits in the 3rd octet and 8 bits in the 4th octet belong to host portion.
2
Calculate the network and broadcast boundaries for 172.16.140.0/22
Network Address: 172.16.140.0; Broadcast Address: 172.16.143.255.
140 is a multiple of 4 (4 * 35 = 140). The next subnet begins at 172.16.144.0, so the broadcast address is 172.16.143.255.
3
Determine the usable host IP address range
Usable host range: 172.16.140.1 through 172.16.143.254.
Usable host addresses are all IPs between the network address and broadcast address.
4
Evaluate candidate options against the usable host range
172.16.141.255 and 172.16.143.254 both fall inside the valid range 172.16.140.1 - 172.16.143.254.
172.16.140.0 is the network IP and 172.16.143.255 is the broadcast IP, leaving 172.16.141.255 and 172.16.143.254 as valid usable host IPs.

Key Concept

IPv4 Usable Host Range in Non-Octet Boundary Subnets (/22 Prefix)
Question 344Question →

Switch SW2 receives an ingress Ethernet frame on interface GigabitEthernet0/5 configured as an access port in VLAN 20. The frame contains a source MAC address of 0010.ab12.3456 and a destination MAC address of 0010.ab99.8877. A review of SW2's MAC address table reveals that neither MAC address currently exists in the table for VLAN 20. Which two actions does the switch take upon processing this frame? (Select two.)

Select all that apply

Show answer & explanation

Answer: Records MAC address 0010.ab12.3456 associated with interface GigabitEthernet0/5 in the VLAN 20 MAC address table.; Floods the frame out all active interfaces assigned to VLAN 20, excluding interface GigabitEthernet0/5.

Answer

The switch records the source MAC address 0010.ab12.3456 dynamically under GigabitEthernet0/5 in VLAN 20, and floods the unknown unicast frame out all active ports in VLAN 20 except the ingress port GigabitEthernet0/5.
When a switch receives a frame, it performs two distinct steps: learning and forwarding. First, it reads the incoming frame's source MAC address and associates it with the receiving port and VLAN in its MAC address table. Second, it checks the destination MAC address. Because the destination is not in the table, it treats the frame as unknown unicast and floods it out all active ports in the same VLAN except the receiving port.

Step-by-Step Solution

1
Inspect the source MAC address of the ingress frame.
The source MAC address 0010.ab12.3456 is learned on port GigabitEthernet0/5 for VLAN 20 and added/updated in the MAC address table.
Switches build and refresh their Layer 2 forwarding table using incoming source MAC addresses.
2
Inspect the destination MAC address of the ingress frame.
The destination MAC address 0010.ab99.8877 is not found in VLAN 20's MAC address table, classifying it as unknown unicast.
A lookup in the MAC table determines whether known unicast forwarding or unknown unicast flooding is required.
3
Forward the unknown unicast frame according to Layer 2 rules.
The frame is flooded to all active ports belonging to VLAN 20, except GigabitEthernet0/5.
Flooding ensures delivery to the unknown destination while avoiding sending traffic back out the receiving interface.

Key Concept

Layer 2 Switch Learning and Unknown Unicast Flooding
Estimated Time:1m 30s
Question 345Question →

A network administrator is designing a two-tier Spine-Leaf (Clos) topology for a data center expansion. Which two physical interconnection rules must be strictly followed in this architectural model? (Select two.)

Select all that apply

Show answer & explanation

Answer: Spine switches must not connect directly to other spine switches.; Every leaf switch must connect directly to every spine switch in the fabric.

Answer

In a standard two-tier Spine-Leaf topology, spine switches must never connect directly to other spine switches, and every leaf switch must connect directly to every spine switch in the fabric.
In a standard Spine-Leaf architecture, every leaf switch connects to every spine switch in a full mesh, while spine switches never connect directly to one another and leaf switches do not connect directly to other leaf switches. This ensures deterministic two-hop forwarding between any hosts connected to the leaf layer.

Step-by-Step Solution

1
Identify the defining structural rules of a two-tier Spine-Leaf (Clos) network topology.
In Spine-Leaf designs, the architecture is a bipartite graph where leaf nodes connect to spine nodes, but nodes within the same tier do not interconnect.
This guarantees equal distance (two hops) between any two leaf-connected endpoints.
2
Evaluate the requirement regarding spine-to-spine connections.
Direct spine-to-spine links violate Clos topology principles.
Spine switches strictly serve as a high-speed forwarding mesh between leaf switches.
3
Evaluate the requirement regarding leaf-to-spine full-mesh connections.
Each leaf switch connects directly to every spine switch.
This structure enables efficient Equal-Cost Multi-Path (ECMP) forwarding across all available upstream links.

Key Concept

Spine-Leaf (Clos) Interconnection Rules
Estimated Time:1m 30s
Question 346Question →

An enterprise network engineer is designing a network topology for a multi-story office campus. The design requires a designated functional layer to handle inter-VLAN routing, access control list (ACL) filtering, broadcast domain boundaries, and policy-based traffic control before passing traffic to the high-speed campus backbone. Which tier of the standard Cisco three-tier hierarchical model is specifically responsible for performing these functions?

Show answer & explanation

Answer: Distribution layer

Answer

The Distribution layer is the functional tier responsible for inter-VLAN routing, security policies, and broadcast boundaries in a three-tier campus architecture.
In the standard Cisco three-tier enterprise campus design, the distribution layer acts as the aggregation point for access layer switches and serves as the Layer 3 routing boundary. It is specifically designed to handle inter-VLAN routing, policy enforcement (such as ACL filtering and QoS marking), address summarization, and boundary isolation for broadcast domains.

Step-by-Step Solution

1
Analyze the functional requirements stated in the scenario
Identified requirements: inter-VLAN routing, ACL security filtering, broadcast domain boundary enforcement, and traffic policy control.
Determining which architectural layer aligns with these policy and routing responsibilities is key to solving the problem.
2
Map requirements to the Cisco three-tier campus model layers
Access layer handles end-host connectivity; Core layer handles high-speed packet transport; Distribution layer aggregates access switches and enforces routing/security policies.
Each hierarchical layer has isolated responsibilities to preserve scalability, predictability, and fault tolerance.
3
Select the layer that performs aggregation and control boundary functions
The Distribution layer matches all required capabilities.
It acts as the Layer 3 boundary between Layer 2 access switches and the Layer 3 core backbone.

Key Concept

Cisco Three-Tier Hierarchical Campus Model - Distribution Layer Responsibilities
Estimated Time:1m 15s
Question 347Question →

A network administrator needs to assign an IPv4 address to the router interface serving a newly established remote VLAN. The subnet is carved from the 10.200.16.0/2010.200.16.0/20 parent address space and must support a maximum of 2828 end-user host devices using the most efficient subnet mask possible. The policy requires configuring the router interface with the last usable host IP address in the first available subnet block. Which IP address and subnet mask combination must be assigned to the router interface?

Show answer & explanation

Answer: 10.200.16.30 255.255.255.224

Answer

10.200.16.30 with subnet mask 255.255.255.224
To support 28 hosts, 5 host bits are required (25−2=302^5 - 2 = 30 usable addresses), yielding a /27 prefix (subnet mask 255.255.255.224). The first subnet derived from 10.200.16.0/20 is 10.200.16.0/27, which spans from 10.200.16.0 (network ID) to 10.200.16.31 (broadcast ID). The last usable host IP address in this range is 10.200.16.30.

Step-by-Step Solution

1
Determine the required number of host bits (nn) for 2828 hosts.
Using the formula 2n−2≥282^n - 2 \ge 28, n=5n = 5 host bits because 25−2=302^5 - 2 = 30 usable addresses (32−5=2732 - 5 = 27 prefix length or subnet mask 255.255.255.224255.255.255.224).
Selecting 4 host bits (/28) provides only 14 usable hosts, which falls short of the 28 required hosts.
2
Calculate the subnet boundaries for the first /27 subnet allocated from 10.200.16.0/2010.200.16.0/20.
Network Address: 10.200.16.0/2710.200.16.0/27; Broadcast Address: 10.200.16.3110.200.16.31.
A /27 subnet has a block size of 32 in the fourth octet (256−224=32256 - 224 = 32).
3
Identify the usable host range and the last usable host IP address.
Usable host range: 10.200.16.110.200.16.1 to 10.200.16.3010.200.16.30. The last usable address is 10.200.16.3010.200.16.30.
The network ID (10.200.16.010.200.16.0) and broadcast ID (10.200.16.3110.200.16.31) cannot be assigned to network interfaces.

Key Concept

Subnet Host Calculation and Usable Address Boundaries
Question 348Question →

A network workstation with MAC address 0010.7b33.44a1 is relocated from switch port GigabitEthernet0/1 to port GigabitEthernet0/2 on the same VLAN. Before the switch's MAC address table aging timer expires, the workstation sends an Ethernet frame out GigabitEthernet0/2 destined for a server with MAC address 0010.7b99.88c2. The server's MAC address is already dynamically mapped to port GigabitEthernet0/4 in the MAC address table. How does the switch process this incoming frame?

Show answer & explanation

Answer: The switch updates the MAC address table entry for 0010.7b33.44a1 to point to GigabitEthernet0/2, then forwards the frame directly out GigabitEthernet0/4.

Answer

The switch updates the MAC address table entry for the moving host to reflect the new ingress port (GigabitEthernet0/2) and forwards the frame directly out the destination port (GigabitEthernet0/4).
When a switch receives a frame, it reads the source MAC address to update its MAC address table. If the source MAC address is already listed but associated with a different interface, the switch instantly updates the entry to point to the new ingress interface (GigabitEthernet0/2). Next, the switch looks up the destination MAC address. Because the destination MAC address (0010.7b99.88c2) is already known and mapped to GigabitEthernet0/4, the switch performs known unicast forwarding directly to GigabitEthernet0/4.

Step-by-Step Solution

1
Examine the source MAC address and ingress interface of the incoming frame.
Source MAC 0010.7b33.44a1 arrives on GigabitEthernet0/2.
Switches inspect the source MAC address of every ingress frame to build and maintain the MAC address table.
2
Update the MAC address table entry if the source MAC is missing or associated with a different port.
The existing entry for 0010.7b33.44a1 is updated from GigabitEthernet0/1 to GigabitEthernet0/2.
Dynamic MAC learning automatically updates port associations when a host moves to a new switch port within the same VLAN.
3
Lookup the destination MAC address in the MAC address table to determine forwarding behavior.
Destination MAC 0010.7b99.88c2 is found mapped to GigabitEthernet0/4.
Known unicast frames are forwarded exclusively out the specific port mapped in the MAC address table.

Key Concept

Layer 2 MAC Address Table Learning and Forwarding Operations
Estimated Time:1m 0s
Question 349Question →

Match each network topology architecture on the left to its defining operational characteristic or primary design focus on the right. Which alignment correctly matches each architecture to its intended implementation traits?

Click a left item, then click its matching right item

Items

Spine-Leaf (Clos) Architecture
Three-Tier Campus Architecture
Collapsed Core Architecture
SOHO Network Topology

Matches

Show answer & explanation

Answer

Spine-Leaf Architecture aligns with delivering deterministic two-hop latency for east-west data center traffic; Three-Tier Campus Architecture aligns with segregating network roles into discrete Access, Distribution, and Core layers; Collapsed Core Architecture aligns with merging distribution routing and core switching into a single switch pair; and SOHO Network Topology aligns with integrating routing, switching, wireless, and firewall services into a single gateway appliance.
Each topology matches its core operational purpose: Spine-Leaf provides deterministic 2-hop latency for east-west data center traffic; Three-Tier segregates access, distribution, and core into separate layers for enterprise scaling; Collapsed Core merges distribution and core into a single switch pair for smaller campuses; and SOHO integrates all network services into one device.

Step-by-Step Solution

1
Analyze the primary operational requirement of Spine-Leaf (Clos) topologies.
Identify that Spine-Leaf topologies use a two-tier full mesh between leaf and spine switches to optimize east-west traffic with predictable 2-hop latency and ECMP.
Spine-Leaf is designed specifically to overcome Spanning Tree bottlenecks in data centers.
2
Evaluate campus architecture options (3-Tier vs. Collapsed Core).
Map 3-Tier to distinct Access, Distribution, and Core layers, and Collapsed Core to designs where Distribution and Core functions share a single device pair.
Collapsed Core optimizes costs for medium sites, whereas 3-Tier scales up to multi-building enterprise environments.
3
Evaluate small office deployment characteristics.
Map SOHO topology to integrated single-appliance gateways combining WAN, LAN switch ports, wireless, and security services.
SOHO environments prioritize low footprint, simplicity, and integrated functions over hierarchical multi-switch scaling.

Key Concept

Enterprise Network Topology Architectures (3-Tier, Collapsed Core, Spine-Leaf, and SOHO)
Question 350Question →

A network administrator is assigned the IPv4 address block 10.50.0.0/2210.50.0.0/22 to provision subnets for an enterprise IP telephony deployment. Each subnet requires support for at least 5050 usable host IP addresses. If the administrator subnets the entire 10.50.0.0/2210.50.0.0/22 block using a uniform prefix length that meets this host requirement with minimal wasted host space, how many total subnets can be created?

Show answer & explanation

Answer: 16

Answer

16 subnets can be created from the block.
To support at least 50 usable host IP addresses per subnet, 6 host bits are needed (26−2=622^6 - 2 = 62 usable IPs), which corresponds to a /26 prefix length (32−6=2632 - 6 = 26). Borrowing bits from the parent /22 prefix to create /26 subnets uses 26−22=426 - 22 = 4 subnet bits, resulting in 24=162^4 = 16 total subnets.

Step-by-Step Solution

1
Calculate the number of host bits required for 50 usable IP addresses
6 host bits are required
The formula for usable host IPs is 2h−22^h - 2. With h=5h = 5, 25−2=302^5 - 2 = 30 hosts (insufficient). With h=6h = 6, 26−2=622^6 - 2 = 62 usable hosts, which satisfies the requirement of at least 50 hosts.
2
Determine the new subnet prefix length
The prefix length is /26
Subtracting the 6 host bits from the 32-bit IPv4 address space yields 32−6=2632 - 6 = 26.
3
Calculate the number of subnets derived from the parent /22 block
16 subnets
Subtracting the original prefix (/22) from the new prefix (/26) gives 26−22=426 - 22 = 4 subnet bits. Borrowing 4 bits produces 24=162^4 = 16 subnets.

Key Concept

Subnet sizing and calculation of total subnets created using FLSM from a CIDR block.
Question 351Question →

An engineer connects a new host to interface FastEthernet0/12 on a Cisco Catalyst switch. The host immediately sends a unicast Ethernet frame intended for a server whose MAC address is already dynamically learned on interface FastEthernet0/18. Arrange the operational steps in the exact chronological order that the switch executes upon receiving this frame.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct operational sequence begins with frame reception on FastEthernet0/12, followed by source MAC address learning, destination MAC lookup in the table, identifying the egress port match on FastEthernet0/18, and finally unicast forwarding out FastEthernet0/18.
Layer 2 switch operation dictates a strict sequence of events upon frame arrival: frame ingress occurs first, followed immediately by source MAC address learning (adding or updating the source MAC and interface association in the MAC address table). Next, the switch performs a table lookup for the destination MAC address, identifies the matching egress interface (FastEthernet0/18), and lastly forwards the unicast frame directly out that specific interface.

Step-by-Step Solution

1
Identify the ingress port event.
The switch receives the raw frame on FastEthernet0/12.
Frame processing cannot occur until the physical interface receives and buffers the frame.
2
Perform ingress source MAC learning.
The switch associates the frame's source MAC address with interface FastEthernet0/12 in its Layer 2 MAC address table.
Cisco switches always process source MAC learning prior to destination forwarding logic.
3
Perform destination MAC table lookup.
The switch checks its MAC address table for the frame's destination address.
The destination MAC determines whether the frame will be unicast forwarded, flooded, or filtered.
4
Evaluate the table lookup result.
An existing mapping points to interface FastEthernet0/18.
Since the server's MAC was previously learned, the destination lookup succeeds.
5
Execute frame forwarding.
The switch transmits the frame solely out interface FastEthernet0/18.
Known unicast frames are forwarded strictly to the matching egress port, preventing unnecessary broadcast domain traffic.

Key Concept

L2 Switch Ingress Processing, MAC Address Learning, and Known Unicast Forwarding
Question 352Question →

A network engineer is configuring interface IP addresses for a cluster of management servers assigned to the 10.100.80.0/2310.100.80.0/23 IPv4 subnet. Which two IPv4 addresses are valid, assignable host addresses within this subnet? (Select two.)

Select all that apply

Show answer & explanation

Answer: 10.100.80.25510.100.80.255; 10.100.81.25410.100.81.254

Answer

The addresses 10.100.80.25510.100.80.255 and 10.100.81.25410.100.81.254 are valid, assignable host addresses in the 10.100.80.0/2310.100.80.0/23 subnet.
For the network block 10.100.80.0/2310.100.80.0/23, the 9 host bits span from 10.100.80.010.100.80.0 to 10.100.81.25510.100.81.255. The network address is 10.100.80.010.100.80.0 (all host bits 0), and the broadcast address is 10.100.81.25510.100.81.255 (all host bits 1). The valid usable host IP range is 10.100.80.110.100.80.1 to 10.100.81.25410.100.81.254. Therefore, both 10.100.80.25510.100.80.255 and 10.100.81.25410.100.81.254 fall within the valid host range.

Step-by-Step Solution

1
Determine the subnet mask and block size for a /23 prefix
A /23 prefix uses a subnet mask of 255.255.254.0255.255.254.0, providing 9 host bits (29=5122^9 = 512 total IP addresses) and a block size of 2 in the third octet.
Calculating block size allows identification of the subnet boundaries.
2
Calculate the network and broadcast addresses for 10.100.80.0/2310.100.80.0/23
Network address is 10.100.80.010.100.80.0. The broadcast address is the last address before the next subnet (10.100.82.010.100.82.0), which is 10.100.81.25510.100.81.255.
Host addresses cannot be assigned the network ID or broadcast ID.
3
Identify the usable host address range
The assignable host range extends from 10.100.80.110.100.80.1 through 10.100.81.25410.100.81.254.
Usable host range includes all IPs between the network ID and broadcast ID.
4
Evaluate the given options against the usable host range
10.100.80.25510.100.80.255 and 10.100.81.25410.100.81.254 both fall strictly within the assignable range [10.100.80.1,10.100.81.254][10.100.80.1, 10.100.81.254].
Addresses within this inclusive range are valid host assignments.

Key Concept

IPv4 Subnet Boundaries and Host Address Ranges for Non-Byte-Aligned Prefixes
Question 353Question →

A network administrator is reviewing the network configuration of a server assigned the IPv4 address 172.28.143.100172.28.143.100 with a subnet mask of 255.255.252.0255.255.252.0 (/22). Which IPv4 address is the broadcast address for this subnet?

Show answer & explanation

Answer: 172.28.143.255172.28.143.255

Answer

The broadcast address for the subnet containing host 172.28.143.100/22172.28.143.100/22 is 172.28.143.255172.28.143.255.
For a /22 subnet mask (255.255.252.0255.255.252.0), the third octet increments in steps of 4 (256−252=4256 - 252 = 4). For the IP address 172.28.143.100172.28.143.100, the containing network boundary starts at 172.28.140.0/22172.28.140.0/22 and ends at 172.28.143.255172.28.143.255. The last address in the subnet range (172.28.143.255172.28.143.255) is reserved as the directed broadcast address.

Step-by-Step Solution

1
Determine the block size in the third octet from the subnet mask
A mask of 255.255.252.0255.255.252.0 (/22) leaves 256−252=4256 - 252 = 4 as the block size in the third octet.
The third octet is the interesting octet where subnetting occurs for a /22 prefix.
2
Find the network address boundary for the host IP address 172.28.143.100
Multiples of 4 in the third octet are 0, 4, 8, ..., 136, 140, 144. Since 143 falls between 140 and 144, the network address is 172.28.140.0172.28.140.0.
The subnet boundary begins at the nearest lower multiple of the block size (140).
3
Calculate the broadcast address of the subnet
The next subnet begins at 172.28.144.0172.28.144.0, so subtracting 1 IP address gives the broadcast address 172.28.143.255172.28.143.255.
The broadcast address is always one address below the network ID of the next consecutive subnet.

Key Concept

IPv4 Subnet Boundaries and Broadcast Address Calculation
Question 354Question →

A network architect is assigned the IPv4 address block 10.116.0.0/1810.116.0.0/18 to provision subnets for remote office sites. Each site requires a dedicated subnet that can support at least 120120 usable host IP addresses. What is the maximum number of subnets of this size that can be created from the allocated block?

Show answer & explanation

Answer: 128

Answer

The maximum number of subnets that can be created is 128.
To support at least 120 usable host IP addresses per subnet, 7 host bits are required because 27−2=1262^7 - 2 = 126 usable hosts (26−2=622^6 - 2 = 62 is insufficient). This results in a /25/25 prefix (32−7=2532 - 7 = 25). Borrowing bits from /18/18 to /25/25 gives 25−18=725 - 18 = 7 subnet bits. The total number of subnets created is 27=1282^7 = 128.

Step-by-Step Solution

1
Calculate the required host bits (hh) for at least 120120 usable host IP addresses.
7 host bits are required (27−2=126≥1202^7 - 2 = 126 \ge 120).
6 host bits provide only 26−2=622^6 - 2 = 62 usable hosts, which is insufficient. 7 host bits provide 126 usable hosts.
2
Determine the prefix length needed for 7 host bits.
The subnet prefix length is /25/25 (32−7=2532 - 7 = 25).
An IPv4 address consists of 32 bits, so subtracting 7 host bits leaves 25 network/subnet bits.
3
Determine the number of subnet bits borrowed from the allocated block.
7 bits are borrowed (25−18=725 - 18 = 7).
The original allocation is /18/18 and the new subnets are /25/25.
4
Compute the total number of valid subnets.
The total number of subnets is 27=1282^7 = 128.
Borrowing 7 bits allows 272^7 distinct subnet combinations.

Key Concept

Subnet Sizing and Subnet Count Calculation
Estimated Time:1m 30s
Question 355Question →

A Cisco Catalyst switch with an empty MAC address table receives an ingress Ethernet frame on interface GigabitEthernet1/0/24 assigned to VLAN 50. The frame contains a source MAC address of 001B.D455.A1B2 and a destination MAC address of 001B.D488.F9E0. Which processing action does the switch perform upon receiving this frame?

Show answer & explanation

Answer: It records source MAC address 001B.D455.A1B2 associated with interface GigabitEthernet1/0/24 in VLAN 50 in its MAC address table and floods the frame out all active interfaces in VLAN 50 except GigabitEthernet1/0/24.

Answer

The switch records the source MAC address 001B.D455.A1B2 associated with interface GigabitEthernet1/0/24 in VLAN 50 in its MAC address table and floods the frame out all active interfaces in VLAN 50 except GigabitEthernet1/0/24.
When a switch receives a frame, it always performs MAC learning by binding the source MAC address to the ingress port and VLAN. If the destination MAC address is not found in the MAC address table for that VLAN, the switch performs an unknown unicast flood, transmitting the frame out all operational ports within the VLAN except the port on which the frame arrived.

Step-by-Step Solution

1
Inspect the ingress frame's source MAC address
The switch reads source MAC address 001B.D455.A1B2 on port GigabitEthernet1/0/24 in VLAN 50.
Layer 2 switches continuously learn source MAC addresses and bind them to their ingress interface and VLAN ID.
2
Update the switch MAC address table
An entry mapping 001B.D455.A1B2 to GigabitEthernet1/0/24 in VLAN 50 is dynamically created with an aging timer.
Populating the table allows future unicast traffic destined for 001B.D455.A1B2 to be forwarded directly out GigabitEthernet1/0/24.
3
Lookup the destination MAC address 001B.D488.F9E0 in the MAC address table
The MAC address table lookup results in a miss (unknown unicast).
The initial state of the MAC address table was completely empty.
4
Execute unknown unicast forwarding logic
The frame is replicated and sent out all active ports assigned to VLAN 50, excluding GigabitEthernet1/0/24.
Flooding ensures the unknown host receives the frame without causing split-horizon interface loops.

Key Concept

Layer 2 MAC Address Table Learning and Unknown Unicast Flooding Logic
Question 356Question →

A network infrastructure team must select a topology for a new data center hosting microservices workloads that generate intensive East-West traffic between virtualized servers. The design requires consistent deterministic latency, equal-cost multi-pathing (ECMP) across all available links, and exactly two Layer 3 forwarding hops between any two leaf nodes. Which network topology architecture best satisfies these technical requirements?

Show answer & explanation

Answer: Spine-Leaf (Clos) architecture

Answer

Spine-Leaf (Clos) architecture
The correct answer describes a Spine-Leaf (Clos) architecture. In a two-tier Spine-Leaf topology, leaf switches connect to end systems and to every spine switch, ensuring that any leaf switch can communicate with any other leaf switch via a single spine switch (two network hops). By running Layer 3 routing between spine and leaf nodes, all inter-spine links remain active and carry traffic simultaneously using Equal-Cost Multi-Pathing (ECMP), providing deterministic low latency ideal for East-West data center workloads.

Step-by-Step Solution

1
Analyze the workload traffic pattern and performance requirements.
Microservices require heavy East-West (server-to-server) communication demanding predictable, low latency and high bandwidth.
Traditional hierarchical networks route traffic north-south through core/distribution layers, causing sub-optimal pathing and latency variance.
2
Evaluate topological hop counts and path redundancy features.
A Spine-Leaf fabric guarantees that any leaf switch is connected directly to every spine switch, resulting in exactly two network hops between any pair of leaf switches.
This full mesh between leaf and spine switches enables routing protocols to utilize all paths simultaneously using Equal-Cost Multi-Pathing (ECMP).

Key Concept

Spine-Leaf (Clos) Topology Characteristics
Estimated Time:1m 15s
Question 357Question →

A Cisco switch receives an Ethernet frame on interface GigabitEthernet0/1 assigned to VLAN 10. The frame features a source MAC address of 0050.56a1.1111 and a destination MAC address of 0050.56b2.2222. The switch MAC address table currently contains an entry for 0050.56b2.2222 associated with interface GigabitEthernet0/3 on VLAN 10, but has no existing entry for 0050.56a1.1111. Which two actions does the switch take upon processing this ingress frame? (Select two.)

Select all that apply

Show answer & explanation

Answer: It records an entry in the MAC address table mapping source MAC address 0050.56a1.1111 to interface GigabitEthernet0/1 on VLAN 10.; It forwards the frame directly out interface GigabitEthernet0/3 without transmitting it on any other port.

Answer

The switch learns the source MAC address by associating 0050.56a1.1111 with ingress port GigabitEthernet0/1 in VLAN 10, and forwards the frame directly out egress port GigabitEthernet0/3 based on the known destination MAC table entry.
Layer 2 switches process Ethernet frames in two distinct phases. First, the switch reads the frame's source MAC address (0050.56a1.1111) and updates its MAC address table by binding that address to ingress port GigabitEthernet0/1 under VLAN 10. Second, the switch checks its MAC table for the destination MAC address (0050.56b2.2222). Since an entry already exists pointing to GigabitEthernet0/3 in VLAN 10, the switch forwards the frame directly out interface GigabitEthernet0/3.

Step-by-Step Solution

1
Examine incoming frame headers for source MAC address processing.
The switch reads source MAC 0050.56a1.1111 on ingress interface GigabitEthernet0/1 (VLAN 10).
Layer 2 switches inspect the source MAC address of every ingress frame to dynamically build and maintain the MAC address table.
2
Update the MAC address table.
A new dynamic MAC address table entry is created: MAC 0050.56a1.1111 → GigabitEthernet0/1 (VLAN 10).
The source MAC was not previously in the table, so a new mapping is recorded to enable future return traffic forwarding.
3
Examine incoming frame headers for destination MAC lookup.
The switch searches the MAC address table for destination MAC 0050.56b2.2222 in VLAN 10.
Destination lookup determines whether the frame requires point-to-point forwarding or unknown unicast flooding.
4
Forward the frame to the destination interface.
The frame is forwarded directly out GigabitEthernet0/3.
Because destination MAC 0050.56b2.2222 is known in the MAC table on port GigabitEthernet0/3, unicast forwarding is performed without flooding.

Key Concept

Layer 2 Ethernet switching operates on a two-step process per frame: (1) Learn source MAC address to ingress port mapping, and (2) Forward based on destination MAC address lookup (known unicast forwarding vs. unknown unicast flooding).
Question 358Question →

An enterprise is deploying a Collapsed Core network topology for a newly acquired medium-sized facility. Which two architectural traits or design implications characterize a Collapsed Core topology when compared to a traditional three-tier campus design? (Select two.)

Select all that apply

Show answer & explanation

Answer: The functional responsibilities of high-speed backbone routing and distribution-layer security/policy enforcement are consolidated into a single physical switch tier.; Capital outlay and operational complexity are minimized by reducing the total required physical switch count and inter-tier cabling.

Answer

A Collapsed Core topology combines the core and distribution layer functions into a single pair of switches, reducing hardware expenditure and operational complexity while combining routing boundaries and policy enforcement.
A Collapsed Core design combines core-layer transport and distribution-layer routing and security policy controls into the same physical switches. This architectural reduction significantly lowers hardware costs, rack space, power requirements, and management overhead for environments that do not require a separate dedicated core tier.

Step-by-Step Solution

1
Analyze the structural difference between Three-Tier and Collapsed Core architectures.
A traditional three-tier network contains distinct Core, Distribution, and Access layers. Collapsed Core merges Core and Distribution functions into one tier.
Understanding tier consolidation clarifies which functional boundaries overlap.
2
Evaluate the operational benefits of tier consolidation.
Eliminating a dedicated core hardware tier saves capital costs and lowers switch count and cabling complexity.
This identifies cost and deployment advantages characteristic of smaller or medium enterprise sites.
3
Evaluate Layer 2 loop considerations in redundant topologies.
Dual-homed uplinks from access switches to two collapsed core switches still form Layer 2 physical loops, requiring STP.
Consolidating tiers does not eliminate broadcast loops on redundant Layer 2 trunks.

Key Concept

Collapsed Core Architecture Traits and Considerations
Estimated Time:1m 15s
Question 359Question →

A Cisco router interface GigabitEthernet0/0/0 has the burned-in MAC address 1866.da34.5678. The interface is configured with the command `ipv6 enable` to automatically derive its IPv6 link-local address using stateless address autoconfiguration (SLAAC) and the IEEE EUI-64 process. Which IPv6 address will be assigned to the interface?

Show answer & explanation

Answer: fe80::1a66:daff:fe34:5678

Answer

fe80::1a66:daff:fe34:5678
The address fe80::1a66:daff:fe34:5678 correctly follows the IEEE EUI-64 interface identifier derivation by inserting FFFE into the middle of MAC address 1866.da34.5678, inverting the 7th bit of the first byte (changing 18 to 1A in hexadecimal), and prefixing it with the link-local scope fe80::.

Step-by-Step Solution

1
Split the 48-bit MAC address into two 24-bit halves and insert FFFE in the middle.
The MAC address 1866.da34.5678 (18:66:DA:34:56:78) becomes 1866:DAFF:FE34:5678.
EUI-64 standard requires expanding a 48-bit MAC address into a 64-bit interface identifier by placing FFFE between the Organizationally Unique Identifier (OUI) and the NIC-assigned extension.
2
Invert the 7th bit (Universal/Local bit) of the first byte.
The first byte 0x18 (binary 00010000) becomes 0x1A (binary 00011010). The expanded interface identifier becomes 1A66:DAFF:FE34:5678.
IEEE EUI-64 rules dictate that the 7th bit of the interface identifier must be inverted (0 indicates local scope, 1 indicates globally unique/universal).
3
Combine the IPv6 link-local prefix with the calculated 64-bit interface identifier.
Prefix fe80:: + 1a66:daff:fe34:5678 = fe80::1a66:daff:fe34:5678.
The `ipv6 enable` command dynamically creates a link-local address starting with the fe80::/10 prefix.

Key Concept

IPv6 EUI-64 Interface Identifier Derivation
Question 360Question →

A Cisco Catalyst switch receives an Ethernet frame on interface FastEthernet0/10. The frame has a source MAC address of 0011.2233.4455 and a destination MAC address of 00aa.bbcc.dd88. The switch MAC address table already contains an active dynamic entry for 0011.2233.4455 mapped to interface FastEthernet0/10, but contains no entry for 00aa.bbcc.dd88. Interfaces FastEthernet0/10 through FastEthernet0/15 are active access ports assigned to VLAN 10, while interface FastEthernet0/16 is an access port assigned to VLAN 20. Which action does the switch perform upon receiving this frame?

Show answer & explanation

Answer: Refreshes the aging timer for MAC 0011.2233.4455 on FastEthernet0/10 and floods the frame out interfaces FastEthernet0/11 through FastEthernet0/15.

Answer

The switch refreshes the aging timer for the existing source MAC entry on interface FastEthernet0/10 and floods the frame out all other active interfaces in VLAN 10 (FastEthernet0/11 through FastEthernet0/15).
When a switch receives a frame, it inspects the source MAC address to update or refresh its MAC address table entry. Next, it looks up the destination MAC address. Because the destination MAC is not in the table, the switch treats the frame as an unknown unicast frame and floods it out all operational interfaces in the same VLAN (VLAN 10), except for the receiving interface (FastEthernet0/10).

Step-by-Step Solution

1
Examine the source MAC address and update the MAC address table.
The source MAC address 0011.2233.4455 is already mapped to interface FastEthernet0/10. The switch resets/refreshes the aging timer for this entry.
MAC learning occurs on every incoming frame. If the entry exists on the correct port, its aging timer is reset.
2
Examine the destination MAC address to determine the forwarding decision.
The destination MAC address 00aa.bbcc.dd88 is not present in the MAC address table (unknown unicast).
Switches use MAC table lookup for unicast forwarding. Absence of an entry triggers unknown unicast flooding.
3
Identify the target egress interfaces for flooding.
The frame is flooded out FastEthernet0/11 through FastEthernet0/15. It is excluded from FastEthernet0/10 (ingress port) and FastEthernet0/16 (different VLAN).
Flooding is constrained within the ingress VLAN boundary and strictly avoids reflecting traffic back out the receiving interface.

Key Concept

Layer 2 Unknown Unicast Flooding and MAC Table Aging
PreviousPage 18 / 20Next
Network Fundamentals Practice Questions — Cisco CCNA — Page 18 | Examkin