Networking Concepts practice questions

A device is configured with IP address 192.168.1.130 and subnet mask 255.255.255.192. What is the network address of this device?

Trap 1: 192.168.1.0

This would be the network address if the subnet were /24, but with /26 it is not correct.

Trap 2: 192.168.1.192

This network address covers 192-255, which does not include 130.

Trap 3: 192.168.1.64

This network address covers 64-127, which does not include 130.

A
192.168.1.0
Why wrong: This would be the network address if the subnet were /24, but with /26 it is not correct.
B
192.168.1.128
Correct: 130 is in the range 128-191, so the network address is 192.168.1.128.
C
192.168.1.192
Why wrong: This network address covers 192-255, which does not include 130.
D
192.168.1.64
Why wrong: This network address covers 64-127, which does not include 130.

Question 2mediummultiple choice

A network technician is assigned a subnet mask of 255.255.255.248. How many usable host addresses are available in this subnet?

Trap 1: B) 8

Incorrect. 8 is the total number of addresses in the subnet, but two are reserved for network and broadcast.

Trap 2: C) 14

Incorrect. 14 usable hosts correspond to a /28 subnet (255.255.255.240), not /29.

Trap 3: D) 30

Incorrect. 30 usable hosts correspond to a /27 subnet (255.255.255.224), not /29.

A
A) 6
Correct. /29 gives 8 total addresses minus 2 = 6 usable hosts.
B
B) 8
Why wrong: Incorrect. 8 is the total number of addresses in the subnet, but two are reserved for network and broadcast.
C
C) 14
Why wrong: Incorrect. 14 usable hosts correspond to a /28 subnet (255.255.255.240), not /29.
D
D) 30
Why wrong: Incorrect. 30 usable hosts correspond to a /27 subnet (255.255.255.224), not /29.

Question 3easymultiple choice

A network engineer is designing a subnet that can support at least 10 hosts. Which subnet mask would provide exactly 14 usable host addresses?

Trap 1: /27

A /27 subnet mask provides 5 host bits (32-27), resulting in 2^5 = 32 total addresses. After accounting for the network and broadcast addresses, this yields 30 usable host addresses. While 30 hosts certainly meets the 'at least 10' requirement, it allocates significantly more addresses than necessary, leading to inefficient use of the IP address space and potential for address exhaustion in larger network designs.

Trap 2: /29

A /29 subnet mask allocates only 3 bits for host addresses (32-29). This results in 2^3 = 8 total IP addresses. Subtracting the mandatory network address and broadcast address leaves only 6 usable host addresses, which is insufficient to meet the requirement of at least 10 hosts for the subnet.

Trap 3: /30

A /30 subnet mask reserves 2 bits for host addresses (32-30), which generates 2^2 = 4 total IP addresses. When the network and broadcast addresses are excluded, only 2 usable host addresses remain. This is far below the minimum requirement of 10 hosts and is typically reserved for point-to-point links between two devices, not for supporting multiple endpoints within a local area network.

A
/27
Why wrong: A /27 subnet mask provides 5 host bits (32-27), resulting in 2^5 = 32 total addresses. After accounting for the network and broadcast addresses, this yields 30 usable host addresses. While 30 hosts certainly meets the 'at least 10' requirement, it allocates significantly more addresses than necessary, leading to inefficient use of the IP address space and potential for address exhaustion in larger network designs.
B
/28
A /28 CIDR prefix designates 4 host bits (32-28), yielding 2^4 = 16 total IP addresses within the subnet. After reserving the network address and broadcast address, this leaves 14 usable host addresses. This perfectly satisfies the requirement of supporting at least 10 hosts, making it the most efficient choice without excessive waste of IP address space.
C
/29
Why wrong: A /29 subnet mask allocates only 3 bits for host addresses (32-29). This results in 2^3 = 8 total IP addresses. Subtracting the mandatory network address and broadcast address leaves only 6 usable host addresses, which is insufficient to meet the requirement of at least 10 hosts for the subnet.
D
/30
Why wrong: A /30 subnet mask reserves 2 bits for host addresses (32-30), which generates 2^2 = 4 total IP addresses. When the network and broadcast addresses are excluded, only 2 usable host addresses remain. This is far below the minimum requirement of 10 hosts and is typically reserved for point-to-point links between two devices, not for supporting multiple endpoints within a local area network.

Question 4easymultiple choice

A network engineer needs to transfer a large database file from one server to another across a WAN link. The transfer must be reliable and guarantee that the data arrives without errors. Which transport layer protocol should the engineer use?

Trap 1: UDP

UDP (User Datagram Protocol) is a connectionless transport protocol that offers no guarantees of delivery, ordering, or error checking. While it provides speed due to its minimal overhead, it lacks the retransmission and flow control mechanisms essential for ensuring that all parts of a large database file arrive intact and in the correct sequence. Using UDP for such a transfer would risk significant data loss and corruption.

Trap 2: IP

IP (Internet Protocol) operates at the Network layer, handling the logical addressing and routing of packets between different networks. Its primary function is to deliver individual packets from a source host to a destination host, but it offers no end-to-end reliability, ordering, or error recovery for the data within those packets. IP relies on higher-layer protocols, like TCP, to provide such transport-layer guarantees for application data.

Trap 3: ICMP

ICMP (Internet Control Message Protocol) operates at the Network layer and is primarily designed for sending error messages and operational information, such as network reachability (ping) or route tracing (traceroute). It does not provide mechanisms for application-level data transfer, nor does it offer any reliability features like retransmission or flow control necessary for moving large files.

A
UDP
Why wrong: UDP (User Datagram Protocol) is a connectionless transport protocol that offers no guarantees of delivery, ordering, or error checking. While it provides speed due to its minimal overhead, it lacks the retransmission and flow control mechanisms essential for ensuring that all parts of a large database file arrive intact and in the correct sequence. Using UDP for such a transfer would risk significant data loss and corruption.
B
IP
Why wrong: IP (Internet Protocol) operates at the Network layer, handling the logical addressing and routing of packets between different networks. Its primary function is to deliver individual packets from a source host to a destination host, but it offers no end-to-end reliability, ordering, or error recovery for the data within those packets. IP relies on higher-layer protocols, like TCP, to provide such transport-layer guarantees for application data.
C
TCP
TCP (Transmission Control Protocol) is a connection-oriented protocol that ensures reliable, ordered, and error-checked delivery of data. For transferring a large database file, TCP establishes a connection, segments the data, and reassembles it at the destination, retransmitting any lost or corrupted segments. This guarantees data integrity and completeness, which is paramount for critical file transfers like databases.
D
ICMP
Why wrong: ICMP (Internet Control Message Protocol) operates at the Network layer and is primarily designed for sending error messages and operational information, such as network reachability (ping) or route tracing (traceroute). It does not provide mechanisms for application-level data transfer, nor does it offer any reliability features like retransmission or flow control necessary for moving large files.

Question 5mediummultiple choice

A network engineer needs to assign IP addresses to a new subnet that will support exactly 25 devices. Which subnet mask would provide the minimum number of usable host addresses while still accommodating the requirement?

Trap 1: 255.255.255.240 (/28)

/28 provides 14 usable hosts, which is insufficient for 25 devices.

Trap 2: 255.255.255.192 (/26)

/26 provides 62 usable hosts, which is more than needed but not the minimum.

Trap 3: 255.255.255.128 (/25)

/25 provides 126 usable hosts, far exceeding the need and not the minimal subnet.

A
255.255.255.240 (/28)
Why wrong: /28 provides 14 usable hosts, which is insufficient for 25 devices.
B
255.255.255.224 (/27)
/27 provides 28 usable hosts, which meets the requirement with minimal waste.
C
255.255.255.192 (/26)
Why wrong: /26 provides 62 usable hosts, which is more than needed but not the minimum.
D
255.255.255.128 (/25)
Why wrong: /25 provides 126 usable hosts, far exceeding the need and not the minimal subnet.

Question 6mediummultiple choice

A company needs a subnet that will support exactly 50 devices. Which subnet mask provides the minimum number of usable host addresses while still accommodating the requirement?

Trap 1: 255.255.255.128 (/25)

A /25 subnet mask allocates 25 bits to the network portion, leaving 7 bits available for host addresses. This provides 2^7 = 128 total IP addresses within the subnet. After accounting for the network and broadcast addresses, 126 usable host addresses are available. While this subnet can accommodate 50 devices, it provides significantly more addresses than needed, leading to inefficient IP address utilization and potential waste of valuable address space.

Trap 2: 255.255.255.224 (/27)

A /27 subnet mask designates 27 bits for the network, leaving only 5 bits for host addresses. This results in 2^5 = 32 total IP addresses for the subnet. Subtracting the network and broadcast addresses leaves only 30 usable host addresses, which is insufficient to support the required 50 devices. Therefore, this subnet is too small for the company's needs.

Trap 3: 255.255.255.240 (/28)

A /28 subnet mask reserves 28 bits for the network, leaving only 4 bits for host addresses. This configuration provides 2^4 = 16 total IP addresses for the subnet. After deducting the network and broadcast addresses, only 14 usable host addresses remain. This is far too few to support the company's requirement for 50 devices, making it an entirely inadequate choice.

A
255.255.255.128 (/25)
Why wrong: A /25 subnet mask allocates 25 bits to the network portion, leaving 7 bits available for host addresses. This provides 2^7 = 128 total IP addresses within the subnet. After accounting for the network and broadcast addresses, 126 usable host addresses are available. While this subnet can accommodate 50 devices, it provides significantly more addresses than needed, leading to inefficient IP address utilization and potential waste of valuable address space.
B
255.255.255.192 (/26)
A /26 subnet mask utilizes 26 bits for the network portion, leaving 6 bits for host addresses. This configuration yields 2^6 = 64 total IP addresses within the subnet. After reserving the network address and broadcast address, 62 usable host addresses remain, which is the smallest subnet that can precisely accommodate the requirement of 50 devices without excessive waste.
C
255.255.255.224 (/27)
Why wrong: A /27 subnet mask designates 27 bits for the network, leaving only 5 bits for host addresses. This results in 2^5 = 32 total IP addresses for the subnet. Subtracting the network and broadcast addresses leaves only 30 usable host addresses, which is insufficient to support the required 50 devices. Therefore, this subnet is too small for the company's needs.
D
255.255.255.240 (/28)
Why wrong: A /28 subnet mask reserves 28 bits for the network, leaving only 4 bits for host addresses. This configuration provides 2^4 = 16 total IP addresses for the subnet. After deducting the network and broadcast addresses, only 14 usable host addresses remain. This is far too few to support the company's requirement for 50 devices, making it an entirely inadequate choice.

Question 7easymultiple choice

A technician needs to connect two different networks and forward traffic based on IP addresses. Which of the following devices operates at Layer 3 of the OSI model and should be used?

Trap 1: Hub

A hub operates at Layer 1 of the OSI model, functioning as a basic multi-port repeater for electrical signals. When a signal arrives on one port, the hub simply regenerates and broadcasts it out to all other connected ports without any intelligent processing or forwarding decisions. Consequently, hubs create a single large collision domain and cannot differentiate traffic based on IP addresses or any other network-layer information.

Trap 2: Switch

A switch operates at Layer 2 (Data Link Layer) of the OSI model, forwarding data frames based on the destination MAC address. Unlike a router, a switch primarily connects devices within the same local area network (LAN) or broadcast domain, learning MAC addresses and building a MAC address table to direct traffic efficiently. It does not inspect or make forwarding decisions based on Layer 3 IP addresses, making it unsuitable for routing traffic between different IP networks.

Trap 3: Bridge

A bridge is a Layer 2 device that connects two or more network segments, typically within the same broadcast domain, by forwarding frames based on MAC addresses. Its primary function is to segment a single large collision domain into smaller ones, improving network performance by filtering traffic between segments. However, like a switch, a bridge operates solely at the data link layer and lacks the intelligence to interpret IP addresses or route traffic between distinct IP networks.

A
Hub
Why wrong: A hub operates at Layer 1 of the OSI model, functioning as a basic multi-port repeater for electrical signals. When a signal arrives on one port, the hub simply regenerates and broadcasts it out to all other connected ports without any intelligent processing or forwarding decisions. Consequently, hubs create a single large collision domain and cannot differentiate traffic based on IP addresses or any other network-layer information.
B
Switch
Why wrong: A switch operates at Layer 2 (Data Link Layer) of the OSI model, forwarding data frames based on the destination MAC address. Unlike a router, a switch primarily connects devices within the same local area network (LAN) or broadcast domain, learning MAC addresses and building a MAC address table to direct traffic efficiently. It does not inspect or make forwarding decisions based on Layer 3 IP addresses, making it unsuitable for routing traffic between different IP networks.
C
Router
Routers are essential Layer 3 network devices specifically designed to connect disparate networks and intelligently forward data packets between them. They achieve this by examining the destination IP address within each packet and consulting their internal routing tables to determine the optimal path. This capability allows them to facilitate communication across different subnets and even the internet, making them fundamental for inter-network connectivity.
D
Bridge
Why wrong: A bridge is a Layer 2 device that connects two or more network segments, typically within the same broadcast domain, by forwarding frames based on MAC addresses. Its primary function is to segment a single large collision domain into smaller ones, improving network performance by filtering traffic between segments. However, like a switch, a bridge operates solely at the data link layer and lacks the intelligence to interpret IP addresses or route traffic between distinct IP networks.

Question 8easymultiple choice

A network administrator needs to create a subnet that supports exactly 50 hosts. Which subnet mask provides the smallest subnet that meets this requirement?

Trap 1: A

A /25 subnet mask dedicates 25 bits to the network, providing 7 bits for host addresses. This results in 2^7 = 128 total IP addresses, which translates to 126 usable host addresses after accounting for the network and broadcast identifiers. While 126 addresses are more than enough for 50 hosts, this option is not the most efficient use of IP space. The question implies selecting the smallest subnet that meets the exact requirement, which /25 exceeds significantly.

Trap 2: C

A /27 subnet mask designates 27 bits for the network, leaving 5 bits for host addresses. This configuration yields 2^5 = 32 total IP addresses. Subtracting the network and broadcast addresses leaves only 30 usable host addresses. This quantity is insufficient to support the required 50 hosts, falling short by 20 addresses.

Trap 3: D

A /28 subnet mask utilizes 28 bits for the network portion, leaving only 4 bits for host addresses. This configuration provides 2^4 = 16 total IP addresses. After subtracting the network and broadcast addresses, only 14 usable host addresses remain. This number is substantially less than the 50 hosts required, rendering this subnet completely inadequate.

A
A
Why wrong: A /25 subnet mask dedicates 25 bits to the network, providing 7 bits for host addresses. This results in 2^7 = 128 total IP addresses, which translates to 126 usable host addresses after accounting for the network and broadcast identifiers. While 126 addresses are more than enough for 50 hosts, this option is not the most efficient use of IP space. The question implies selecting the smallest subnet that meets the exact requirement, which /25 exceeds significantly.
B
B
A /26 subnet mask allocates 26 bits for the network portion, leaving 6 bits for host addresses. This configuration provides 2^6 = 64 total IP addresses. After reserving two addresses for the network identifier and broadcast address, 62 usable host addresses remain. This is the smallest subnet size that can accommodate exactly 50 hosts, making it the most efficient choice.
C
C
Why wrong: A /27 subnet mask designates 27 bits for the network, leaving 5 bits for host addresses. This configuration yields 2^5 = 32 total IP addresses. Subtracting the network and broadcast addresses leaves only 30 usable host addresses. This quantity is insufficient to support the required 50 hosts, falling short by 20 addresses.
D
D
Why wrong: A /28 subnet mask utilizes 28 bits for the network portion, leaving only 4 bits for host addresses. This configuration provides 2^4 = 16 total IP addresses. After subtracting the network and broadcast addresses, only 14 usable host addresses remain. This number is substantially less than the 50 hosts required, rendering this subnet completely inadequate.

Review the full OSPF breakdown →

Question 9hardmultiple choice

A network engineer configures OSPF on two routers with a primary link (1 Gbps) and a backup link (100 Mbps). The engineer expects traffic to always use the primary link unless it fails, but the router is sending traffic over the backup link. What is the most likely cause?

Trap 1: The backup link has a lower OSPF cost than the primary link.

While it is true that OSPF will always prefer a path with a lower cost, stating that 'the backup link has a lower OSPF cost than the primary link' describes the outcome rather than the root cause from the primary link's perspective. The issue arises because the *intended primary* link's cost is comparatively higher, leading OSPF to select the backup path. This option merely reiterates the observed behavior without pinpointing the specific configuration error on the primary link that led to this preference.

Trap 2: The reference bandwidth is set to 1 Gbps on the primary router.

OSPF calculates interface cost using the formula 'reference bandwidth divided by interface bandwidth'. The default reference bandwidth is 100 Mbps. If the reference bandwidth on the primary router were set to 1 Gbps (1000 Mbps), a 1 Gbps primary link would have a cost of 1 (1000/1000). A typical 100 Mbps backup link would then have a cost of 10 (1000/100). This configuration would cause the primary link to be preferred due to its lower cost, which directly contradicts the problem statement where traffic is flowing over the backup link.

Trap 3: The bandwidth command is misconfigured on the backup interface.

The `bandwidth` command on an interface directly influences its OSPF cost, as OSPF uses this value in its cost calculation. While a misconfiguration on the backup interface could certainly alter its OSPF cost, the central problem described is the *primary* link not being preferred. If the backup interface's bandwidth were misconfigured in a way that made its OSPF cost lower than the primary, it would contribute to the issue. However, the most direct cause for the primary link being ignored is a higher OSPF cost configured on the primary link itself, making it less attractive than the backup.

A
The backup link has a lower OSPF cost than the primary link.
Why wrong: While it is true that OSPF will always prefer a path with a lower cost, stating that 'the backup link has a lower OSPF cost than the primary link' describes the outcome rather than the root cause from the primary link's perspective. The issue arises because the *intended primary* link's cost is comparatively higher, leading OSPF to select the backup path. This option merely reiterates the observed behavior without pinpointing the specific configuration error on the primary link that led to this preference.
B
The OSPF cost of the primary link is higher than the backup link.
OSPF utilizes cost as its primary metric for path selection, always favoring routes with the lowest cumulative cost to a destination. If the network engineer intends a specific link to be primary, its OSPF cost must be explicitly configured to be lower than any intended backup paths. In this scenario, if the designated primary link possesses a higher OSPF cost than the backup link, OSPF will logically install the backup link into the routing table as the preferred path, diverting traffic away from the intended primary.
C
The reference bandwidth is set to 1 Gbps on the primary router.
Why wrong: OSPF calculates interface cost using the formula 'reference bandwidth divided by interface bandwidth'. The default reference bandwidth is 100 Mbps. If the reference bandwidth on the primary router were set to 1 Gbps (1000 Mbps), a 1 Gbps primary link would have a cost of 1 (1000/1000). A typical 100 Mbps backup link would then have a cost of 10 (1000/100). This configuration would cause the primary link to be preferred due to its lower cost, which directly contradicts the problem statement where traffic is flowing over the backup link.
D
The bandwidth command is misconfigured on the backup interface.
Why wrong: The `bandwidth` command on an interface directly influences its OSPF cost, as OSPF uses this value in its cost calculation. While a misconfiguration on the backup interface could certainly alter its OSPF cost, the central problem described is the *primary* link not being preferred. If the backup interface's bandwidth were misconfigured in a way that made its OSPF cost lower than the primary, it would contribute to the issue. However, the most direct cause for the primary link being ignored is a higher OSPF cost configured on the primary link itself, making it less attractive than the backup.

Question 10easymultiple choice

Which of the following devices operates at Layer 3 of the OSI model and makes forwarding decisions based on destination IP addresses?

Trap 1: Switch

A network switch primarily operates at Layer 2 (Data Link Layer) of the OSI model. Its core function is to forward data frames within a single local area network (LAN) segment by learning and using Media Access Control (MAC) addresses. When a frame arrives, the switch consults its MAC address table to determine the correct outgoing port, thereby reducing unnecessary traffic and improving network efficiency compared to hubs. It does not inspect or route based on Layer 3 IP addresses.

Trap 2: Bridge

A network bridge is a Layer 2 (Data Link Layer) device designed to connect two separate network segments, effectively creating a single, larger collision domain. Like a switch, it learns MAC addresses of devices on each segment and forwards frames selectively based on the destination MAC address. Bridges are typically simpler than switches, often having fewer ports, but they perform the same fundamental function of segmenting traffic at the MAC address level, not at the IP layer.

Trap 3: Hub

A network hub operates at Layer 1 (Physical Layer) of the OSI model, functioning as a multi-port repeater. When a signal arrives at one port, the hub simply regenerates it and broadcasts it out to all other connected ports, regardless of the intended recipient. This lack of intelligence means all devices connected to a hub share a single collision domain, leading to inefficient network traffic and potential performance bottlenecks, as it does not perform any MAC or IP address-based forwarding.

A
Switch
Why wrong: A network switch primarily operates at Layer 2 (Data Link Layer) of the OSI model. Its core function is to forward data frames within a single local area network (LAN) segment by learning and using Media Access Control (MAC) addresses. When a frame arrives, the switch consults its MAC address table to determine the correct outgoing port, thereby reducing unnecessary traffic and improving network efficiency compared to hubs. It does not inspect or route based on Layer 3 IP addresses.
B
Bridge
Why wrong: A network bridge is a Layer 2 (Data Link Layer) device designed to connect two separate network segments, effectively creating a single, larger collision domain. Like a switch, it learns MAC addresses of devices on each segment and forwards frames selectively based on the destination MAC address. Bridges are typically simpler than switches, often having fewer ports, but they perform the same fundamental function of segmenting traffic at the MAC address level, not at the IP layer.
C
Router
Routers are fundamental network devices operating at Layer 3 (Network Layer) of the OSI model. They are responsible for forwarding IP packets between different networks or subnets by examining the destination IP address in the packet header. To make intelligent forwarding decisions, a router consults its routing table, which contains information about known network paths and their associated next-hop interfaces. This capability enables communication across disparate networks, forming the backbone of the internet.
D
Hub
Why wrong: A network hub operates at Layer 1 (Physical Layer) of the OSI model, functioning as a multi-port repeater. When a signal arrives at one port, the hub simply regenerates it and broadcasts it out to all other connected ports, regardless of the intended recipient. This lack of intelligence means all devices connected to a hub share a single collision domain, leading to inefficient network traffic and potential performance bottlenecks, as it does not perform any MAC or IP address-based forwarding.

Study the full IPv6 explanation →

Question 11mediummultiple choice

A network engineer is configuring IPv6 on a router interface. Which of the following is a valid global unicast IPv6 address?

Trap 1: fe80::1

The address fe80::1 belongs to the fe80::/10 range, which is reserved for IPv6 link-local addresses. These addresses are automatically configured on every IPv6-enabled interface and are strictly confined to a single network segment or "link." Link-local addresses are essential for neighbor discovery, stateless address autoconfiguration (SLAAC), and other on-link communication but are explicitly not routable beyond that local link by any router.

Trap 2: ff02::1

The address ff02::1 is an IPv6 multicast address, specifically the "all-nodes link-local multicast" address. Multicast addresses are used to send a single packet to multiple destinations simultaneously, rather than to a single unique host. This particular address targets all IPv6-enabled devices on the local link, primarily for discovery protocols, and is not a unicast address that can be assigned to a router interface for point-to-point communication.

Trap 3: ::1

The address ::1 is the IPv6 loopback address, analogous to 127.0.0.1 in IPv4. This special address is exclusively used for a device to send traffic to itself, primarily for testing network applications and services running on the local host. It is not designed for communication with other devices on a network segment or for global routing, making it unsuitable for assignment to a physical router interface.

A
fe80::1
Why wrong: The address fe80::1 belongs to the fe80::/10 range, which is reserved for IPv6 link-local addresses. These addresses are automatically configured on every IPv6-enabled interface and are strictly confined to a single network segment or "link." Link-local addresses are essential for neighbor discovery, stateless address autoconfiguration (SLAAC), and other on-link communication but are explicitly not routable beyond that local link by any router.
B
2001:db8::1
The address 2001:db8::1 is a valid IPv6 global unicast address. It falls within the 2000::/3 range designated for global unicast addresses, making it globally routable across the internet. Specifically, 2001:db8::/32 is reserved by RFC 3849 for use in documentation and example code, ensuring that these addresses will not conflict with actual deployed networks. Therefore, it is a perfectly valid address to assign to a router interface for external communication.
C
ff02::1
Why wrong: The address ff02::1 is an IPv6 multicast address, specifically the "all-nodes link-local multicast" address. Multicast addresses are used to send a single packet to multiple destinations simultaneously, rather than to a single unique host. This particular address targets all IPv6-enabled devices on the local link, primarily for discovery protocols, and is not a unicast address that can be assigned to a router interface for point-to-point communication.
D
::1
Why wrong: The address ::1 is the IPv6 loopback address, analogous to 127.0.0.1 in IPv4. This special address is exclusively used for a device to send traffic to itself, primarily for testing network applications and services running on the local host. It is not designed for communication with other devices on a network segment or for global routing, making it unsuitable for assignment to a physical router interface.

Question 12easymultiple choice

Which of the following best describes the primary function of a subnet mask in IPv4 networking?

Trap 1: It dynamically assigns IP addresses to devices on the network

IP address assignment is performed by DHCP, not by the subnet mask.

Trap 2: It encrypts data packets to ensure secure communication

Encryption is provided by protocols such as IPsec or TLS, not by a subnet mask.

Trap 3: It specifies the IP address of the router that forwards traffic to…

The default gateway is configured manually or via DHCP, and is a separate parameter from the subnet mask.

A
It identifies which part of an IP address is the network address and which part is the host address
The subnet mask is used to divide the IP address into network and host bits, allowing devices to determine if a destination is on the same local network or requires routing.
B
It dynamically assigns IP addresses to devices on the network
Why wrong: IP address assignment is performed by DHCP, not by the subnet mask.
C
It encrypts data packets to ensure secure communication
Why wrong: Encryption is provided by protocols such as IPsec or TLS, not by a subnet mask.
D
It specifies the IP address of the router that forwards traffic to other networks
Why wrong: The default gateway is configured manually or via DHCP, and is a separate parameter from the subnet mask.

Read the full wireless explanation →

Question 13easymultiple choice

A network engineer is explaining to a manager why wireless networks often have lower throughput than wired Ethernet. Which of the following best describes the primary reason for this difference?

Trap 1: Wireless uses a different MAC method that requires waiting for an…

While wireless protocols often incorporate acknowledgments (ACKs) to confirm successful data reception due to the unreliable nature of the wireless medium, this is not the primary mechanism responsible for the significant reduction in available bandwidth compared to wired Ethernet. The main overhead stems from the proactive collision avoidance mechanisms of CSMA/CA, such as mandatory listening periods (DIFS, SIFS), random backoff timers, and optional RTS/CTS handshakes, which consume substantial airtime even in the absence of actual data transmission.

Trap 2: Wireless operates at half-duplex, while wired Ethernet typically…

It is true that wireless networks typically operate in half-duplex mode, meaning a device cannot transmit and receive simultaneously on the same frequency, largely due to self-interference and the shared medium nature. Conversely, modern wired Ethernet connections commonly operate in full-duplex, allowing simultaneous transmission and reception, which significantly boosts theoretical throughput. However, the half-duplex nature of wireless is a consequence of the physical layer and medium, whereas the primary reason for lower effective throughput is the extensive MAC layer overhead introduced by CSMA/CA's collision avoidance mechanisms.

Trap 3: Wireless signals are subject to interference, but the MAC protocol…

This option is incorrect because the Medium Access Control (MAC) protocols for wireless and wired Ethernet are fundamentally different. Wired Ethernet uses CSMA/CD (Collision Detection), while wireless networks use CSMA/CA (Collision Avoidance). While wireless signals are indeed susceptible to various forms of interference (e.g., from other Wi-Fi networks, microwaves, Bluetooth), which can degrade throughput, the core architectural difference in their MAC layer protocols is the primary reason for their inherent performance disparity.

A
Wireless uses a different MAC method that requires waiting for an acknowledgment, reducing available bandwidth.
Why wrong: While wireless protocols often incorporate acknowledgments (ACKs) to confirm successful data reception due to the unreliable nature of the wireless medium, this is not the primary mechanism responsible for the significant reduction in available bandwidth compared to wired Ethernet. The main overhead stems from the proactive collision avoidance mechanisms of CSMA/CA, such as mandatory listening periods (DIFS, SIFS), random backoff timers, and optional RTS/CTS handshakes, which consume substantial airtime even in the absence of actual data transmission.
B
Wireless operates at half-duplex, while wired Ethernet typically operates at full-duplex.
Why wrong: It is true that wireless networks typically operate in half-duplex mode, meaning a device cannot transmit and receive simultaneously on the same frequency, largely due to self-interference and the shared medium nature. Conversely, modern wired Ethernet connections commonly operate in full-duplex, allowing simultaneous transmission and reception, which significantly boosts theoretical throughput. However, the half-duplex nature of wireless is a consequence of the physical layer and medium, whereas the primary reason for lower effective throughput is the extensive MAC layer overhead introduced by CSMA/CA's collision avoidance mechanisms.
C
Wireless uses CSMA/CA which involves a collision avoidance mechanism that adds overhead, whereas wired Ethernet uses CSMA/CD which only responds after collision.
This statement is accurate because wireless networks employ Carrier Sense Multiple Access with Collision Avoidance (CSMA/CA). This protocol mandates a station to listen for a clear channel, wait for a Distributed Interframe Space (DIFS), and potentially use Request To Send/Clear To Send (RTS/CTS) before transmitting, all of which introduce significant latency and overhead. In contrast, wired Ethernet utilizes CSMA/CD, where stations transmit immediately upon sensing a clear channel and only react to collisions if they occur, making it more efficient in a controlled, detectable collision domain.
D
Wireless signals are subject to interference, but the MAC protocol is identical to Ethernet.
Why wrong: This option is incorrect because the Medium Access Control (MAC) protocols for wireless and wired Ethernet are fundamentally different. Wired Ethernet uses CSMA/CD (Collision Detection), while wireless networks use CSMA/CA (Collision Avoidance). While wireless signals are indeed susceptible to various forms of interference (e.g., from other Wi-Fi networks, microwaves, Bluetooth), which can degrade throughput, the core architectural difference in their MAC layer protocols is the primary reason for their inherent performance disparity.

Question 14easymultiple choice

Which of the following is a characteristic of a connectionless protocol at the transport layer?

Trap 1: It establishes a session before sending data

Establishing a session, often involving a multi-step handshake process like TCP's three-way handshake, is a hallmark of connection-oriented protocols. This setup phase creates a logical connection and synchronizes sequence numbers and window sizes between the sender and receiver. Connectionless protocols, by their very nature, bypass this session establishment entirely, immediately sending data without any prior negotiation or state synchronization, which reduces latency but sacrifices reliability.

Trap 2: It guarantees delivery using acknowledgments

Connectionless protocols do not guarantee delivery because they lack the necessary mechanisms, such as acknowledgments. An acknowledgment (ACK) is a message sent by the receiver back to the sender to confirm successful receipt of data. Without these acknowledgments, the sender has no confirmation that the data arrived, and therefore cannot ensure its delivery, making the service "best-effort" rather than guaranteed.

Trap 3: It retransmits lost segments

Connectionless protocols, like UDP, do not implement mechanisms for detecting or retransmitting lost data segments. Since they do not maintain connection state or use acknowledgments, they have no way of knowing if a segment reached its destination or if it was dropped en route. This lack of retransmission contributes to their lower overhead and faster transmission, but also means applications relying on them must handle reliability at a higher layer if needed.

A
It establishes a session before sending data
Why wrong: Establishing a session, often involving a multi-step handshake process like TCP's three-way handshake, is a hallmark of connection-oriented protocols. This setup phase creates a logical connection and synchronizes sequence numbers and window sizes between the sender and receiver. Connectionless protocols, by their very nature, bypass this session establishment entirely, immediately sending data without any prior negotiation or state synchronization, which reduces latency but sacrifices reliability.
B
It guarantees delivery using acknowledgments
Why wrong: Connectionless protocols do not guarantee delivery because they lack the necessary mechanisms, such as acknowledgments. An acknowledgment (ACK) is a message sent by the receiver back to the sender to confirm successful receipt of data. Without these acknowledgments, the sender has no confirmation that the data arrived, and therefore cannot ensure its delivery, making the service "best-effort" rather than guaranteed.
C
It does not require a virtual circuit
Connectionless protocols, such as UDP at the transport layer, operate by sending individual data units (datagrams) independently. They do not establish a dedicated, end-to-end logical path, known as a virtual circuit, between the sender and receiver prior to data transmission. Each datagram is treated as a standalone entity, routed based solely on its destination address without any pre-negotiated state or connection setup overhead. This design prioritizes speed and efficiency over reliability.
D
It retransmits lost segments
Why wrong: Connectionless protocols, like UDP, do not implement mechanisms for detecting or retransmitting lost data segments. Since they do not maintain connection state or use acknowledgments, they have no way of knowing if a segment reached its destination or if it was dropped en route. This lack of retransmission contributes to their lower overhead and faster transmission, but also means applications relying on them must handle reliability at a higher layer if needed.

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Question 15mediumdrag order

Drag and drop the steps to configure a static route on a network router into the correct order.

Drag steps to the numbered slots on the right, or tap a step then tap a slot.

Steps

Order

1Step 1

2Step 2

3Step 3

4Step 4

5Step 5

Trap 1: Step 1: Access the router's configuration mode

This order is incorrect because attempting to verify the static route using `show ip route` (Step 2) before it has actually been configured (Step 3) will not yield the desired result. At the point of verification, the new static route does not exist in the router's running configuration or routing table. Therefore, `show ip route` would only display existing routes, making this verification step premature and ineffective for confirming the intended new route.

Trap 2: Step 1: Define the static route with the destination network,…

This option is incorrect because the `ip route` command, which is used to configure a static route, is a global configuration command. It cannot be executed directly from user EXEC mode or privileged EXEC mode. Attempting to enter the `ip route` command (Step 1) without first entering global configuration mode (`configure terminal`) (Step 2) would result in a command rejection error, preventing any configuration from taking place.

Trap 3: Step 1: Access the router's configuration mode

This sequence is incorrect because saving the configuration (Step 2) occurs prematurely, before the static route has actually been configured (Step 3). When `copy running-config startup-config` is executed, the `running-config` does not yet contain the new static route, meaning the `startup-config` would be updated without the intended change. Consequently, the static route would not be saved, and verifying it after this incorrect save but before configuration would be meaningless.

A
Step 1: Access the router's configuration mode. Step 2: Define the static route with the destination network, subnet mask, and next hop. Step 3: Verify the route appears in the routing table. Step 4: Save the configuration to persistent storage.
This sequence correctly outlines the necessary steps to implement a static route on a Cisco device. First, entering global configuration mode (`configure terminal`) is essential as the `ip route` command is a global configuration command. Next, the `ip route` command is executed to add the static route to the router's running configuration. Subsequently, `show ip route` is used to verify that the route has been successfully added to the routing table and is operational. Finally, `copy running-config startup-config` saves the changes from volatile RAM to non-volatile memory, ensuring the static route persists across reboots.
B
Step 1: Access the router's configuration mode. Step 2: Verify the route appears in the routing table. Step 3: Define the static route with the destination network, subnet mask, and next hop. Step 4: Save the configuration to persistent storage.
Why wrong: This order is incorrect because attempting to verify the static route using `show ip route` (Step 2) before it has actually been configured (Step 3) will not yield the desired result. At the point of verification, the new static route does not exist in the router's running configuration or routing table. Therefore, `show ip route` would only display existing routes, making this verification step premature and ineffective for confirming the intended new route.
C
Step 1: Define the static route with the destination network, subnet mask, and next hop. Step 2: Access the router's configuration mode. Step 3: Verify the route appears in the routing table. Step 4: Save the configuration to persistent storage.
Why wrong: This option is incorrect because the `ip route` command, which is used to configure a static route, is a global configuration command. It cannot be executed directly from user EXEC mode or privileged EXEC mode. Attempting to enter the `ip route` command (Step 1) without first entering global configuration mode (`configure terminal`) (Step 2) would result in a command rejection error, preventing any configuration from taking place.
D
Step 1: Access the router's configuration mode. Step 2: Save the configuration to persistent storage. Step 3: Define the static route with the destination network, subnet mask, and next hop. Step 4: Verify the route appears in the routing table.
Why wrong: This sequence is incorrect because saving the configuration (Step 2) occurs prematurely, before the static route has actually been configured (Step 3). When `copy running-config startup-config` is executed, the `running-config` does not yet contain the new static route, meaning the `startup-config` would be updated without the intended change. Consequently, the static route would not be saved, and verifying it after this incorrect save but before configuration would be meaningless.

Review the full routing breakdown →

Question 16mediumdrag order

Drag and drop the steps to recover a lost router password on a Cisco device into the correct order.

Drag steps to the numbered slots on the right, or tap a step then tap a slot.

Steps

Order

1Step 1

2Step 2

3Step 3

4Step 4

5Step 5

Trap 1: 1

This is incorrect because reloading before changing the register means the router will boot with the original register value, ignoring the registers change, so the startup config will be loaded and you cannot modify it.

Trap 2: 1

Attempting to change the configuration register with the "confreg" command as the first step is incorrect because this command is only accessible from ROMmon mode. To enter ROMmon, the boot sequence must first be interrupted. Without this initial interruption, the router will proceed with its normal boot process, making the "confreg" command unavailable and resulting in an unrecognized command error, thus preventing any progress in the password recovery procedure.

Trap 3: 1

This sequence is flawed from the outset because the "confreg" command, used to modify the configuration register, is a utility exclusive to the ROM Monitor (ROMmon) environment. It is impossible to execute this command from the normal operating system or during the initial boot phase before ROMmon has been accessed. Therefore, attempting to change the register as the very first action, prior to interrupting the boot sequence to enter ROMmon, will fail, rendering the entire recovery process ineffective.

A
1. Interrupt boot sequence, 2. Change configuration register to 0x2142, 3. Reload router, 4. Copy startup-config to running-config, 5. Change enable password, 6. Reset configuration register to 0x2102, 7. Save running-config, 8. Reload router
This order is correct because you must first halt the boot to access ROMmon, then set the register to ignore startup config, reload to load the OS without config, copy the config to modify it, change the password, save, reset the register to boot normally, and final reload.
B
1. Interrupt boot sequence, 2. Reload router, 3. Change configuration register to 0x2142, 4. Copy startup-config to running-config, 5. Change enable password, 6. Save running-config, 7. Reset configuration register to 0x2102, 8. Reload router
Why wrong: This is incorrect because reloading before changing the register means the router will boot with the original register value, ignoring the registers change, so the startup config will be loaded and you cannot modify it.
C
1. Change configuration register to 0x2142, 2. Reload router, 3. Interrupt boot sequence, 4. Copy startup-config to running-config, 5. Change enable password, 6. Save running-config, 7. Reset configuration register to 0x2102, 8. Reload router
Why wrong: Attempting to change the configuration register with the "confreg" command as the first step is incorrect because this command is only accessible from ROMmon mode. To enter ROMmon, the boot sequence must first be interrupted. Without this initial interruption, the router will proceed with its normal boot process, making the "confreg" command unavailable and resulting in an unrecognized command error, thus preventing any progress in the password recovery procedure.
D
1. Change configuration register to 0x2142, 2. Interrupt boot sequence, 3. Reload router, 4. Copy startup-config to running-config, 5. Change enable password, 6. Save running-config, 7. Reset configuration register to 0x2102, 8. Reload router
Why wrong: This sequence is flawed from the outset because the "confreg" command, used to modify the configuration register, is a utility exclusive to the ROM Monitor (ROMmon) environment. It is impossible to execute this command from the normal operating system or during the initial boot phase before ROMmon has been accessed. Therefore, attempting to change the register as the very first action, prior to interrupting the boot sequence to enter ROMmon, will fail, rendering the entire recovery process ineffective.

Question 17easymultiple choice

A network administrator is troubleshooting a connectivity issue and suspects the problem is related to the physical cabling. At which layer of the OSI model should the administrator begin their investigation?

Trap 1: Transport layer

The Transport layer (Layer 4) provides end-to-end communication services between applications, including segmentation, reassembly, flow control, and error recovery (for TCP). It operates logically between two host processes, ensuring data delivery across the network. A problem at this layer would typically involve issues like blocked ports, incorrect firewall rules preventing application communication, or application-level protocol errors, assuming the underlying network connectivity (including physical cabling) is already established and functional.

Trap 2: Data Link layer

The Data Link layer (Layer 2) is responsible for reliable data transfer across a direct link, handling framing, error detection within frames, and Media Access Control (MAC) addressing. While it interacts with the Physical layer, its concerns are logical link establishment and managing access to the shared medium, not the physical integrity of the cable itself. Problems at this layer typically involve MAC address conflicts, incorrect frame formatting, or issues with switches and network interface cards (NICs) that prevent proper frame delivery, assuming the underlying physical connection is sound.

Trap 3: Network layer

The Network layer handles logical addressing and routing decisions, not physical cabling faults. The administrator suspects a cable issue, which manifests as signal degradation or electrical problems at the Physical layer (Layer 1). This option is tempting because network-layer troubleshooting often involves connectivity failures, but those stem from IP misconfiguration or routing loops, not from the physical medium itself.

A
Transport layer
Why wrong: The Transport layer (Layer 4) provides end-to-end communication services between applications, including segmentation, reassembly, flow control, and error recovery (for TCP). It operates logically between two host processes, ensuring data delivery across the network. A problem at this layer would typically involve issues like blocked ports, incorrect firewall rules preventing application communication, or application-level protocol errors, assuming the underlying network connectivity (including physical cabling) is already established and functional.
B
Data Link layer
Why wrong: The Data Link layer (Layer 2) is responsible for reliable data transfer across a direct link, handling framing, error detection within frames, and Media Access Control (MAC) addressing. While it interacts with the Physical layer, its concerns are logical link establishment and managing access to the shared medium, not the physical integrity of the cable itself. Problems at this layer typically involve MAC address conflicts, incorrect frame formatting, or issues with switches and network interface cards (NICs) that prevent proper frame delivery, assuming the underlying physical connection is sound.
C
Physical layer
The Physical layer (Layer 1) is fundamentally responsible for the raw bit stream transmission over the physical medium, defining the electrical, mechanical, and procedural specifications for transmitting data signals. Therefore, issues such as faulty cables, damaged connectors, incorrect cable types, or signal degradation due to attenuation or electromagnetic interference directly manifest as problems at this foundational layer, preventing any higher-layer communication from occurring reliably or at all.
D
Network layer
Why wrong: The Network layer handles logical addressing and routing decisions, not physical cabling faults. The administrator suspects a cable issue, which manifests as signal degradation or electrical problems at the Physical layer (Layer 1). This option is tempting because network-layer troubleshooting often involves connectivity failures, but those stem from IP misconfiguration or routing loops, not from the physical medium itself.

Question 18mediummultiple choice

A network technician is explaining the concept of encapsulation to a junior technician. At which OSI layer does a packet get encapsulated with a source and destination IP address?

Trap 1: Layer 2

Layer 2, the Data Link layer, is responsible for encapsulating the Layer 3 IP packet into a frame for transmission over a local network segment. While it adds its own header and trailer, which include MAC addresses for physical addressing within that segment and a Frame Check Sequence for error detection, it does not add IP addresses. IP addressing is a higher-level function handled exclusively by Layer 3.

Trap 2: Layer 4

The Transport layer (Layer 4) performs encapsulation by adding a header to application data, forming segments (TCP) or datagrams (UDP). This header includes source and destination port numbers, which are essential for directing data to the correct application process on the host, as well as managing flow control and reliability. However, Layer 4 operates above the network layer and is not involved in adding IP addresses, which are logical network identifiers.

Trap 3: Layer 1

Layer 1, the Physical layer, is concerned solely with the physical transmission of raw bit streams across the network medium, such as cables or wireless signals. It defines the electrical, mechanical, procedural, and functional specifications for activating, maintaining, and deactivating physical links. This layer does not perform any form of logical encapsulation or add any headers or trailers; it simply converts the frame's bits into a physical signal.

A
Layer 2
Why wrong: Layer 2, the Data Link layer, is responsible for encapsulating the Layer 3 IP packet into a frame for transmission over a local network segment. While it adds its own header and trailer, which include MAC addresses for physical addressing within that segment and a Frame Check Sequence for error detection, it does not add IP addresses. IP addressing is a higher-level function handled exclusively by Layer 3.
B
Layer 3
The Network layer (Layer 3) is precisely where the IP header is added to the Transport layer segment or datagram, creating an IP packet. This header contains the crucial source and destination IP addresses, enabling logical addressing across disparate networks. This encapsulation is fundamental for routing decisions, allowing data to traverse multiple routers and reach its ultimate destination beyond the local network segment.
C
Layer 4
Why wrong: The Transport layer (Layer 4) performs encapsulation by adding a header to application data, forming segments (TCP) or datagrams (UDP). This header includes source and destination port numbers, which are essential for directing data to the correct application process on the host, as well as managing flow control and reliability. However, Layer 4 operates above the network layer and is not involved in adding IP addresses, which are logical network identifiers.
D
Layer 1
Why wrong: Layer 1, the Physical layer, is concerned solely with the physical transmission of raw bit streams across the network medium, such as cables or wireless signals. It defines the electrical, mechanical, procedural, and functional specifications for activating, maintaining, and deactivating physical links. This layer does not perform any form of logical encapsulation or add any headers or trailers; it simply converts the frame's bits into a physical signal.

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Question 19hardmultiple choice

A network engineer is designing a new IPv6 addressing scheme. The company has been assigned a /48 prefix and needs to support up to 250 subnets. Which subnet size should be used to minimize waste while meeting the requirement?

Trap 1: /52

Selecting a /52 prefix for subnetting from a /48 global routing prefix allocates only 4 bits for subnet identification (52 - 48 = 4). This results in 2^4, or 16, available subnets. This quantity is significantly less than the required 250 subnets, rendering /52 an inadequate choice for the specified network design.

Trap 2: /64

A /64 prefix uses 16 bits for subnetting (48+16=64), providing 65,536 subnets. While it works, it wastes many subnet IDs and is not minimal for 250 subnets. Also, /64 is typically used for end-user subnets, not as a subnetting strategy from a /48.

Trap 3: /60

Utilizing a /60 prefix for subnetting from a /48 global routing prefix dedicates 12 bits for subnet identification (60 - 48 = 12). This provides 2^12, or 4,096, distinct subnets. While this technically satisfies the requirement for 250 subnets, it represents a substantial over-allocation and inefficient use of the available address space, making it a less optimal choice compared to a /56.

A
/52
Why wrong: Selecting a /52 prefix for subnetting from a /48 global routing prefix allocates only 4 bits for subnet identification (52 - 48 = 4). This results in 2^4, or 16, available subnets. This quantity is significantly less than the required 250 subnets, rendering /52 an inadequate choice for the specified network design.
B
/56
A /48 global routing prefix is typically assigned to an organization. To create 250 subnets, additional bits beyond the /48 must be used for subnetting. A /56 prefix length means 8 bits are available for subnetting (56 - 48 = 8). This yields 2^8, or 256, possible subnets, which perfectly accommodates the requirement for 250 subnets with minimal address space waste.
C
/64
Why wrong: A /64 prefix uses 16 bits for subnetting (48+16=64), providing 65,536 subnets. While it works, it wastes many subnet IDs and is not minimal for 250 subnets. Also, /64 is typically used for end-user subnets, not as a subnetting strategy from a /48.
D
/60
Why wrong: Utilizing a /60 prefix for subnetting from a /48 global routing prefix dedicates 12 bits for subnet identification (60 - 48 = 12). This provides 2^12, or 4,096, distinct subnets. While this technically satisfies the requirement for 250 subnets, it represents a substantial over-allocation and inefficient use of the available address space, making it a less optimal choice compared to a /56.

Question 20easymultiple choice

A network technician is explaining the difference between a hub and a switch to a junior technician. Which statement correctly describes a key difference between these devices?

Trap 1: A hub operates at Layer 2, while a switch operates at Layer 1.

This statement incorrectly assigns the OSI model layers to the devices. A hub is a Layer 1 (Physical Layer) device, meaning it only deals with the physical transmission and reception of raw bit streams without any understanding of data frames or addresses. Conversely, a switch operates at Layer 2 (Data Link Layer), where it can interpret MAC addresses, build a MAC address table, and make intelligent forwarding decisions based on these addresses, providing more advanced network functionality.

Trap 2: A hub uses MAC addresses to make forwarding decisions, while a…

This statement is incorrect because hubs are Layer 1 devices and operate solely on electrical signals, lacking any intelligence to interpret MAC addresses for forwarding decisions; they simply broadcast all received data. Conversely, a standard Ethernet switch operates at Layer 2 (Data Link Layer) and makes its forwarding decisions based on destination MAC addresses, which it learns and stores in its MAC address table. While Layer 3 switches exist, a typical Layer 2 switch does not use IP addresses for basic frame forwarding.

Trap 3: A hub creates a separate collision domain for each port, while a…

This statement is factually reversed regarding collision domains. A hub operates as a shared medium, meaning all devices connected to it are part of a single, large collision domain. Any transmission from one device is seen by all others, and simultaneous transmissions result in collisions, degrading performance. In contrast, a switch creates a separate collision domain for each of its ports, effectively isolating traffic and preventing collisions between devices connected to different ports, thereby enhancing network throughput.

A
A hub operates at Layer 2, while a switch operates at Layer 1.
Why wrong: This statement incorrectly assigns the OSI model layers to the devices. A hub is a Layer 1 (Physical Layer) device, meaning it only deals with the physical transmission and reception of raw bit streams without any understanding of data frames or addresses. Conversely, a switch operates at Layer 2 (Data Link Layer), where it can interpret MAC addresses, build a MAC address table, and make intelligent forwarding decisions based on these addresses, providing more advanced network functionality.
B
A hub sends frames out all ports except the incoming port; a switch sends frames only to the port with the matching MAC address.
A hub is a Layer 1 device that simply regenerates electrical signals, broadcasting incoming data frames to all connected ports except the one from which the frame originated, without any intelligence. In contrast, a switch is a Layer 2 device that intelligently learns the MAC addresses of connected devices and stores them in its MAC address table. When a switch receives a frame, it examines the destination MAC address and forwards the frame only to the specific port associated with that MAC address, significantly reducing unnecessary network traffic and improving efficiency.
C
A hub uses MAC addresses to make forwarding decisions, while a switch uses IP addresses.
Why wrong: This statement is incorrect because hubs are Layer 1 devices and operate solely on electrical signals, lacking any intelligence to interpret MAC addresses for forwarding decisions; they simply broadcast all received data. Conversely, a standard Ethernet switch operates at Layer 2 (Data Link Layer) and makes its forwarding decisions based on destination MAC addresses, which it learns and stores in its MAC address table. While Layer 3 switches exist, a typical Layer 2 switch does not use IP addresses for basic frame forwarding.
D
A hub creates a separate collision domain for each port, while a switch creates a single collision domain.
Why wrong: This statement is factually reversed regarding collision domains. A hub operates as a shared medium, meaning all devices connected to it are part of a single, large collision domain. Any transmission from one device is seen by all others, and simultaneous transmissions result in collisions, degrading performance. In contrast, a switch creates a separate collision domain for each of its ports, effectively isolating traffic and preventing collisions between devices connected to different ports, thereby enhancing network throughput.