D-NWG-FN-23 Sample Questions & Answers
Ethernet addressing, LLDP and VLANs make up nearly half the weighting, alongside IPv4 and IPv6 addressing schemes, switch access security and ACLs, device purposes and initial configuration, file transfer and troubleshooting, and Smart Fabric automation.
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- Question 1Advanced
Ethernet Technologies · STP (Spanning Tree Protocol)
A financial services company is setting up a new branch office. The network design includes two core Dell PowerSwitch switches (Core-1 and Core-2) and two access switches (Access-1 and Access-2). To provide redundancy, each access switch is connected to both core switches, creating a square topology. The primary requirement is to ensure the most efficient and predictable Layer 2 traffic path while preventing loops. The core switches have higher processing power than the access switches.
The network architect has decided that Core-1 should be the primary root bridge for the Spanning Tree Protocol (STP) domain, and Core-2 should be the secondary root bridge. All traffic from the access layer should ideally flow through Core-1. The default STP priorities are in place on all switches.
Which configuration action will most effectively achieve this design goal?
graph TD subgraph Core Layer C1(Core-1) C2(Core-2) end subgraph Access Layer A1(Access-1) A2(Access-2) end C1 --- C2 A1 --- C1 A1 --- C2 A2 --- C1 A2 --- C2Show answer & explanation
Correct answer: B
In STP, the switch with the lowest bridge priority becomes the root bridge. The default priority is 32768. To make Core-1 the primary root, its priority must be set to a value lower than all other switches. Setting Core-2's priority to the next lowest value ensures it will become the root bridge if Core-1 fails. Setting Core-1 to 4096 and Core-2 to 8192 accomplishes this, as both values are lower than the default and 4096 is lower than 8192.
- Question 2IntermediateSelect 2
Ethernet Technologies · LAG (Link Aggregation Group)
A network engineer is configuring a connection between a server with two 10 GbE NICs and a Dell PowerSwitch. What are the primary benefits of configuring a Link Aggregation Group (LAG) using LACP for this connection? (Select TWO)
Show answer & explanation
Correct answers: A, C
- Question 3Beginner
Ethernet Technologies · Storage Networking Features
When optimizing a network segment dedicated to iSCSI storage traffic, an administrator considers enabling jumbo frames. What is the primary benefit of using jumbo frames for this type of traffic?
Show answer & explanation
Correct answer: B
Jumbo frames increase the Ethernet frame size from the standard 1500 bytes to around 9000 bytes. This means that for the same amount of data, fewer frames (and thus fewer headers) need to be processed. This reduction in frame processing leads to lower CPU utilization on both the sending server and the receiving storage array, improving overall throughput for large block storage traffic like iSCSI.
- Question 4Intermediate
Ethernet Technologies · VLANs (Virtual Local Area Networks)
A user's workstation is connected to a port assigned to VLAN 10. A server is connected to a different port on the same Dell PowerSwitch, and this port is assigned to VLAN 20. Both devices are configured with correct IP addresses in their respective subnets. The user cannot access the server. What is required to enable communication between these two devices?
Show answer & explanation
Correct answer: C
VLANs create separate Layer 2 broadcast domains. By design, devices in different VLANs cannot communicate directly. To enable communication, a Layer 3 device (like a router or a Layer 3 switch) is needed to route traffic between the different VLANs/subnets. This process is called inter-VLAN routing.
- Question 5Advanced
Ethernet Technologies · STP (Spanning Tree Protocol)
In Rapid Spanning Tree Protocol (RSTP), what is the port state that combines the STP states of Blocking, Listening, and Disabled into a single state?
Show answer & explanation
Correct answer: C
RSTP (IEEE 802.1w) simplifies the STP port states for faster convergence. It combines the original STP states of Disabled, Blocking, and Listening into a single 'Discarding' state. In the Discarding state, the port does not forward frames or learn MAC addresses.
- Question 6Intermediate
Ethernet Technologies · LLDP (Link Layer Discovery Protocol)
An engineer needs to identify the device name, port ID, and capabilities of a directly connected switch. Which Dell OS10 command should be used to display this information for interface
ethernet 1/1/5?Show answer & explanation
Correct answer: C
The
show lldp neighborscommand is used to display information about devices discovered via LLDP. Specifyinginterface ethernet 1/1/5filters the output to show only the neighbor connected to that specific port, providing details like Chassis ID (device name), Port ID, and system capabilities. - Question 7Intermediate
Ethernet Technologies · LAG (Link Aggregation Group)
A new server is connected to a Dell PowerSwitch port that is part of a 4-port LACP-based LAG. The administrator connects all four network cables from the server to the switch. However, the LAG fails to become active and only one link shows as operational. Which of the following is the most likely cause for this issue?
Show answer & explanation
Correct answer: B
For a dynamic LACP LAG to form, both ends of the connection (the switch and the server) must be configured for LACP and agree on the parameters. If the server side is not configured for LACP teaming/bonding, the switch will not receive LACP PDUs and will not bundle the ports into an active LAG, often resulting in only one link becoming active due to STP.
- Question 8Beginner
IP Networking · Private and Public IP Addressing
Which of the following IP addresses is defined by RFC 1918 for use in private networks?
Show answer & explanation
Correct answer: B
RFC 1918 specifies three blocks of IP addresses for private internets: 10.0.0.0/8, 172.16.0.0/12 (i.e., 172.16.0.0 to 172.31.255.255), and 192.168.0.0/16. The address 192.168.1.100 falls within the 192.168.0.0/16 range. 172.32.10.5 is outside the private range.
- Question 9Intermediate
IP Networking · IPv4 Addressing
A network administrator is given the IP address 192.168.50.133 and a subnet mask of 255.255.255.224. What is the broadcast address for this subnet?
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Correct answer: B
A subnet mask of 255.255.255.224 (/27) has a block size of 32 (256-224=32). The subnets are .0, .32, .64, .96, .128, .160, etc. The IP address 192.168.50.133 falls into the subnet that starts at 192.168.50.128. The next subnet starts at .160, so the broadcast address for the .128 subnet is the address just before that, which is 192.168.50.159.
- Question 10Intermediate
IP Networking · IPv6 Addressing
According to IPv6 address compression rules, which is the correct way to shorten the address 2001:0db8:0000:0000:a1b2:00ff:fe34:5678?
Show answer & explanation
Correct answer: A
IPv6 compression rules allow two things: 1) removing leading zeros in each hextet, and 2) using a double colon (::) once to represent the longest consecutive string of all-zero hextets. In this address, '0db8' becomes 'db8', the two '0000' hextets become '::', and '00ff' becomes 'ff'. This results in 2001:db8::a1b2:ff:fe34:5678.
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