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Network Implementation

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Question 141Question →

A network administrator is configuring an IEEE 802.1Q trunk link between two managed switches to carry traffic for multiple VLANs over a single physical cable. Which of the following statements correctly describe the default operation and frame structure of IEEE 802.1Q trunking? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: Frames associated with standard tagged VLANs have a 4-byte header containing the VLAN ID inserted into the Ethernet frame.; Frames belonging to the native VLAN are transmitted across the trunk link without an 802.1Q header tag.

Answer

IEEE 802.1Q trunking inserts a 4-byte tag into frames for tagged VLANs and transmits native VLAN frames without an added VLAN header tag.
IEEE 802.1Q trunking operates by inserting a 4-byte header tag into standard Ethernet frames to identify VLAN membership across switches. Traffic belonging to the native VLAN is transmitted untagged by default.

Step-by-Step Solution

1
Identify how IEEE 802.1Q maintains VLAN separation across a trunk link.
A 4-byte VLAN header tag is inserted into Ethernet frames belonging to tagged VLANs.
The receiving switch reads this tag to determine which VLAN broadcast domain the frame belongs to.
2
Identify how IEEE 802.1Q handles native VLAN traffic.
Frames on the native VLAN are sent across the trunk link without a tag.
Native VLAN support allows untagged legacy traffic or control frames to pass across trunk interfaces.

Key Concept

802.1Q Frame Tagging and Native VLAN Operation
Question 142Question →

A network engineer is configuring switch interface `GigabitEthernet1/0/12` on `SW-DIST-05` to connect a multi-SSID wireless access point. Management traffic from the access point must travel untagged on VLAN 15, while client traffic is tagged on VLAN 45 (Corporate) and VLAN 85 (Guest).

Initially, the interface was configured as follows:
text switchport mode trunk switchport trunk native vlan 15 switchport trunk allowed vlan 15,45,85
During a maintenance window, a junior technician updates the interface configuration by executing:
text switchport trunk allowed vlan 45,85
Which of the following describes the immediate operational outcome of this configuration change?

Show answer & explanation

Answer: Untagged management traffic on VLAN 15 will be dropped at the switchport because VLAN 15 is missing from the allowed VLAN list.

Answer

Untagged management traffic on VLAN 15 will be dropped at the switchport because VLAN 15 is missing from the allowed VLAN list.
On an 802.1Q trunk, the interface filters both tagged and untagged traffic using the allowed VLAN list. When the native VLAN (VLAN 15) is removed from the allowed VLAN list, untagged management frames arriving from the wireless access point are dropped by the switch interface because VLAN 15 is no longer permitted on the link.

Step-by-Step Solution

1
Analyze the original trunk configuration.
Interface `GigabitEthernet1/0/12` was configured to use VLAN 15 as the native VLAN, allowing untagged frames from the AP to be processed in VLAN 15. The allowed list explicitly permitted VLANs 15, 45, and 85.
By default, an 802.1Q trunk processes untagged ingress frames into the designated native VLAN.
2
Evaluate the impact of the modification command.
The technician ran `switchport trunk allowed vlan 45,85`, overwriting the allowed list and removing VLAN 15.
On Cisco switches, `switchport trunk allowed vlan <list>` replaces the current allowed VLAN set rather than appending to it unless the `add` keyword is used.
3
Determine the forwarding behavior for native VLAN traffic when excluded from the allowed list.
Because VLAN 15 is no longer in the allowed list, any untagged ingress or egress frames associated with native VLAN 15 are pruned/dropped at the trunk port.
An 802.1Q trunk interface only forwards traffic for VLANs present in its active allowed VLAN list, regardless of the native VLAN setting.

Key Concept

802.1Q Trunk Allowed VLAN Pruning and Native VLAN Processing
Question 143Question →

A network technician is deploying multiple wireless access points across an office floor using the 2.4 GHz frequency band in North America. To prevent co-channel interference between adjacent access points, the technician needs to select channels that do not overlap. Which combination of channels should the technician assign to the access points?

Show answer & explanation

Answer: Channels 1, 6, and 11

Answer

Channels 1, 6, and 11 are the three standard non-overlapping channels in the 2.4 GHz wireless spectrum for North America.
In 2.4 GHz wireless deployments in North America, standard Wi-Fi signals occupy 20 MHz of channel bandwidth. Because center frequencies are spaced 5 MHz apart, selecting channels 1, 6, and 11 provides a 25 MHz separation between center frequencies, ensuring complete frequency isolation with zero overlap.

Step-by-Step Solution

1
Analyze the bandwidth and channel spacing in the 2.4 GHz spectrum.
The 2.4 GHz Wi-Fi spectrum uses 20 MHz wide channels spaced 5 MHz apart from center-to-center.
Understanding spacing vs bandwidth is essential to determine frequency overlap.
2
Determine the required separation between non-overlapping channels.
Channels require a spacing of at least 5 channel numbers (25 MHz) between center frequencies to prevent spectral overlap.
A 25 MHz separation ensures that signal sidebands do not interfere with neighbouring transmissions.
3
Identify the standard non-overlapping channel set.
Channels 1, 6, and 11 provide the necessary 25 MHz separation across the 11 available channels in North America.
Using channels 1, 6, and 11 allows multi-AP deployment without co-channel interference.

Key Concept

2.4 GHz Wireless Non-Overlapping Channels
Question 144Question →

A core router receives an IPv4 packet destined for a remote network host. The router's routing table contains multiple paths learned via different dynamic routing protocols, static routes, and directly connected subnets. Place the decision-making steps in the exact sequential order that the router's control plane uses to evaluate the candidate paths and forward the packet to the next hop.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence for packet forwarding decision logic is: 1) Apply Longest Prefix Match to filter candidates by subnet specificity, 2) Compare Administrative Distance to select the most reliable route source among matching prefix lengths, 3) Compare metric values among routes from the same routing source to identify the lowest-cost path, and 4) Resolve the next-hop Layer 2 destination address for frame encapsulation on the egress interface.
Route lookup logic follows a strict hierarchy. First, Longest Prefix Match (LPM) determines subnet specificity. Second, among candidate routes with equal prefix length, Administrative Distance (AD) selects the most reliable protocol source. Third, for ties within the same protocol source, metric values isolate the optimal cost path. Finally, Layer 2 ARP resolution encapsulates the packet onto the egress link.

Step-by-Step Solution

1
Evaluate subnet mask specificity (Longest Prefix Match)
The router identifies all routes matching the destination IP and isolates those with the longest prefix length (e.g., /28 over /24).
Prefix length takes ultimate precedence in IP route lookup; a more specific route is always chosen over a less specific one.
2
Compare Administrative Distance
If multiple routing sources (e.g., OSPF vs BGP) advertise the exact same prefix length, the router selects the source with the lowest AD.
Administrative Distance ranks the believability of different route sources when prefix lengths match.
3
Evaluate Protocol Metrics
If multiple paths exist from the same routing protocol with equal prefix length and AD, the router selects the path with the lowest metric (or uses ECMP).
Metrics measure path cost within a single routing protocol instance.
4
Perform Data Link Layer Encapsulation
The router looks up the next-hop IP in the ARP table to map the destination MAC address and forwards the frame out the designated physical/logical interface.
Forwarding decision is complete, allowing physical Layer 2 frame transmission.

Key Concept

Route Selection Precedence (Longest Prefix Match -> Administrative Distance -> Metric -> Egress Encapsulation)
Question 145Question →

A network engineer is troubleshooting interconnectivity between two switches, Switch-Alpha and Switch-Beta, over a dedicated GigabitEthernet link. The switchport configuration for interface GigabitEthernet1/0/24 on both switches is shown below:

Switch-Alpha:
interface GigabitEthernet1/0/24
switchport mode trunk
switchport trunk native vlan 50
switchport trunk allowed vlan 10,20,50

Switch-Beta:
interface GigabitEthernet1/0/24
switchport mode trunk
switchport trunk native vlan 1
switchport trunk allowed vlan 10,20,50

A host in VLAN 10 on Switch-Alpha sends data to a host in VLAN 10 on Switch-Beta. Simultaneously, an untagged broadcast frame is generated by a device in VLAN 50 on Switch-Alpha and transmitted across the link.

Which statement accurately describes how these frames are handled across the trunk link?

Show answer & explanation

Answer: Tagged frames belonging to VLAN 10 cross the trunk link normally, while untagged frames sent from VLAN 50 on Switch-Alpha are placed into VLAN 1 upon arrival at Switch-Beta.

Answer

Tagged frames belonging to VLAN 10 cross the trunk link normally, while untagged frames sent from VLAN 50 on Switch-Alpha are placed into VLAN 1 upon arrival at Switch-Beta.
Under 802.1Q encapsulation, frames belonging to allowed tagged VLANs (such as VLAN 10) are explicitly tagged and delivered correctly across the trunk. However, frames originating from VLAN 50 on Switch-Alpha are egressed untagged because VLAN 50 is Switch-Alpha's native VLAN. When Switch-Beta receives these untagged frames, it classifies them according to its own native VLAN setting (VLAN 1), leading to native VLAN traffic leakage between VLAN 50 and VLAN 1.

Step-by-Step Solution

1
Analyze the handling of tagged VLAN traffic across the trunk link.
VLAN 10 is configured in the allowed list on both ends of the trunk. Switch-Alpha adds an 802.1Q tag (VLAN ID 10) to the frame. Switch-Beta reads the tag, accepts the frame, and forwards it to VLAN 10.
802.1Q trunks maintain tag headers for non-native allowed VLANs regardless of native VLAN configuration.
2
Analyze the egress behavior for VLAN 50 on Switch-Alpha.
Because VLAN 50 is Switch-Alpha's configured native VLAN on GigabitEthernet1/0/24, frames originating in VLAN 50 are transmitted out the trunk port untagged (without an 802.1Q header).
By default in IEEE 802.1Q specification, traffic matching the local interface's native VLAN is sent unencapsulated.
3
Analyze the ingress behavior for untagged frames arriving at Switch-Beta.
Switch-Beta receives the untagged frame and associates it with its own configured native VLAN, which is VLAN 1.
When a trunk port receives an untagged frame, it automatically assigns that frame to whichever VLAN is defined as native on its local interface.

Key Concept

802.1Q Native VLAN Ingress/Egress Behavior during a Trunk Mismatch
Question 146Question →

A network administrator is configuring DNS infrastructure for an enterprise web application hosted behind a cloud provider's elastic load balancer. The administrator must map the organization's root domain apex (`example.com`) to the load balancer's domain name (`lb-prod-948201.cloudprovider.net`). When attempting to create a CNAME record for `example.com`, the primary DNS server fails to commit the change. The administrator executes the following diagnostic query against the local authoritative DNS server:

text
$ dig @192.168.1.10 example.com SOA +noall +answer +authority
;; ANSWER SECTION:
example.com. 3600 IN SOA ns1.example.com. hostmaster.example.com. 2026072701 7200 3600 1209600 3600

;; AUTHORITY SECTION:
example.com. 3600 IN NS ns1.example.com.

Which of the following DNS record implementations will successfully map the root domain apex to the cloud load balancer hostname without violating standard DNS specifications?

Show answer & explanation

Answer: An ALIAS (or ANAME) record at the zone apex, which dynamically resolves the target hostname to IP addresses for incoming client queries.

Answer

An ALIAS (or ANAME) record at the zone apex, which dynamically resolves the target hostname to IP addresses for incoming client queries.
The DNS specification (RFC 1034 section 3.6.2) mandates that if a CNAME record is present at a host label, no other resource records (such as MX, A, SOA, or NS) can exist for that same label. Because every DNS zone root (apex) must contain SOA and NS records, a standard CNAME record at `example.com` is invalid. Modern DNS providers offer virtual or synthetic record types—commonly called ALIAS or ANAME records—that simulate CNAME functionality at the zone apex by dynamically resolving the target domain's IP addresses and returning A/AAAA records to querying clients.

Step-by-Step Solution

1
Analyze the DNS RFC requirement for CNAME coexistence at the zone apex.
According to RFC 1034, a CNAME record cannot coexist with any other record types for the same label. Because a zone apex MUST contain SOA and NS records, a standard CNAME record cannot be placed at `example.com`.
Placing a standard CNAME at the apex creates structural conflicts with the SOA and NS records essential for domain operation.
2
Evaluate technical mechanisms designed to overcome zone apex CNAME limitations.
DNS providers implement synthetic or virtual record types (such as ALIAS, ANAME, or CNAME flattening) that allow a hostname target at the root apex.
The authoritative DNS server resolves the target hostname (`lb-prod-948201.cloudprovider.net`) to its underlying IP addresses on the backend and responds to clients with standard A or AAAA records.
3
Select the compliant record type that satisfies the application requirements.
Deploying an ALIAS (or ANAME) record maps `example.com` to the load balancer hostname without violating DNS specifications.
This maintains RFC compliance for SOA/NS records while fulfilling the dynamic host mapping requirement.

Key Concept

DNS Zone Apex Restrictions and ALIAS/ANAME Records
Estimated Time:2m 0s
Question 147Question →

A network administrator is designing an IPv4 addressing scheme for a enterprise WLAN segment that must support at least 500500 usable host devices per subnet. What is the maximum CIDR prefix length (integer value of the mask bits) that can be assigned to meet this requirement without wasting IP address space?

Show answer & explanation

Answer: 23

Answer

The maximum CIDR prefix length that supports at least 500 usable host addresses is 23.
To host at least 500500 usable devices, a subnet requires 9 host bits because 29−2=5102^9 - 2 = 510 usable IP addresses. Subtracting 9 host bits from the 32 total bits in an IPv4 address results in a prefix length of 23.

Step-by-Step Solution

1
Calculate the required number of host bits
9 host bits are required.
Using the formula 2h−22^h - 2, 8 host bits yield only 254 usable addresses, whereas 9 host bits yield 510510 usable addresses, which satisfies the 500500 host minimum requirement.
2
Calculate the network prefix length in CIDR notation
The prefix length is 23.
An IPv4 address consists of 32 bits in total. Subtracting the 9 host bits (32−932 - 9) leaves 23 bits for the network and subnet identifier.

Key Concept

IPv4 Host Capacity and CIDR Prefix Calculation
Question 148Question →

A network operations team is subnetting the enterprise block 10.45.128.0/1810.45.128.0/18 to create equal-sized subnets capable of supporting at least 500500 usable host IP addresses per subnet. If subnets are allocated sequentially starting from the lowest address in the block, what is the network address (Network ID) of the fifth allocated subnet?

Show answer & explanation

Answer: 10.45.136.0; 10.45.136.0/23

Answer

10.45.136.0
To accommodate at least 500500 usable hosts per subnet, 99 host bits are required because 29−2=5102^9 - 2 = 510 usable addresses. This results in a prefix length of /23/23 (32−9=2332 - 9 = 23), which increments the third octet by 22 per subnet (256−254=2256 - 254 = 2). Starting sequentially from 10.45.128.010.45.128.0, the subnets are: 1st: 10.45.128.010.45.128.0, 2nd: 10.45.130.010.45.130.0, 3rd: 10.45.132.010.45.132.0, 4th: 10.45.134.010.45.134.0, and 5th: 10.45.136.010.45.136.0.

Step-by-Step Solution

1
Determine the number of host bits required for 500 usable host IP addresses.
9 host bits are required.
The formula for usable hosts is 2h−2≥5002^h - 2 \ge 500. For h=8h=8, 28−2=2542^8 - 2 = 254 hosts (insufficient). For h=9h=9, 29−2=5102^9 - 2 = 510 hosts (sufficient).
2
Calculate the subnet prefix length and block size.
Prefix length is /23/23 and the block size in the third octet is 2.
Subtracting 99 host bits from 3232 total bits gives 32−9=2332 - 9 = 23 (/23/23). The block size per subnet is 224−23=22^{24-23} = 2 in the third octet.
3
Calculate the network ID of the fifth subnet allocated sequentially from 10.45.128.010.45.128.0.
The network address is 10.45.136.0.
The first subnet is 10.45.128.010.45.128.0. Each subsequent subnet increments the third octet by 2. The 5th subnet corresponds to 4 increments (4×2=84 \times 2 = 8): 128+8=136128 + 8 = 136, yielding 10.45.136.0/2310.45.136.0/23.

Key Concept

Host bit calculation and sequential IPv4 subnet allocation
Estimated Time:2m 0s
Question 149Question →

A network architect is evaluating dynamic routing protocols for an enterprise environment with redundant paths. The architect needs to ensure optimal path selection and fast convergence using Interior Gateway Protocols (IGPs). Which of the following statements accurately describe the operational characteristics of OSPF and EIGRP? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: EIGRP calculates backup routes known as feasible successors and maintains them in its topology table to enable rapid failover without recalculating paths.; OSPF routers construct a synchronized Link-State Database (LSDB) representing the network map and run the Shortest Path First algorithm to determine the shortest tree to every destination.

Answer

The statements correctly describing OSPF and EIGRP are: EIGRP maintains precalculated backup routes (feasible successors) in its topology table for fast convergence, and OSPF maintains a synchronized Link-State Database (LSDB) to compute shortest path trees using Dijkstra's algorithm.
EIGRP uses the DUAL algorithm to identify feasible successors in the topology table, guaranteeing loop-free backup routes for instantaneous failover. Additionally, OSPF routers build a synchronized Link-State Database (LSDB) within an area and run Dijkstra's SPF algorithm to calculate the shortest path to every destination subnetwork.

Step-by-Step Solution

1
Analyze EIGRP convergence mechanisms.
EIGRP calculates primary routes (successors) and backup routes (feasible successors) stored in the topology table, enabling fast convergence via DUAL.
If the primary route fails, EIGRP immediately promotes the feasible successor without needing a full recalculation.
2
Analyze OSPF operation and OSI layer encapsulation.
OSPF is a link-state protocol using Dijkstra's algorithm over an identical LSDB, and its packets are encapsulated in IP (Layer 3) headers using protocol number 89.
OSPF is a Layer 3 protocol, not a Layer 2 frame-level protocol, and builds a complete map of the network within an area.
3
Evaluate NAT/PAT interaction with dynamic routing protocols.
PAT translates private sockets to public IP/port combinations at the edge device and does not inject static host routes into RIB tables via dynamic routing protocols.
Routing protocols manage network prefix reachability, whereas PAT manages address translation tables.

Key Concept

Link-State vs Distance-Vector / Advanced Distance-Vector Routing Protocol Operations
Question 150Question →

A network administrator is designing a Variable Length Subnet Mask (VLSM) scheme starting from the base address block 10.100.128.0/2110.100.128.0/21. Subnets must be allocated contiguously without unallocated gaps, prioritized from largest host requirement to smallest as follows:

- Segment 1: Requires at least 500500 usable host IP addresses
- Segment 2: Requires at least 250250 usable host IP addresses
- Segment 3: Requires at least 120120 usable host IP addresses
- Segment 4: Requires a point-to-point link accommodating exactly 22 usable host IP addresses

Which of the following statements regarding this VLSM allocation scheme are correct? (Select THREE.)

Select all that apply

Show answer & explanation

Answer: The broadcast address for Segment 1 is 10.100.129.25510.100.129.255.; The valid usable host IP address range for Segment 3 is 10.100.131.110.100.131.1 through 10.100.131.12610.100.131.126.; The first usable IP address assigned to Segment 4 is 10.100.131.12910.100.131.129.

Answer

The correct statements are: the broadcast address for Segment 1 is 10.100.129.25510.100.129.255, the valid usable host IP address range for Segment 3 is 10.100.131.110.100.131.1 through 10.100.131.12610.100.131.126, and the first usable IP address assigned to Segment 4 is 10.100.131.12910.100.131.129.
The statement regarding Segment 1's broadcast address being 10.100.129.25510.100.129.255 is correct because a /23/23 subnet spanning 10.100.128.010.100.128.0 ends at 10.100.129.25510.100.129.255. The statement regarding Segment 3's usable range (10.100.131.110.100.131.1–10.100.131.12610.100.131.126) is correct because its /25/25 allocation spans 10.100.131.010.100.131.0 to 10.100.131.12710.100.131.127. The statement specifying Segment 4's first usable IP as 10.100.131.12910.100.131.129 is correct because Segment 4's /30/30 block begins at Network ID 10.100.131.12810.100.131.128.

Step-by-Step Solution

1
Calculate subnet requirement and bounds for Segment 1
Segment 1 needs 500 usable hosts. 29−2=510≥5002^9 - 2 = 510 \ge 500, requiring a /23/23 block (255.255.254.0255.255.254.0). Starting at 10.100.128.0/2310.100.128.0/23, the network ID is 10.100.128.010.100.128.0 and the broadcast address is 10.100.129.25510.100.129.255. Total usable hosts is 510 (512 total IPs minus 2).
VLSM allocations must satisfy host requirements using the smallest matching block (2n−22^n - 2).
2
Calculate subnet requirement and bounds for Segment 2
Segment 2 needs 250 usable hosts. 28−2=254≥2502^8 - 2 = 254 \ge 250, requiring a /24/24 block (255.255.255.0255.255.255.0). Contiguously following Segment 1, Segment 2 starts at 10.100.130.0/2410.100.130.0/24. Network ID is 10.100.130.010.100.130.0, broadcast address is 10.100.130.25510.100.130.255.
The next available IP after 10.100.129.25510.100.129.255 is 10.100.130.010.100.130.0.
3
Calculate subnet requirement and bounds for Segment 3
Segment 3 needs 120 usable hosts. 27−2=126≥1202^7 - 2 = 126 \ge 120, requiring a /25/25 block (255.255.255.128255.255.255.128). Contiguously following Segment 2, Segment 3 starts at 10.100.131.0/2510.100.131.0/25. Network ID is 10.100.131.010.100.131.0, broadcast address is 10.100.131.12710.100.131.127, usable range is 10.100.131.110.100.131.1 to 10.100.131.12610.100.131.126.
The next available IP after 10.100.130.25510.100.130.255 is 10.100.131.010.100.131.0.
4
Calculate subnet requirement and bounds for Segment 4
Segment 4 needs 2 usable hosts. 22−2=22^2 - 2 = 2, requiring a /30/30 block (255.255.255.252255.255.255.252). Contiguously following Segment 3, Segment 4 starts at 10.100.131.128/3010.100.131.128/30. Network ID is 10.100.131.12810.100.131.128, first usable IP is 10.100.131.12910.100.131.129, second usable IP is 10.100.131.13010.100.131.130, broadcast address is 10.100.131.13110.100.131.131.
The next available IP after 10.100.131.12710.100.131.127 is 10.100.131.12810.100.131.128.

Key Concept

Variable Length Subnet Masking (VLSM) and IPv4 Usable Host Calculations
Question 151Question →

A network administrator configures a DHCP relay agent on a Layer 3 switch interface to support host workstations on VLAN 25 (10.25.0.0/2410.25.0.0/24, default gateway 10.25.0.110.25.0.1). The centralized DHCP server resides on VLAN 100 (10.100.0.510.100.0.5). When an unconfigured client on VLAN 25 broadcasts a DHCPDISCOVER packet, which TWO of the following actions does the DHCP relay agent execute to ensure the DHCP server correctly processes the request?

Select all that apply

Show answer & explanation

Answer: It converts the incoming client broadcast packet into a Layer 3 unicast packet destined for 10.100.0.510.100.0.5.; It populates the Gateway IP Address (`giaddr`) field in the DHCP header with 10.25.0.110.25.0.1.

Answer

The DHCP relay agent converts the host broadcast request into a unicast packet directed to the DHCP server (10.100.0.510.100.0.5) and populates the `giaddr` field with the receiving gateway interface IP address (10.25.0.110.25.0.1).
When a DHCP relay agent processes a client's broadcast request, it encapsulates the message into a Layer 3 unicast packet destined for the centralized DHCP server's IP address (10.100.0.510.100.0.5). To inform the DHCP server which subnet scope must be used for address assignment, the relay agent places the IP address of its receiving interface (10.25.0.110.25.0.1) into the Gateway IP Address (`giaddr`) field of the DHCP header.

Step-by-Step Solution

1
Analyze client broadcast message reception
The host on VLAN 25 broadcasts a DHCPDISCOVER message using source IP 0.0.0.00.0.0.0, destination IP 255.255.255.255255.255.255.255, UDP source port 68, and UDP destination port 67.
Unconfigured DHCP clients cannot route traffic outside their local broadcast domain.
2
Examine DHCP relay agent packet processing
The Layer 3 switch interface intercepts the broadcast, writes its own interface IP (10.25.0.110.25.0.1) into the `giaddr` header field, and converts the destination IP to the unicast address of the remote DHCP server (10.100.0.510.100.0.5).
The server requires the `giaddr` value to select the correct address pool (10.25.0.0/2410.25.0.0/24) and unicast addressing to cross routed boundaries.

Key Concept

DHCP Relay Agent and GIADDR Operation
Estimated Time:1m 30s
Question 152Question →

A network technician is configuring an isolated IPv4 subnet for a small cluster of servers using the subnet mask 255.255.255.248255.255.255.248 (/29/29). What is the maximum number of usable host IP addresses available on this subnet?

Show answer & explanation

Answer: 66

Answer

The maximum number of usable host IP addresses on a 255.255.255.248255.255.255.248 (/29/29) subnet is 66.
A subnet mask of 255.255.255.248255.255.255.248 corresponds to a /29/29 CIDR prefix. This leaves 3 host bits (32−29=332 - 29 = 3). The total number of IP addresses is 23=82^3 = 8. Subtracting 2 (one for the network address and one for the broadcast address) leaves 66 usable host IP addresses.

Step-by-Step Solution

1
Determine the number of host bits in the /29/29 subnet mask.
There are 32−29=332 - 29 = 3 host bits.
An IPv4 address has 32 total bits. Subtracting the network prefix bits leaves the bits reserved for hosts.
2
Calculate the total number of IP addresses in the subnet block.
23=82^3 = 8 total IP addresses.
The total number of IP addresses in a subnet is calculated using the formula 2host bits2^{\text{host bits}}.
3
Subtract the reserved network and broadcast addresses to find usable host IP addresses.
8−2=68 - 2 = 6 usable host IP addresses.
The first address is reserved as the network address and the last address is reserved as the broadcast address, neither of which can be assigned to host devices.

Key Concept

IPv4 Subnet Usable Host Calculation
Question 153Question →

An enterprise enterprise router has learned routes to the destination network 172.16.10.0/24 from multiple dynamic routing sources and static configurations. The routing table currently evaluates path determination during active traffic forwarding. Which of the following operational behaviors correctly describe how the router selects and processes these routes? (Select TWO).

Select all that apply

Show answer & explanation

Answer: The router compares Administrative Distance (AD) first when multiple routing protocols advertise the exact same destination prefix, selecting the route with the lowest AD to populate the routing table.; When forwarding a packet, the router evaluates Longest Prefix Match (LPM) prior to Administrative Distance, preferring a /26 route over a /24 route regardless of protocol trust rating.

Answer

The correct selections are the statements regarding Administrative Distance selection for identical prefixes and Longest Prefix Match precedence during packet forwarding.
Administrative Distance is used to choose the most reliable protocol path when identical prefixes are advertised by different protocols. However, during packet forwarding, Longest Prefix Match takes absolute precedence, meaning a more specific prefix (/26) will always be chosen over a less specific one (/24) regardless of how the route was learned.

Step-by-Step Solution

1
Analyze route selection hierarchy for identical prefixes learned via different sources.
Administrative Distance (AD) determines which protocol's route is populated into the routing table when identical destination subnets exist.
Lower AD values indicate higher protocol trustworthiness.
2
Analyze packet forwarding decision logic for populated routes of varying prefix lengths.
The router applies Longest Prefix Match (LPM) to select the most specific subnet mask.
LPM supersedes AD because AD only determines entry into the routing table, whereas LPM rules active packet forwarding across all table entries.

Key Concept

Route Selection Precedence (Longest Prefix Match vs Administrative Distance vs Metric)
Question 154Question →

A network engineer is analyzing an 802.1Q trunk connection established between `SW-Access-01` and `SW-Core-01` on interface `GigabitEthernet0/1`. Interface configurations are as follows:

SW-Access-01:
text
interface GigabitEthernet0/1
switchport mode trunk
switchport trunk native vlan 1
switchport trunk allowed vlan 10,20,30

SW-Core-01:
text
interface GigabitEthernet0/1
switchport mode trunk
switchport trunk native vlan 10
switchport trunk allowed vlan 10,20,30

Assuming both switches are active and exchanging data, which TWO of the following statements accurately describe how untagged and tagged frames are processed across this link? (Select TWO)

Select all that apply

Show answer & explanation

Answer: Untagged frames transmitted from VLAN 1 on SW-Access-01 will be forwarded into VLAN 10 upon arriving at SW-Core-01.; VLAN 10 traffic sent from SW-Core-01 will egress untagged and be assigned to VLAN 1 when received by SW-Access-01.

Answer

Untagged frames transmitted from VLAN 1 on SW-Access-01 will be forwarded into VLAN 10 upon arriving at SW-Core-01, and VLAN 10 traffic sent from SW-Core-01 will egress untagged and be assigned to VLAN 1 when received by SW-Access-01.
On an 802.1Q trunk link, frames belonging to the configured native VLAN are transmitted without an 802.1Q tag, while frames from all other allowed VLANs are tagged. When a native VLAN mismatch exists between two switches, untagged frames sent by one switch are received and placed into the receiving switch's configured native VLAN. Specifically, untagged frames from VLAN 1 on the access switch arrive at the core switch and are assigned to VLAN 10. Conversely, VLAN 10 frames originating on the core switch are sent untagged because VLAN 10 is native on that switch; upon arrival at the access switch, these untagged frames are placed into VLAN 1.

Step-by-Step Solution

1
Analyze frame encapsulation on SW-Access-01 for VLAN 1 and VLAN 10
Native VLAN on SW-Access-01 is VLAN 1. Therefore, VLAN 1 traffic is sent untagged, while VLAN 10 traffic is encapsulated with an 802.1Q tag bearing VLAN ID 10.
802.1Q trunks strip tags for traffic belonging to the local native VLAN and add tags for all other allowed VLANs.
2
Analyze ingress frame processing on SW-Core-01
Native VLAN on SW-Core-01 is VLAN 10. Any untagged frame arriving at SW-Core-01 is implicitly assigned to VLAN 10. Tagged frames with VLAN ID 10 are placed directly into VLAN 10.
Switches assign incoming untagged frames to whichever native VLAN is configured on the receiving trunk port.
3
Analyze egress frame processing from SW-Core-01 towards SW-Access-01
VLAN 10 frames leaving SW-Core-01 are stripped of tags because VLAN 10 is native on SW-Core-01. SW-Access-01 receives untagged frames and places them into its native VLAN 1.
Native VLAN mismatch causes asymmetrical traffic leakage between the two configured native VLANs across opposing switch ports.

Key Concept

802.1Q Native VLAN Processing & Tagging Behavior
Question 155Question →

A network administrator is auditing 802.1Q trunking and access port behavior on enterprise switches. Match each switchport operational scenario on the left with its corresponding 802.1Q frame processing or tagging outcome on the right.

Click a left item, then click its matching right item

Items

An access port configured with 'switchport access vlan 20' receives an untagged Ethernet frame.
A trunk port configured with 'switchport trunk native vlan 50' receives an untagged ingress frame.
A trunk port configured with 'switchport trunk native vlan 50' transmits an egress frame belonging to VLAN 50.
A trunk port configured with 'switchport trunk allowed vlan 10,20' receives an ingress frame tagged with VLAN 30.

Matches

Show answer & explanation

Answer

Each scenario matches as follows: Access port receiving untagged traffic maps to internal VLAN 20 processing; Trunk port receiving untagged traffic implicitly maps to native VLAN 50; Trunk port transmitting native VLAN 50 egresses untagged without an 802.1Q header; Trunk port receiving VLAN 30 on an allowed list of 10,20 drops the frame due to VLAN filtering.
The pairings correctly demonstrate fundamental 802.1Q trunking and access port behavior: access ports map untagged ingress frames to their access VLAN, trunk ports map untagged ingress frames to their native VLAN, native VLAN egress traffic is transmitted without an 802.1Q tag header, and frames tagged with VLANs outside the trunk allowed list are dropped at ingress.

Step-by-Step Solution

1
Analyze access port frame ingestion behavior.
Untagged ingress frames on an access port configured for VLAN 20 are logically mapped to VLAN 20.
Access ports only accept untagged frames (or frames with the matching access VLAN if voice VLAN is configured) and place them into the assigned access VLAN.
2
Analyze trunk port native VLAN ingress processing.
Untagged frames arriving on an 802.1Q trunk link are implicitly assigned to the native VLAN (VLAN 50).
IEEE 802.1Q design mandates that any untagged ingress frame arriving on a trunk interface belongs to the native VLAN configured on that interface.
3
Analyze trunk port native VLAN egress header tagging.
Frames assigned to VLAN 50 exit the trunk port untagged.
To maintain backward compatibility with non-802.1Q aware devices, 802.1Q trunks strip the 4-byte VLAN tag field for any frames belonging to the native VLAN prior to transmission.
4
Analyze trunk allowed VLAN list filtering rules.
A frame tagged with VLAN 30 arriving on a trunk restricted to VLANs 10 and 20 is dropped at ingress.
The 'switchport trunk allowed vlan' command defines the allowed list; any tagged frame carrying a VLAN ID not present in the allowed list is pruned/dropped by switch hardware.

Key Concept

802.1Q Frame Tagging, Native VLAN Ingress/Egress Rules, and Trunk Allowed List Pruning
Question 156Question →

A network administrator needs to connect an enterprise network to two different Internet Service Providers (ISPs) located in separate Autonomous Systems (AS). Which dynamic routing protocol should be implemented to exchange routing information between these distinct Autonomous Systems?

Show answer & explanation

Answer: Border Gateway Protocol (BGP)

Answer

Border Gateway Protocol (BGP) is an Exterior Gateway Protocol (EGP) designed specifically to exchange prefix routing information between different Autonomous Systems across the internet.
Border Gateway Protocol (BGP) is the standardized Exterior Gateway Protocol (EGP) created to manage routing decisions and exchange reachability information between distinct Autonomous Systems across the global internet.

Step-by-Step Solution

1
Identify the scenario requirement
Routing must occur between two different organizations/ISPs operating in distinct Autonomous Systems (AS).
Routing protocols are classified as either Interior Gateway Protocols (IGP) for intra-domain routing or Exterior Gateway Protocols (EGP) for inter-domain routing.
2
Evaluate protocol categories
Border Gateway Protocol (BGP) is the primary Exterior Gateway Protocol used for routing between separate Autonomous Systems.
Protocols like OSPF and EIGRP operate inside a single AS, whereas BGP manages policy-based routing between different Autonomous Systems across the internet core.

Key Concept

Exterior Gateway Protocol (EGP) vs. Interior Gateway Protocol (IGP)
Question 157Question →

A network administrator allocates the IPv4 network block 192.168.10.0/27192.168.10.0/27 to a remote branch office subnet. What is the maximum number of usable host IP addresses that can be assigned to host devices on this subnet?

Show answer & explanation

Answer: 30 usable host addresses

Answer

The maximum number of usable host IP addresses on a /27 subnet is 30.
A CIDR prefix of /27 leaves 5 bits for host addresses (32−27=532 - 27 = 5). Calculating 252^5 gives 32 total IP addresses in the subnet block. Subtracting 2 for the Network ID and Broadcast address (32−232 - 2) gives 30 usable host IP addresses.

Step-by-Step Solution

1
Calculate the number of available host bits from the CIDR prefix.
Host bits = 32−27=532 - 27 = 5 bits.
An IPv4 address consists of 32 total bits. Subtracting the 27 network bits yields the remaining host bits.
2
Calculate the total number of IP addresses in the block.
Total addresses = 25=322^5 = 32 addresses.
The total capacity of a subnet is equal to 2 raised to the power of the number of host bits.
3
Subtract the reserved network and broadcast addresses to determine assignable host IPs.
Usable addresses = 32−2=3032 - 2 = 30 host addresses.
The first address serves as the Network ID and the final address serves as the Subnet Broadcast Address; neither can be assigned to individual host network interfaces.

Key Concept

Calculating usable host IP addresses using CIDR subnet notation
Estimated Time:45s
Question 158Question →

A network technician provisions an IEEE 802.1Q trunk link on switch `SW-Floor2` interface `GigabitEthernet0/48` to connect to `SW-Core1`. Workstations on VLAN 10 (Finance) immediately lose connectivity to remote resources on the same VLAN, while devices on VLAN 20 (IT) continue communicating normally. The running configuration for interface `GigabitEthernet0/48` displays the following output:

text
interface GigabitEthernet0/48
switchport mode trunk
switchport trunk native vlan 1
switchport trunk allowed vlan 20,30

Which of the following statements accurately identify the cause of the connectivity loss and specify the correct corrective action? (Select TWO.)

Select all that apply

Show answer & explanation

Answer: VLAN 10 frames are dropped at the trunk port because VLAN 10 is omitted from the allowed VLAN pruning list on interface GigabitEthernet0/48.; The administrator must execute `switchport trunk allowed vlan add 10` on interface GigabitEthernet0/48 to permit VLAN 10 traffic across the link.

Answer

The outage is caused by VLAN 10 being excluded from the trunk port's explicit allowed list. The issue is resolved by running `switchport trunk allowed vlan add 10` on the trunk interface.
On an 802.1Q trunk, configuring `switchport trunk allowed vlan` creates an explicit permit list. Because VLAN 10 was not listed, its frames were filtered at the trunk boundary. Executing `switchport trunk allowed vlan add 10` appends VLAN 10 to the allowed list, enabling frame transmission across switches for that broadcast domain.

Step-by-Step Solution

1
Analyze the trunk configuration snippet on interface GigabitEthernet0/48.
The line `switchport trunk allowed vlan 20,30` limits trunking strictly to VLANs 20 and 30.
VLAN 10 is omitted from the allowed list, which causes the switch port to prune and drop all ingress and egress 802.1Q frames belonging to VLAN 10.
2
Determine the CLI command required to modify the allowed list safely.
Applying `switchport trunk allowed vlan add 10` updates the allowed list to 10,20,30.
Using the `add` parameter prevents overwriting existing allowed VLANs, ensuring VLAN 20 and VLAN 30 remain operational while restoring VLAN 10.

Key Concept

802.1Q Trunk Allowed VLAN Filtering
Question 159Question →

Match each IEEE 802.1Q trunking component or configuration parameter on the left with its correct operational behavior or definition on the right.

Click a left item, then click its matching right item

Items

Native VLAN
802.1Q Header
Allowed VLAN List
Access Port

Matches

Show answer & explanation

Answer

Native VLAN matches with carrying untagged frame traffic across an 802.1Q trunk link. 802.1Q Header matches with inserting a 4-byte field containing a VLAN ID into the Ethernet frame. Allowed VLAN List matches with specifying which VLAN traffic is permitted to traverse a trunk link. Access Port matches with belonging to a single VLAN and transmitting untagged frames to end-user devices.
Each key trunking component maps directly to its operational definition: the Native VLAN handles untagged trunk frames, the 802.1Q header inserts a 4-byte tagging field, the allowed VLAN list restricts allowed VLAN IDs on a trunk, and an access port connects end devices to a single untagged VLAN domain.

Step-by-Step Solution

1
Identify the function of Native VLAN on 802.1Q trunks.
Native VLAN carries untagged traffic over trunk links without modification.
By definition in 802.1Q, frames originating on the native VLAN do not receive a 4-byte VLAN tag when crossing a trunk link.
2
Identify the structure and purpose of the 802.1Q header.
802.1Q inserts a 4-byte tag into the frame header to identify the destination VLAN ID.
Trunk interfaces use tagging to multiplex frames from multiple VLANs over a single physical link.
3
Identify the purpose of the allowed VLAN list configuration.
Allowed VLAN list restricts which VLANs can pass traffic across the trunk link.
Pruning or specifying allowed VLANs prevents unnecessary broadcast and unicast traffic for unneeded VLANs from traversing the trunk.
4
Identify the characteristics of an access port.
Access ports belong to a single VLAN and carry untagged frames for endpoint connectivity.
End-user devices such as PCs or printers do not understand 802.1Q VLAN tags, so access ports strip tags prior to frame egress.

Key Concept

VLAN Configuration and 802.1Q Trunking Fundamentals
Estimated Time:1m 0s
Question 160Question →

An enterprise network topology consists of four switches operating standard IEEE 802.1w Rapid Spanning Tree Protocol (RSTP) using default path costs (1 Gbps=41\text{ Gbps} = 4, 100 Mbps=19100\text{ Mbps} = 19).

- Switch-R is configured with a bridge priority of 4096 (MAC address `00:11:22:33:44:55`).
- Switch-X, Switch-Y, and Switch-Z all maintain the default bridge priority of 32768.
- Switch-X connects to Switch-R via a 1 Gbps1\text{ Gbps} trunk link.
- Switch-Y connects to Switch-R via a 100 Mbps100\text{ Mbps} trunk link.
- Switch-Y also connects to Switch-X via a 1 Gbps1\text{ Gbps} trunk link.

Assuming all links are operational and STP convergence is complete, which statement accurately describes the Root Port selection and port status on Switch-Y?

Show answer & explanation

Answer: Switch-Y selects its interface connected to Switch-X as its Root Port, while its direct interface connected to Switch-R enters the Alternate (Discarding) state.

Answer

Switch-Y selects its interface connected to Switch-X as its Root Port, while its direct interface connected to Switch-R enters the Alternate (Discarding) state.
Root Port election on a non-root switch prioritizes the path with the lowest cumulative Root Path Cost to the Root Bridge. Switch-Y has two paths to Switch-R: a direct 100 Mbps100\text{ Mbps} link (cost 1919) and an indirect path via Switch-X (1 Gbps+1 Gbps1\text{ Gbps} + 1\text{ Gbps}, cost 4+4=84 + 4 = 8). Since 8<198 < 19, Switch-Y selects the link toward Switch-X as its Root Port. On the direct link between Switch-R and Switch-Y, Switch-R is the Designated Switch because it advertises a Root Path Cost of 00. Consequently, Switch-Y's direct port to Switch-R assumes the Alternate role and enters the Discarding state.

Step-by-Step Solution

1
Determine the Root Bridge of the topology.
Switch-R is elected Root Bridge because it has the lowest Bridge Priority (4096 vs 32768).
The switch with the lowest Bridge ID (Priority + MAC address) becomes the Root Bridge.
2
Calculate the cumulative Root Path Cost (RPC) for all candidate paths from Switch-Y to Switch-R.
Direct path via 100 Mbps100\text{ Mbps} link: RPC=19\text{RPC} = 19.
Indirect path via Switch-X (1 Gbps+1 Gbps1\text{ Gbps} + 1\text{ Gbps}): RPC=4+4=8\text{RPC} = 4 + 4 = 8.
Root Path Cost is the sum of path costs along the path to the Root Bridge.
3
Select the Root Port for Switch-Y based on the lowest Root Path Cost.
The interface connecting Switch-Y to Switch-X is elected as Switch-Y's Root Port because its path cost (88) is strictly less than the direct path cost (1919).
Each non-root switch selects a single Root Port corresponding to the path with the lowest cumulative cost to the Root Bridge.
4
Determine the role and state of Switch-Y's direct interface connected to Switch-R.
On the Switch-R to Switch-Y link, Switch-R is the Designated Bridge (advertising cost 00). Switch-Y's port on this segment receives superior BPDUs from Switch-R, so Switch-Y's local port becomes an Alternate Port in the Discarding state.
An interface that is neither a Root Port nor a Designated Port transitions to the Alternate role (Discarding/Blocking state) to prevent Layer 2 loops.

Key Concept

Spanning Tree Protocol Root Port Election and Root Path Cost Evaluation
Estimated Time:2m 0s
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