Tüm alıştırma soruları

1987 soru

Soru 361Soru

A network engineer is configuring the default gateway interface for a enterprise VLAN assigned the IPv4 network block 172.24.128.0/21172.24.128.0/21. According to network policy, the default gateway must be assigned the highest usable IPv4 address within the subnet. Which IPv4 address must be assigned to the gateway interface?

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Cevap: 172.24.135.254172.24.135.254

Cevap

The IPv4 address 172.24.135.254172.24.135.254 must be assigned to the gateway interface.
For a /21/21 network (255.255.248.0255.255.248.0), the third octet increments by 8. Given the network ID 172.24.128.0172.24.128.0, the subnet range encompasses 172.24.128.0172.24.128.0 through 172.24.135.255172.24.135.255. The broadcast address is 172.24.135.255172.24.135.255, making 172.24.135.254172.24.135.254 the highest usable host IPv4 address.

Adım Adım Çözüm

1
Determine the subnet mask and block size for a /21/21 prefix
A /21/21 prefix corresponds to a subnet mask of 255.255.248.0255.255.248.0. The block size in the third octet is 256248=8256 - 248 = 8.
Calculating the block size identifies the boundary for the network increment.
2
Calculate the network ID and broadcast address
Starting at network ID 172.24.128.0172.24.128.0, adding the block size of 88 gives the next subnet starting at 172.24.136.0172.24.136.0. Therefore, the broadcast address for this subnet is 172.24.135.255172.24.135.255.
The broadcast address is the last address in the subnet before the start of the next subnet.
3
Identify the highest usable host IP address
Subtracting 1 from the broadcast address 172.24.135.255172.24.135.255 yields 172.24.135.254172.24.135.254 as the highest usable host address.
The network ID (172.24.128.0172.24.128.0) and broadcast address (172.24.135.255172.24.135.255) are reserved and cannot be assigned to network interfaces.

Anahtar Kavram

IPv4 Subnet Boundaries and Usable Host Range Calculation
Soru 362Soru

Match each Virtual Private Network (VPN) technology or protocol component on the left with its corresponding operational characteristic or security capability on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Encapsulating Security Payload (ESP)
Authentication Header (AH)
Internet Key Exchange Version 2 (IKEv2)
Clientless SSL/TLS VPN

Eşleşmeler

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Cevap

Encapsulating Security Payload (ESP) matches payload encryption and integrity; Authentication Header (AH) matches unencrypted integrity for the entire IP packet; Internet Key Exchange Version 2 (IKEv2) matches control-plane SA negotiation with EAP/NAT-T support; Clientless SSL/TLS VPN matches web browser portal access to corporate applications without client software.
Each technology is paired with its precise functional domain: ESP delivers encrypted payload security; AH provides unencrypted authentication across the header; IKEv2 negotiates control-plane SAs with EAP and NAT-T capabilities; and Clientless SSL VPN provides browser-restricted access without endpoint software requirements.

Adım Adım Çözüm

1
Analyze IPsec Data Plane Protocols
Identify that ESP provides payload encryption and authentication (protocol 50), whereas AH provides packet-wide authentication without encryption (protocol 51).
Differentiating between ESP and AH is fundamental to IPsec security design.
2
Evaluate Key Management Control Plane Protocols
Recognize that IKEv2 negotiates SAs dynamically and includes native improvements over IKEv1, such as EAP integration and reliable NAT-T execution.
IKEv2 manages tunnel setup and identity authentication in modern VPN implementations.
3
Distinguish Remote Access Deployment Models
Match Clientless SSL/TLS VPN to browser-based application access that requires no endpoint client software deployment.
Clientless SSL VPNs operate at the application layer to provide constrained portal access.

Anahtar Kavram

IPsec and SSL VPN Protocol Frameworks
Soru 363Soru

A network administrator is modifying syslog settings on a Cisco IOS router. The router currently has `logging trap errors` configured. The administrator needs to change the configuration so that all events up to severity level 4 (warnings) are forwarded to the remote syslog host, and log messages up to severity level 4 are displayed live during the active SSH session. Which set of commands accomplishes these requirements?

Cevabı ve açıklamayı göster

Cevap: Configure `logging trap warnings` in global configuration mode and execute `terminal monitor` in privileged EXEC mode.

Cevap

Configure `logging trap warnings` in global configuration mode and execute `terminal monitor` in privileged EXEC mode.
The option specifying `logging trap warnings` in global configuration mode and `terminal monitor` in privileged EXEC mode correctly sets the syslog trap threshold to severity level 4 (which includes levels 0 through 4) and enables real-time logging output on the user's active SSH terminal session.

Adım Adım Çözüm

1
Identify the numeric severity level and keyword for syslog warnings.
Syslog severity level 4 corresponds to the keyword 'warnings'. The levels range from 0 (Emergency) to 7 (Debugging).
The 'logging trap' command defines which messages are sent to remote syslog servers based on severity.
2
Determine the global configuration command to send messages up to severity level 4 to the syslog server.
Executing `logging trap warnings` (or `logging trap 4`) configures the router to forward severity levels 0 (Emergency), 1 (Alert), 2 (Critical), 3 (Error), and 4 (Warning) to the remote syslog host.
Syslog severity thresholds include all levels numerically equal to or lower than the specified severity.
3
Identify the operational command required to enable log message output on a remote SSH line.
Executing `terminal monitor` in privileged EXEC mode enables logging output for the current VTY line.
By default, console logging is enabled, but VTY (SSH/Telnet) sessions require `terminal monitor` to receive logging output in real time.

Anahtar Kavram

Syslog Operations, Severity Levels, and Log Destination Configuration
Tahmini Süre:1m 30s
Soru 364Soru

A network administrator is troubleshooting an OSPFv2 neighbor relationship between two directly connected Cisco routers. Command-line interface output indicates that the neighbor relationship is persistently stuck in the ExStart state. Which configuration mismatch is the primary cause of this specific neighbor state condition?

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Cevap: Interface Maximum Transmission Unit (MTU) size

Cevap

Interface Maximum Transmission Unit (MTU) size
During OSPF adjacency negotiation, routers pass through Init and 2-Way before entering ExStart. In ExStart, master/slave roles are determined using Database Description (DBD) packets, which contain the interface MTU size. If the interface MTUs between neighbor interfaces do not match, the router receiving DBDs with a higher MTU drops them, causing the neighbor state to remain stuck in ExStart.

Adım Adım Çözüm

1
Analyze OSPF Hello packet verification parameters.
Parameters verified in Hello packets include Area ID, Hello/Dead timers, Subnet Mask, and Authentication. A mismatch in any of these parameters causes Hello packets to be ignored, leaving neighbors in the Down state.
Interface MTU is not checked during the initial Hello packet exchange.
2
Trace adjacency progression into the ExStart state.
Because initial Hello parameters match, the routers establish two-way communication (2-Way) and progress to ExStart to begin Database Description (DBD) packet negotiation.
ExStart is the state where master/slave roles and initial Database Description sequence numbers are negotiated.
3
Identify the parameter evaluated during DBD packet exchange.
DBD packets carry the configured interface MTU value. If one router receives a DBD packet specifying an MTU larger than its own configured interface MTU, it drops the packet.
Dropping DBD packets prevents sequence number acknowledgment, trapping the neighbor relationship in the ExStart/Exchange state.

Anahtar Kavram

OSPF Neighbor States and MTU Mismatch Impact
Tahmini Süre:1m 0s
Soru 365Soru

A network administrator is troubleshooting neighbor discovery on a Cisco Catalyst switch connected to a multi-vendor IP phone on interface GigabitEthernet1/0/5. Global LLDP has already been enabled with the `lldp run` command. However, output from `show lldp neighbors` indicates that while the switch successfully processes incoming packets from the IP phone, the IP phone is not receiving any LLDP advertisements from the switch.

Which interface configuration command must be applied under interface GigabitEthernet1/0/5 to resolve this issue and allow the switch to send LLDP frames to the phone?

Cevabı ve açıklamayı göster

Cevap: lldp transmit

Cevap

The command 'lldp transmit' must be configured under interface GigabitEthernet1/0/5 to allow outbound LLDP advertisements.
The correct answer specifies 'lldp transmit'. Unlike Cisco Discovery Protocol (CDP), which uses a single interface command ('cdp enable'), Link Layer Discovery Protocol (LLDP) provides granular directional control on interfaces. The 'lldp transmit' interface submode command explicitly instructs the Cisco switch port to generate and send outbound LLDP frames to adjacent devices.

Adım Adım Çözüm

1
Identify the protocol and scope required
The target protocol is LLDP (IEEE 802.1AB) operating at the interface level on a Cisco Catalyst switch.
Global LLDP is active, but outbound transmission on GigabitEthernet1/0/5 is currently disabled or missing.
2
Differentiate CDP and LLDP interface command syntax
CDP uses 'cdp enable' on interfaces, whereas LLDP uses separate direction-specific submode commands: 'lldp transmit' and 'lldp receive'.
LLDP allows directional control over transmission and reception per port.
3
Select the correct directional interface submode command
Applying 'lldp transmit' enables outbound advertisement of local switch Type-Length-Value (TLV) attributes.
The issue specifically states that incoming packets are received but outbound packets are not being sent.

Anahtar Kavram

LLDP Interface Configuration Commands
Tahmini Süre:1m 0s
Soru 366Soru

An administrator executes the command `show ip route` on a Cisco router and observes the following routing table entry:

`O 10.40.0.0/24 [110/50] via 172.16.1.2, 00:15:30, GigabitEthernet0/0`

Which component of the routing table entry is represented by the numerical value 50 inside the brackets?

Cevabı ve açıklamayı göster

Cevap: The metric (cost) of the route to the destination network

Cevap

The numerical value 50 inside the brackets represents the metric (cost) of the route to the destination network.
In Cisco IOS routing tables, bracketed values following a destination network prefix use the standard format `[Administrative Distance / Metric]`. In the given output `[110/50]`, `110` is the default Administrative Distance for OSPF routes, and `50` is the calculated cost metric to reach network `10.40.0.0/24`.

Adım Adım Çözüm

1
Identify the standard bracket notation in Cisco IPv4 routing table output.
The bracketed syntax is defined as `[Administrative Distance / Metric]`.
Cisco routers display the trustworthiness (AD) and path cost (metric) together inside square brackets immediately after the network prefix.
2
Parse the bracketed values `[110/50]` from the OSPF route entry.
The first number `110` is the OSPF Administrative Distance, and the second number `50` is the route metric.
Separating the position of each number identifies `50` as the metric/cost value calculated by OSPF.

Anahtar Kavram

Routing Table Components: Administrative Distance vs. Metric
Tahmini Süre:45s
Soru 367Soru

A network administrator is configuring static routing on router R1 to forward traffic to the destination network 2001:DB8:CAFE:5::/642001:\text{DB8}:\text{CAFE}:5::/64. The egress interface on R1 is GigabitEthernet0/0/0, which connects directly to router R2. Router R2's connected interface has a global unicast address of 2001:DB8:LAB:1::2/642001:\text{DB8}:\text{LAB}:1::2/64 and a link-local address of FE80::2\text{FE80}::2. Which command correctly configures a static route on R1 using R2's link-local address as the next hop?

Cevabı ve açıklamayı göster

Cevap: ipv6 route 2001:DB8:CAFE:5::/64 GigabitEthernet0/0/0 FE80::2

Cevap

The command 'ipv6 route 2001:DB8:CAFE:5::/64 GigabitEthernet0/0/0 FE80::2' correctly specifies both the local exit interface and the neighbor's link-local next-hop address.
In Cisco IOS, IPv6 static routes using a link-local address as the next-hop MUST be fully specified by including both the local exit interface and the neighbor's link-local IPv6 address. The command specifying both GigabitEthernet0/0/0 and FE80::2 meets this requirement.

Adım Adım Çözüm

1
Identify the destination prefix and prefix length.
The target destination network is 2001:DB8:CAFE:5::/642001:\text{DB8}:\text{CAFE}:5::/64.
The first argument of an IPv6 static route in Cisco IOS is always the destination IPv6 prefix/prefix-length.
2
Determine the next-hop requirements when using a link-local address.
Link-local addresses (FE80::/10) are bound strictly to a specific link (interface).
Because the same link-local address can exist on multiple interfaces of the same router, the router cannot determine which egress interface to use unless the exit interface is explicitly stated in the command.
3
Construct the complete fully specified IPv6 static route command.
'ipv6 route 2001:DB8:CAFE:5::/64 GigabitEthernet0/0/0 FE80::2'
This includes the destination prefix, the local interface (GigabitEthernet0/0/0), and the neighbor's link-local address (FE80::2).

Anahtar Kavram

IPv6 Static Route Link-Local Next-Hop Syntax Requirement
Tahmini Süre:1m 30s
Soru 368Soru

A network host generates its IPv6 link-local address using stateless address autoconfiguration (SLAAC) and the standard IEEE EUI-64 process. If the Ethernet interface has a burned-in MAC address of `0800.27a4.12b3`, what is the resulting IPv6 link-local address assigned to the interface?

Cevabı ve açıklamayı göster

Cevap: fe80::a00:27ff:fea4:12b3

Cevap

The resulting link-local address is fe80::a00:27ff:fea4:12b3.
The correct option correctly converts the MAC address `0800.27a4.12b3` into an EUI-64 interface identifier by inserting `FFFE` into the middle (`0800:27FF:FEa4:12b3`) and toggling the 7th bit of the first byte (`0x08` to `0x0A`). Prepending the `fe80::` link-local prefix results in `fe80::a00:27ff:fea4:12b3`.

Adım Adım Çözüm

1
Split the 48-bit MAC address into two 24-bit halves and insert FFFE into the midpoint.
The MAC address `08-00-27-a4-12-b3` becomes `0800:27FF:FEA4:12B3`.
EUI-64 requires expanding a 48-bit MAC address into a 64-bit interface identifier by placing FFFE between the OUI and the vendor-assigned payload.
2
Invert the 7th bit (Universal/Local bit) of the first byte.
First byte `0x08` (`0000 1000` in binary) has its 7th bit inverted to become `0000 1010` (`0x0A`).
IEEE EUI-64 rules dictate that the 7th bit of the first byte must be inverted (0 for global uniqueness becomes 1 for local scope in IPv6).
3
Prepend the standard link-local network prefix fe80::/10.
The complete IPv6 link-local address is `fe80::a00:27ff:fea4:12b3` (leading zeroes in hex blocks are omitted per standard IPv6 notation).
Link-local addresses generated via SLAAC automatically use the prefix fe80::/10 followed by 54 zero bits and the 64-bit EUI-64 interface identifier.

Anahtar Kavram

EUI-64 Interface Identifier Generation and IPv6 Link-Local Addressing
Soru 369Soru

A Cisco router receives an IPv4 unicast packet destined for 10.150.45.6710.150.45.67. The router's Routing Information Base (RIB) and Forwarding Information Base (FIB) contain multiple overlapping routes, static entries, and dynamic routing protocol paths. In what exact sequence does the router evaluate routing parameters and process the packet from ingress lookup to physical egress forwarding? Arrange the steps in the correct chronological order from first to last.

Öğeleri doğru sıraya koymak için sürükleyin

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Cevap

The correct evaluation sequence for router forwarding decision logic is: 1) Longest Prefix Match evaluation, 2) Administrative Distance tie-breaking for identical prefixes across different protocols, 3) Metric tie-breaking for identical prefixes within the same protocol, 4) ARP lookup for next-hop MAC address resolution, and 5) Layer 2 encapsulation rewrite, TTL decrement, checksum update, and frame transmission.
The router forwarding process follows a strict hierarchy: Longest Prefix Match (LPM) is evaluated first across all routes in the routing table. If multiple routes from different protocol sources share the exact same prefix length, Administrative Distance (AD) breaks the tie. If multiple routes from the same protocol share the exact same prefix length, protocol metric breaks the tie. Once the best route and next-hop are chosen, the router resolves the next-hop MAC via ARP/NDP, updates IP header fields (TTL decrement by 1 and checksum update), rewrites the Layer 2 header, and transmits the frame out the egress interface.

Adım Adım Çözüm

1
Perform destination IP lookup using Longest Prefix Match (LPM).
The router identifies all matching subnet routes and selects the route entry with the highest number of matching network bits (most specific mask length).
LPM supersedes lower Administrative Distance values or routing metrics. A /28/28 route will always be chosen over a /24/24 route regardless of whether the /24/24 is static or connected.
2
Compare Administrative Distance (AD) if multiple sources present identical prefix masks.
The route from the source with the lowest AD (e.g., Static 11 vs. OSPF 110110) is preferred.
AD measures route source believability and only arbitrates between identical prefix masks from different routing processes.
3
Compare routing protocol metrics for ties within the same protocol.
The path with the lowest metric (cost, hop count, composite metric) is selected for RIB insertion.
Metrics only compare routes of the exact same prefix length learned via the same routing protocol algorithm.
4
Perform Layer 2 address resolution for the resolved next-hop address.
The router locates the next-hop IPv4 address in the ARP cache to retrieve its target MAC address.
Layer 3 forwarding decision requires binding the IP next-hop to a physical Layer 2 destination MAC address.
5
Execute packet header modifications and physical transmission.
The IPv4 TTL is decremented by 11, header checksum recalculated, source MAC updated to egress interface MAC, destination MAC updated to next-hop MAC, and the frame is serialized onto the wire.
Routers act as Layer 3 boundaries and must rebuild the Layer 2 frame header for each routed hop.

Anahtar Kavram

Router Forwarding Decision Logic and Longest Prefix Match Hierarchical Precedence
Soru 370Soru

A network administrator is setting up a new wireless network using the Cisco Wireless LAN Controller (WLC) web GUI. After defining the WLAN profile name and SSID on the General tab, the administrator needs to configure WPA2 Pre-Shared Key (PSK) authentication for client connectivity. Under which tab and sub-tab within the WLANs > Edit menu is the PSK passphrase field configured?

Cevabı ve açıklamayı göster

Cevap: Security tab, under the Layer 2 sub-tab

Cevap

The PSK passphrase is configured under the Security tab on the Layer 2 sub-tab.
In the Cisco WLC web GUI, Layer 2 wireless security protocols (WPA2/WPA3) and key management methods (PSK vs 802.1X) are located under WLANs > Edit > Security > Layer 2. Enabling PSK under Auth Key Management displays the Pre-Shared Key entry box directly on the Layer 2 sub-tab.

Adım Adım Çözüm

1
Navigate to WLANs > Edit for the target WLAN in the WLC GUI.
Access the configuration tabs for the specific WLAN.
WLAN-specific parameters are modified in this section.
2
Click on the Security tab, then select the Layer 2 sub-tab.
Layer 2 security options (such as WPA+WPA2) are displayed.
Wi-Fi authentication mechanisms (PSK and 802.1X) operate at Layer 2.
3
Select PSK under Auth Key Management and enter the passphrase in the Pre-Shared Key box.
The PSK passphrase is applied directly within the Layer 2 sub-tab.
Cisco AireOS WLC GUI integrates PSK key entry directly into the Layer 2 security policy page.

Anahtar Kavram

WLC GUI WLAN Security Configuration & Navigation
Tahmini Süre:1m 15s
Soru 371Soru

An infrastructure team is provisioning a new subnet from the IPv4 address block 172.20.96.0/23172.20.96.0/23 for a server farm that requires support for up to 250250 usable host interfaces. If the team applies the most efficient prefix length that satisfies this requirement, which IPv4 address is the last usable host address in the first subnet created?

Cevabı ve açıklamayı göster

Cevap: 172.20.96.254

Cevap

172.20.96.254
To support 250250 usable host interfaces efficiently, 88 host bits are required (282=2542^8 - 2 = 254 usable hosts), corresponding to a /24/24 subnet prefix. The first /24/24 subnet carved from the parent block 172.20.96.0/23172.20.96.0/23 has a network address of 172.20.96.0172.20.96.0 and a broadcast address of 172.20.96.255172.20.96.255. The last usable host IP address is one position below the broadcast address, which is 172.20.96.254172.20.96.254.

Adım Adım Çözüm

1
Determine the required host bits and prefix length for 250 hosts.
Using 2n22502^n - 2 \ge 250, n=8n = 8 host bits are needed (282=2542^8 - 2 = 254 usable addresses). The corresponding prefix length is 328=/2432 - 8 = /24.
Selecting 8 host bits fulfills the requirement of at least 250 usable hosts with minimum wasted IP space.
2
Calculate the boundaries for the first /24 subnet from 172.20.96.0/23.
The first subnet is 172.20.96.0/24, spanning network address 172.20.96.0 to broadcast address 172.20.96.255.
A /24 subnet increments in the third octet by 1, starting at the base network address 172.20.96.0.
3
Identify the last usable host address of the 172.20.96.0/24 subnet.
Subtracting 1 from the broadcast address (172.20.96.255 - 1) gives 172.20.96.254.
Usable host IP addresses lie strictly between the network address and the broadcast address.

Anahtar Kavram

IPv4 Subnetting, Prefix Derivation, and Host Range Boundaries
Soru 372Soru

Match each Layer 2 discovery protocol characteristic or configuration parameter on the left with its correct operational description or standard on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

IEEE 802.1AB
CDP Default Holdtime
LLDP-MED
lldp transmit

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

IEEE 802.1AB matches the vendor-neutral open standard specification for Layer 2 discovery. CDP Default Holdtime matches 180 seconds before purging a neighbor entry. LLDP-MED matches the extension providing endpoint attributes such as VoIP policy. lldp transmit matches the interface-level command to enable sending outgoing discovery frames.
Each match correctly aligns the discovery protocol attribute, standards body designation, timer setting, or interface command syntax to its operational definition in Cisco IOS networks.

Adım Adım Çözüm

1
Identify the standard associated with LLDP
IEEE 802.1AB defines the open-standard LLDP protocol across heterogenous network devices.
CDP is Cisco proprietary, whereas LLDP was developed under IEEE 802.1AB.
2
Recall CDP default timer values
CDP sends messages every 60 seconds and retains neighbor information for a holdtime of 180 seconds.
Knowing default timers is crucial for troubleshooting layer 2 neighbor relationships in Cisco networks.
3
Differentiate LLDP base protocol from LLDP-MED extensions
LLDP-MED extends base LLDP capability to advertise endpoint information like voice VLAN IDs, call control details, and PoE parameters.
LLDP-MED operates specifically between network infrastructure devices and media endpoints.
4
Analyze interface syntax for directional LLDP control
'lldp transmit' and 'lldp receive' control outgoing and incoming frames per interface.
Unlike CDP which is enabled bidirectionally when turned on, LLDP allows fine-grained directional enabling per interface.

Anahtar Kavram

Layer 2 Discovery Protocol Capabilities, Standards, and CLI Syntax
Soru 373Soru

Refer to the following partial routing table output from a Cisco router:

text
Codes: C - connected, S - static, R - RIP, M - mobile, B - BGP
D - EIGRP, EX - EIGRP external, O - OSPF, IA - OSPF inter area

Gateway of last resort is 10.4.4.4 to network 0.0.0.0

O 172.16.32.0/20 [110/20] via 10.1.1.1, 00:10:05, GigabitEthernet0/0
D 172.16.35.64/28 [90/307200] via 10.2.2.2, 00:05:12, GigabitEthernet0/1
S 172.16.35.64/26 [1/0] via 10.3.3.3
C 172.16.35.0/24 is directly connected, GigabitEthernet0/2
S* 0.0.0.0/0 [1/0] via 10.4.4.4

The router receives an ingress IPv4 packet with a destination address of 172.16.35.73. Which statements accurately describe the router forwarding decision for this packet? (Select TWO.)

Geçerli olan tümünü seçin

Cevabı ve açıklamayı göster

Cevap: The router forwards the packet to next-hop IP 10.2.2.2 via GigabitEthernet0/1 because the route to 172.16.35.64/28 provides the longest prefix match.; The router bypasses the static route to 172.16.35.64/26 despite its lower Administrative Distance because Administrative Distance is only compared among routes with identical prefix lengths.

Cevap

The router forwards the packet to next-hop IP 10.2.2.2 via GigabitEthernet0/1 because 172.16.35.64/28 is the longest prefix match, and it ignores lower administrative distances on shorter prefixes because AD is only evaluated when prefix lengths are equal.
When forwarding an IPv4 packet, a Cisco router evaluates the Forwarding Information Base using Longest Prefix Match (LPM) as the primary criteria. For destination address 172.16.35.73, the route 172.16.35.64/28 (host range 172.16.35.64 to 172.16.35.79) has a 28-bit prefix length, making it more specific than the /26 static route, the /24 directly connected subnet, or the /20 OSPF route. Administrative Distance (AD) is only evaluated when multiple routing protocols install identical network prefixes into the routing table, so the router does not consider the lower AD of the static route.

Adım Adım Çözüm

1
Identify all candidate routes matching destination IP 172.16.35.73
Candidate routes are 172.16.32.0/20, 172.16.35.0/24, 172.16.35.64/26, 172.16.35.64/28, and 0.0.0.0/0.
The destination host IP 172.16.35.73 falls within the IP address range of all these configured route subnets.
2
Determine the Longest Prefix Match (LPM) among candidates
172.16.35.64/28 has the longest prefix mask length (/28 = 28 bits).
Cisco IOS forwarding logic always selects the route with the most specific subnet mask matching the target destination.
3
Evaluate Administrative Distance and Metric applicability
Administrative Distance and Metric are not compared across different prefix lengths.
AD is only used when multiple routing protocols offer routes to the exact same network prefix length.

Anahtar Kavram

Router Forwarding Decision Logic (Longest Prefix Match Rule)
Soru 374Soru

A router learns the destination network 10.20.30.0/2410.20.30.0/24 dynamically via RIPv2 with a default Administrative Distance of 120120. A network engineer needs to configure a floating static backup route for this network via next-hop IP address 192.168.12.2192.168.12.2. The backup route must only enter the routing table if the primary RIP route fails. Complete the Cisco IOS configuration command to assign an Administrative Distance of 130130 to the floating static route.

Aşağıdaki boşlukları doldurun

Router(config)# ip route 10.20.30.0 255.255.255.0 192.168.12.2
Cevabı ve açıklamayı göster

Cevap

130
A floating static route serves as a backup path and must be configured with an Administrative Distance (AD) strictly higher than that of the primary route. Since RIPv2 has a default AD of 120, specifying 130 as the trailing metric in the 'ip route' command ensures that the static route remains out of the routing table until the RIP route fails.

Adım Adım Çözüm

1
Identify the Administrative Distance (AD) of the primary dynamic routing protocol.
RIPv2 has a default Administrative Distance of 120.
Routing decisions prioritize routes with lower Administrative Distance values.
2
Determine the required Administrative Distance for the backup floating static route.
The backup route must have an AD greater than 120 (specifically 130 as requested).
A higher AD keeps the static route inactive (floating) in the routing table as long as the primary route is active.
3
Complete the Cisco IOS command by appending the administrative distance integer at the end.
The trailing argument in 'ip route 10.20.30.0 255.255.255.0 192.168.12.2 130' sets the AD to 130.
In Cisco IOS syntax, an optional integer at the end of the 'ip route' command overrides the default static route AD of 1.

Anahtar Kavram

Floating Static Route Administrative Distance Configuration
Tahmini Süre:1m 0s
Soru 375Soru

A network administrator is configuring an internal host interface and must assign an IP address within the RFC 1918 Class B private address space. Which of the following IP addresses is a valid RFC 1918 private IPv4 address?

Cevabı ve açıklamayı göster

Cevap: 172.20.50.1

Cevap

172.20.50.1
The address 172.20.50.1 is a valid RFC 1918 private IPv4 address because it falls within the 172.16.0.0/12 block, which ranges from 172.16.0.0 to 172.31.255.255.

Adım Adım Çözüm

1
Identify the RFC 1918 Class B private address boundaries
RFC 1918 defines Class B private space as 172.16.0.0 through 172.31.255.255 (/12 prefix).
Understanding RFC 1918 ranges is essential for properly distinguishing private non-routable internal addresses from globally routable public addresses.
2
Evaluate the candidate addresses against the RFC 1918 boundaries
The address 172.20.50.1 falls between 172.16.0.0 and 172.31.255.255. The other addresses (172.32.1.10, 192.169.100.1, 11.10.0.1) fall outside RFC 1918 ranges.
Only addresses within the designated RFC 1918 ranges are non-routable on the public Internet and reserved for private network deployment.

Anahtar Kavram

Private IPv4 Address Ranges (RFC 1918)
Tahmini Süre:45s
Soru 376Soru

An enterprise network engineering team is upgrading a data center fabric to support high-density East-West microsegmentation traffic using a Spine-Leaf architecture. Which of the following statements accurately characterize the topology rules and traffic handling of a standard two-tier Spine-Leaf design? (Select TWO.)

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Cevap: Every leaf switch connects to every spine switch, and spine switches do not connect directly to each other.; Traffic traveling between endpoints connected to different leaf switches experiences a consistent, predictable hop count across the fabric.

Cevap

The correct statements are: 1) Every leaf switch connects to every spine switch, and spine switches do not connect directly to each other; and 2) Traffic traveling between endpoints connected to different leaf switches experiences a consistent, predictable hop count across the fabric.
In a standard two-tier Spine-Leaf architecture, every leaf switch connects to every spine switch, and spine switches do not connect directly to each other. This creates a uniform bipartite topology where any endpoint on a leaf switch reaches an endpoint on another leaf switch in exactly three hops (Leaf -> Spine -> Leaf), yielding consistent and predictable latency for East-West traffic.

Adım Adım Çözüm

1
Analyze Spine-Leaf physical connectivity rules
Confirm that Spine-Leaf fabrics require a bipartite connection pattern where leaf switches connect to all spine switches, while spine-to-spine and leaf-to-leaf direct links are omitted.
This structural pattern allows Equal-Cost Multi-Pathing (ECMP) to load-balance traffic dynamically across all available spine paths.
2
Evaluate traffic path consistency for East-West data flows
Determine that any leaf-to-leaf communication crosses exactly one spine switch.
Uniform path lengths prevent latency variance for inter-server communication.
3
Evaluate attachment points for endpoints and hypervisors
Identify that servers, hypervisors, and security appliances attach strictly to leaf switches (access layer of the fabric).
Attaching hosts to spine switches breaks the bipartite Clos architecture and compromises ECMP forwarding.

Anahtar Kavram

Spine-Leaf Topology Connectivity & Traffic Dynamics
Tahmini Süre:1m 30s
Soru 377Soru

Match each Cisco Lightweight Access Point (AP) operational mode to its precise functional capability and operational behavior within a centralized or branch wireless architecture.

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Öğeler

FlexConnect Mode
Monitor Mode
Rogue Detector Mode
Sniffer Mode

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

FlexConnect Mode pairs with local VLAN switching and WAN offline authentication fallback; Monitor Mode pairs with dedicated radio scanning for location tracking and wIDS without client association; Rogue Detector Mode pairs with wired interface monitoring of ARP/MAC tables to identify rogue wired connections; Sniffer Mode pairs with capturing raw 802.11 frames on a designated channel for remote packet analysis.
Each Cisco AP mode serves a specialized operational role: FlexConnect is tailored for remote branch local switching and WAN survivability; Monitor mode provides dedicated wireless security scanning without client association; Rogue Detector mode checks wired ARP/MAC tables to find rogue infrastructure connected to wired switches; and Sniffer mode captures raw wireless traffic on a fixed channel for deep packet inspection.

Adım Adım Çözüm

1
Analyze FlexConnect Mode capabilities
Identified local switching and local authentication fallback behavior during WLC WAN unreachable states.
FlexConnect APs are specifically designed for branch/WAN deployments to prevent branch data traffic from bottlenecking over the WAN while supporting standalone survivability.
2
Analyze Monitor Mode operational parameters
Matched with dedicated multi-channel scanning for wIDS, Rogue AP detection, and location tracking.
Monitor mode APs do not transmit SSIDs or allow client associations, freeing the radios to continuously scan tuned channels for security metrics.
3
Differentiate Rogue Detector Mode from wireless scanning modes
Matched with wired switch ARP/MAC inspection.
Rogue Detector APs leverage the wired network infrastructure rather than wireless radios to correlate MAC addresses reported as rogues with connected switch ports.
4
Analyze Sniffer Mode functionality
Matched with channel-specific raw 802.11 frame capturing forwarded to remote analysis tools.
Sniffer mode tunes an AP radio to a fixed channel and streams captured wireless headers and payloads to an IP endpoint running a frame analyzer.

Anahtar Kavram

Cisco Access Point Operational Modes
Tahmini Süre:2m 30s
Soru 378Soru

A network engineering team is evaluating configuration management tools to automate Cisco IOS XE switch deployments. The architecture mandates that managed network devices must not require a persistent local management daemon or software agent, and all configuration task execution must be driven from a central control node using native management protocols and human-readable YAML files.

Which of the following operational characteristics correctly describe the selected configuration management tool? (Select TWO.)

Geçerli olan tümünü seçin

Cevabı ve açıklamayı göster

Cevap: The framework utilizes a push-based communication model where the control node initiates management sessions directly to target network devices.; Configuration tasks are defined in playbooks and executed across endpoints over standard SSH or NETCONF transport protocols.

Cevap

The configuration management mechanism described is Ansible, which uses a push-based communication model from a control node over standard transport protocols (SSH/NETCONF) with YAML playbooks and requires no agent software on target endpoints.
The scenario describes Ansible, which is unique among the three major configuration management tools for being completely agentless on target network devices. Ansible relies on a push-based model where the control node opens standard transport sessions (such as SSH or NETCONF) to execute tasks defined in YAML playbooks.

Adım Adım Çözüm

1
Analyze the architectural requirements stated in the scenario.
The requirements specify an agentless architecture, central control node driving execution, native management protocols (SSH/NETCONF), and YAML files.
Identifying these key attributes narrows the tool down to Ansible, distinguishing it from agent-based pull systems like Puppet and Chef.
2
Evaluate the delivery mechanism and transport method of Ansible.
Ansible uses a push model where the control node initiates connections to managed devices using SSH or NETCONF.
Because no persistent agent daemon is installed on managed network switches, Ansible relies on native network management transport protocols.
3
Verify file structure and contrast with alternative mechanisms.
Ansible uses YAML playbooks. In contrast, Puppet uses Puppet DSL manifests over TCP 8140 (pull), and Chef uses Ruby recipes/cookbooks over TCP 443 (pull).
This confirms that options describing push-based execution over SSH/NETCONF and YAML playbooks are correct, while options referring to TCP 8140 pull operations or Ruby cookbooks describe Puppet and Chef.

Anahtar Kavram

Ansible Agentless Push Architecture and Protocol Capabilities
Tahmini Süre:2m 0s
Soru 379Soru

A network engineering team is evaluating the operational differences between a traditional distributed architecture and a centralized controller-based network deployment. Which operational characteristic correctly distinguishes the control plane handling in a controller-based network from that of a traditional distributed network?

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Cevap: Control plane intelligence is centralized within a software controller that dictates forwarding policies across the network, whereas traditional network devices each run independent control plane processes locally.

Cevap

Control plane intelligence is centralized within a software controller that dictates forwarding policies across the network, whereas traditional network devices each run independent control plane processes locally.
The correct answer highlights the fundamental shift from distributed to centralized architecture. In traditional networking, every device operates its own control plane protocols to learn routes and switching paths. In a controller-based architecture, a centralized controller offloads or manages these control plane duties and programs the forwarding tables on underlying data plane devices.

Adım Adım Çözüm

1
Analyze how traditional networks handle control plane operations
Identify that traditional network infrastructure relies on distributed intelligence where each switch or router runs its own control plane algorithms (e.g., OSPF, EIGRP, STP) locally.
Understanding traditional architecture requires recognizing that every node operates autonomously to build its routing and MAC tables.
2
Analyze how controller-based networks handle control plane operations
Identify that controller-based networks shift control plane intelligence to a centralized software controller.
Centralization allows the controller to possess global visibility of the network topology and programmatically direct device forwarding behaviors via Southbound APIs.
3
Compare the two models to select the correct architectural distinction
Confirm that centralizing control plane intelligence on a software controller versus running independent local control plane algorithms per device is the primary architectural difference.
This fundamental shift enables automated management, intent-based policies, and consistent network-wide provisioning.

Anahtar Kavram

Centralized Control Plane vs. Distributed Control Plane
Soru 380Soru

When an interface automatically constructs an IPv6 EUI-64 interface identifier using the MAC address 00:1A:2B:3C:4D:5E, which string represents the correctly generated 64-bit interface ID?

Cevabı ve açıklamayı göster

Cevap: 021a:2bff:fe3c:4d5e

Cevap

The correct interface identifier is 021a:2bff:fe3c:4d5e.
To create a 64-bit EUI-64 interface ID from a 48-bit MAC address, the 16-bit value FFFE is inserted between the third and fourth bytes of the MAC address. In addition, the 7th bit (universal/local bit) of the first byte must be inverted. Inverting the 7th bit of 0x00 results in 0x02, producing the complete interface ID 021a:2bff:fe3c:4d5e.

Adım Adım Çözüm

1
Split the 48-bit MAC address into two 24-bit halves.
The left half is 00:1A:2B and the right half is 3C:4D:5E.
EUI-64 generation requires inserting a specific 16-bit hex value between the OUI and NIC specific bytes.
2
Insert the 16-bit hexadecimal value FFFE between the two halves.
The combined value becomes 00:1A:2B:FF:FE:3C:4D:5E.
FFFE expands the 48-bit MAC address into a 64-bit structure.
3
Invert the 7th bit (Universal/Local bit) of the first byte (00).
Hexadecimal 00 (0000 0000 in binary) becomes hexadecimal 02 (0000 0010 in binary).
IEEE EUI-64 rules require modifying the 7th bit to represent universal scope.
4
Format the resulting 64 bits into standard IPv6 colon-separated hextets.
021a:2bff:fe3c:4d5e
IPv6 interface identifiers are presented as four 16-bit hexadecimal hextets.

Anahtar Kavram

EUI-64 Interface ID Generation
Tahmini Süre:45s
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