Tüm alıştırma soruları

1150 soru

Soru 1Soru

A network administrator upgrades several core backbone links within an Open Shortest Path First (OSPF) routing domain from 1 Gbps to 10 Gbps. Despite the physical bandwidth upgrade, the core routers assign an identical metric cost of 1 to both the 1 Gbps and 10 Gbps interfaces, resulting in unwanted equal-cost multi-path (ECMP) load balancing across asymmetric links. Which configuration change must be applied to ensure OSPF correctly assigns a lower metric cost to the 10 Gbps interfaces?

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Cevap: Increase the OSPF auto-cost reference bandwidth setting across all participating routers in the routing domain.

Cevap

Increase the OSPF auto-cost reference bandwidth setting across all participating routers in the routing domain.
The correct answer highlights that default OSPF implementations use a reference bandwidth of 100 Mbps (10810^8 bps) to calculate link cost. As a result, both 1 Gbps and 10 Gbps interfaces evaluate to a cost less than 1, which rounds up to the minimum metric value of 1. Increasing the auto-cost reference bandwidth across all routers allows OSPF to assign distinct, lower cost metrics to faster links, resolving unwanted equal-cost load balancing.

Adım Adım Çözüm

1
Analyze the default OSPF interface cost calculation formula.
Recognize that OSPF calculates cost as Cost=Reference BandwidthInterface Bandwidth\text{Cost} = \frac{\text{Reference Bandwidth}}{\text{Interface Bandwidth}}, where default reference bandwidth equals 100 Mbps (10810^8 bps).
For interfaces with bandwidth equal to or greater than 100 Mbps (such as 1 Gbps and 10 Gbps), the formula yields values less than 1, which are rounded up to the minimum metric integer of 1.
2
Determine how to restore metric differentiation for high-speed interfaces.
Adjust the auto-cost reference bandwidth parameter to a higher threshold (such as 100,000 Mbps or 100 Gbps).
With a 100,000 Mbps reference bandwidth, a 1 Gbps link has a metric cost of 100 (100,0001,000\frac{100,000}{1,000}), whereas a 10 Gbps link has a metric cost of 10 (100,00010,000\frac{100,000}{10,000}), successfully preferring the faster path.

Anahtar Kavram

OSPF Auto-Cost Reference Bandwidth Adjustment
Soru 2Soru

A network engineer is analyzing traffic traversing an enterprise Layer 3 switch configured for inter-VLAN routing. A client host on VLAN 10 sends a request to a destination server on VLAN 20. When the switch receives the frame on its VLAN 10 interface and routes it out through the VLAN 20 interface, which of the following best describes the de-encapsulation and re-encapsulation process performed by the switch?

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Cevap: The switch strips the incoming Layer 2 Ethernet header and trailer, inspects the Layer 3 destination IP address to determine the egress interface, and encapsulates the IP packet into a new Layer 2 Ethernet frame with updated source and destination MAC addresses.

Cevap

The switch strips the incoming Layer 2 Ethernet header and trailer, inspects the Layer 3 destination IP address to determine the egress interface, and encapsulates the IP packet into a new Layer 2 Ethernet frame with updated source and destination MAC addresses.
When a Layer 3 device (such as a router or Layer 3 switch) routes traffic between subnets/VLANs, it de-encapsulates the incoming Layer 2 Ethernet frame to extract the enclosed Layer 3 IP packet. It inspects the destination IP address to determine the appropriate egress interface. Before transmitting onto the destination VLAN, the device re-encapsulates the IP packet into a brand new Layer 2 Ethernet frame. The new frame uses the router's egress interface MAC as the source MAC address and the destination host's MAC (or next hop MAC) as the destination MAC address. The original IP packet payload and end-to-end IP addresses remain unchanged (except for TTL decrementing and checksum updating).

Adım Adım Çözüm

1
Analyze the incoming frame at Layer 2
The Layer 3 switch receives the Ethernet frame on VLAN 10, verifies the frame check sequence (FCS) trailer, and verifies that the destination MAC matches its own gateway MAC address.
Layer 2 processing requires verifying frame integrity and confirming the frame is addressed to the default gateway.
2
De-encapsulate the packet at Layer 3
The switch strips the Layer 2 Ethernet header and trailer to expose the enclosed IP packet.
To route traffic between different subnets/VLANs, the network layer header must be inspected.
3
Perform routing lookup and update IP header fields
The switch reads the destination IP address, checks its routing table, decrements the Time to Live (TTL) value by 1, and recalculates the IP header checksum.
The Layer 3 header provides forwarding instructions across network boundaries.
4
Re-encapsulate the packet into a new Layer 2 frame
The switch appends a new Layer 2 Ethernet header containing its egress interface MAC address as the source MAC, the destination server's MAC address as the destination MAC, and recalculates the FCS trailer for VLAN 20.
Each physical or logical link (VLAN) requires local Layer 2 addressing for delivery to the destination or next-hop device.

Anahtar Kavram

Hop-by-hop encapsulation and de-encapsulation during Layer 3 routing
Tahmini Süre:1m 30s
Soru 3Soru

During data transmission across an IEEE 802.1Q trunk link, a switch inserts a 4-byte header tag into Ethernet frames for non-native VLANs. At which specific location within the standard Ethernet II frame header is this 802.1Q tag inserted?

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Cevap: Between the Source MAC Address and the Type/Length fields

Cevap

The 802.1Q tag is inserted between the Source MAC Address and the Type/Length fields.
IEEE 802.1Q trunking inserts a 4-byte VLAN header directly into the Layer 2 Ethernet frame between the Source MAC Address field and the original EtherType/Length field. This tag includes the Tag Protocol Identifier (TPID set to 0x8100) and Tag Control Information (TCI, containing the VLAN ID and priority bits).

Adım Adım Çözüm

1
Analyze standard Ethernet II frame structure
An Ethernet II frame consists of Preamble/SFD, Destination MAC, Source MAC, EtherType, Payload, and FCS.
Understanding standard frame layout is required to identify encapsulation insertion points.
2
Identify 802.1Q trunking tag placement rules
IEEE 802.1Q modifies the Layer 2 header by inserting a 4-byte VLAN tag field (TPID 0x8100 + TCI) directly following the 6-byte Source MAC address.
This location allows receiving switches to parse the Tag Protocol Identifier (TPID) in place of the original EtherType.
3
Evaluate frame check sequence recalculation
Because the Layer 2 header is modified, the sending switch recalculates the Frame Check Sequence (FCS) trailer before transmission.
Ensures the frame passes integrity checks on the receiving switch trunk port.

Anahtar Kavram

IEEE 802.1Q Frame Tagging and Header Structure
Soru 4Soru

An enterprise branch office has 500 internal workstations configured on a private IPv4 network (10.1.0.0/2210.1.0.0/22). Outbound internet access is currently handled by an edge router using Port Address Translation (PAT / NAT Overload) mapped to a single public IP address (203.0.113.5203.0.113.5). During peak operational hours, users report that while existing web sessions remain active, new outbound connection attempts to cloud services begin timing out consistently. A network administrator inspects the router's active translation table and discovers over 64,00064,000 active translation entries associated with 203.0.113.5203.0.113.5, while CPU, memory, and interface bandwidth utilization on the edge router remain below 25%25\%. Which of the following implementation changes is the most appropriate solution to resolve the connection failures?

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Cevap: Configure a dynamic NAT pool containing multiple public IP addresses and bind PAT to the pool.

Cevap

The correct action is to configure a dynamic NAT pool containing multiple public IP addresses and bind PAT to the pool.
Port Address Translation (PAT) uses 16-bit source port numbers to multiplex multiple private IP streams over a single public IP address. Because a 16-bit port field yields a maximum of 65,535 ports, a high-density subnet can exhaust the ephemeral port pool of a single public IP address. Configuring a NAT pool of public IP addresses and applying PAT across the pool multiplies the available transport port pool by the number of IP addresses in the pool, resolving exhaustion without sacrificing simultaneous connectivity for internal hosts.

Adım Adım Çözüm

1
Analyze the symptom and diagnostic evidence.
Router hardware resources (CPU, RAM, bandwidth) are normal (<25%), but active NAT translation entries have reached over 64,000 on a single public IP address.
PAT tracks outbound connections by mapping private IP addresses and source port numbers to a single public IP address and unique external source ports. Since TCP/UDP port fields are 16-bit, the maximum theoretical port count per IP is 65,535 (with around 1,024 reserved/well-known ports, leaving ~64,500 usable ephemeral ports).
2
Identify the underlying bottleneck.
The network is suffering from PAT transport layer source port exhaustion.
Because 500 hosts are generating thousands of concurrent short-lived or long-lived sessions, the single public IP address has exhausted its available unique source port pool.
3
Evaluate candidate solutions to increase socket capacity.
Binding PAT to a pool of public IP addresses increases capacity linearly (e.g., 5 public IPs × ~64,000 ports = ~320,000 available concurrent translations).
Adding additional public IPs into a NAT pool tied to PAT overload expands the total available source port space while allowing all 500 internal hosts to maintain simultaneous outbound connectivity.

Anahtar Kavram

PAT Source Port Exhaustion and NAT Pool Scaling
Tahmini Süre:2m 0s
Soru 5Soru

A network administrator needs to add a DNS entry that resolves the hostname `appserver.corp.local` directly to its assigned IPv6 address `2001:db8:85a3::8a2e:370:7334`. Which DNS resource record type must be configured for this resolution?

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Cevap: AAAA record

Cevap

An AAAA record is used to map a domain hostname directly to an IPv6 address.
An AAAA record (quad-A record) stores a 128-bit IPv6 address and maps it to a host domain name. To resolve hostnames to IPv6 destinations, DNS authoritative servers must be configured with AAAA records.

Adım Adım Çözüm

1
Identify the required mapping
The scenario requires mapping a hostname (`appserver.corp.local`) to an IPv6 address (`2001:db8:85a3::8a2e:370:7334`).
Determining the network protocol version (IPv4 vs. IPv6) dictates the correct DNS record type.
2
Select the corresponding DNS record type
AAAA records are designated for 128-bit IPv6 address resolution.
Standard DNS taxonomy uses A records for IPv4 addresses and AAAA records for IPv6 addresses.

Anahtar Kavram

DNS IPv6 Host Mapping (AAAA Records)
Soru 6Soru

A network engineer is reviewing an active translation table on a corporate edge gateway to verify Port Address Translation (PAT) functionality. An internal client workstation with the private IP address 172.16.10.50172.16.10.50 initiates an HTTPS session to an external web server at 198.51.100.25:443198.51.100.25:443. The gateway translates all outbound traffic using its single public interface IP address of 203.0.113.1203.0.113.1. As the outgoing IP packet leaves the firewall's public interface toward the destination server, which address and port combination accurately represents the Inside Global identifier for this session?

Cevabı ve açıklamayı göster

Cevap: 203.0.113.1203.0.113.1 paired with a dynamically assigned high-numbered source port (such as 5214452144)

Cevap

The Inside Global address and port combination is the firewall's public IP address (203.0.113.1203.0.113.1) combined with a dynamic high-numbered source port (such as 5214452144).
The correct answer identifies that in NAT/PAT terminology, the Inside Global address represents the internal host as it appears to the outside network. Under Port Address Translation (PAT), the firewall replaces the private source IP with its own public WAN interface IP (203.0.113.1203.0.113.1) and substitutes the original source port with a uniquely allocated high-numbered ephemeral port (such as 5214452144) to maintain connection state mapping.

Adım Adım Çözüm

1
Identify the standard CompTIA NAT terminology definitions.
Inside Local is the private client IP/port (172.16.10.50:ephemeral172.16.10.50:ephemeral). Inside Global is the public translated IP/port representing the client to the internet (203.0.113.1:translated_port203.0.113.1:translated\_port). Outside Global is the target public IP/port (198.51.100.25:443198.51.100.25:443).
CompTIA Network+ standard definitions differentiate internal local private addresses from internal global public translated addresses.
2
Analyze the impact of Port Address Translation (PAT / NAT Overload).
PAT tracks individual outbound client connections by swapping the original private source IP and source port with the public gateway IP and a newly assigned high-numbered ephemeral port.
Multiple internal hosts share a single public IP address by keeping sessions distinct at Layer 4 using unique source port numbers.
3
Determine the specific Inside Global entry for the packet exiting the WAN interface.
The packet on the public network displays the source IP 203.0.113.1203.0.113.1 and a dynamically generated source port (e.g., 5214452144).
This unique combination allows the gateway to map return traffic back to the originating client (172.16.10.50172.16.10.50).

Anahtar Kavram

NAT Terminology and PAT Port Assignment Mechanics
Tahmini Süre:2m 0s
Soru 7Soru

A network technician needs to plan the installation of a new patch panel and power distribution unit (PDU) inside an existing equipment cabinet in an IDF. Which document should the technician consult to check the available vertical space and unit height positions inside the cabinet?

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Cevap: Rack diagram

Cevap

The technician should consult a rack diagram.
A rack diagram provides a visual elevation view of an equipment cabinet, detailing the exact rack unit (U) positions, device heights, and available mounting slots required for physical installation planning.

Adım Adım Çözüm

1
Analyze the technician's requirement.
The requirement is to determine physical unit spacing and hardware elevation inside a server/network cabinet.
Installing new physical hardware requires verified vertical space (rack unit counts) inside the rack.
2
Select the corresponding network documentation type.
A rack diagram visualizes 19-inch equipment enclosures, rack unit (U) positions, and existing hardware placement.
Rack diagrams specifically capture physical height allocations and spatial layout within equipment racks.

Anahtar Kavram

Rack Diagrams and Physical Space Planning
Tahmini Süre:45s
Soru 8Soru

During a routine internal security audit, a network analyst discovers that administrators are using Telnet to remotely configure network infrastructure devices, exposing authentication credentials in unencrypted cleartext across internal subnets. Security policy requires immediately replacing this practice with Secure Shell (SSH) to secure all interactive administrative sessions. Which default port and transport layer protocol combination must be permitted on internal firewalls to allow SSH traffic?

Cevabı ve açıklamayı göster

Cevap: TCP port 22

Cevap

TCP port 22
Secure Shell (SSH) operates over TCP port 22 by default. SSH provides encrypted terminal management sessions and relies on TCP's connection-oriented delivery to maintain session integrity.

Adım Adım Çözüm

1
Identify the target secure administration protocol.
The scenario requires implementing Secure Shell (SSH) to replace unencrypted Telnet sessions.
SSH encrypts management traffic, preserving confidentiality for administrative credentials and commands.
2
Determine the transport layer protocol required by SSH.
SSH requires Transmission Control Protocol (TCP) to ensure reliable, connection-oriented data delivery.
Interactive shell sessions depend on ordered delivery and error recovery provided by TCP.
3
Identify the standard IANA assigned port for SSH.
SSH operates on default port 22.
Standard network firewall rules reference IANA default port allocations for service filtering.

Anahtar Kavram

Common Ports and Protocols - SSH (TCP 22) vs Telnet (TCP 23)
Soru 9Soru

A network administrator is documenting a host configuration and needs to compress the IPv6 address fe80:0000:0000:0000:0202:b3ff:fe1e:8329 using standard RFC 5952 rules. Which of the following is the correctly compressed IPv6 address?

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Cevap: fe80::202:b3ff:fe1e:8329

Cevap

The correctly compressed IPv6 address is fe80::202:b3ff:fe1e:8329.
The notation fe80::202:b3ff:fe1e:8329 follows RFC 5952 standards by replacing three contiguous zero hextets with a single double colon (::) and dropping the leading zero in the fifth hextet (0202 to 202).

Adım Adım Çözüm

1
Identify contiguous blocks of 16-bit zero fields (hextets).
The block 0000:0000:0000 contains three consecutive zero hextets.
RFC 5952 mandates shortening the longest run of consecutive zero hextets using double-colon (::) notation.
2
Suppress all leading zeros in each 16-bit field.
The hextet 0202 becomes 202, while 0000 hextets reduce to single zeros before double-colon compression.
Leading zeros in an IPv6 hextet carry no numerical value and must be omitted.
3
Replace the single longest contiguous run of zero fields with a single double colon (::).
fe80:0000:0000:0000:0202:b3ff:fe1e:8329 compresses to fe80::202:b3ff:fe1e:8329.
Using double colon exactly once creates a unique, fully deterministic 128-bit IPv6 representation.

Anahtar Kavram

IPv6 Address Compression Rules (RFC 5952)
Tahmini Süre:45s
Soru 10Soru

A network administrator is performing Variable Length Subnet Masking (VLSM) on an enterprise network segment. A router interface is currently assigned the IP address 10.140.77.218/2110.140.77.218/21. Upon inspecting the existing IP allocations, the administrator determines that subnets 10.140.72.0/2310.140.72.0/23 and 10.140.74.0/2310.140.74.0/23 are already assigned to production VLANs. What is the maximum number of additional /26/26 subnets that can be created using only the remaining unallocated address space within this parent /21/21 block?

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Cevap: 16

Cevap

16 additional /26 subnets can be created.
The host IP address 10.140.77.218/2110.140.77.218/21 resides in the network block 10.140.72.0/2110.140.72.0/21, spanning 10.140.72.010.140.72.0 through 10.140.79.25510.140.79.255. The two allocated /23/23 subnets consume IPs 10.140.72.010.140.72.0 through 10.140.75.25510.140.75.255. This leaves 10.140.76.010.140.76.0 through 10.140.79.25510.140.79.255 unallocated, which equals a contiguous /22/22 block (10.140.76.0/2210.140.76.0/22). Dividing a /22/22 space into /26/26 subnets yields 2(2622)=24=162^{(26-22)} = 2^4 = 16 subnets.

Adım Adım Çözüm

1
Determine the network boundary of the parent /21 subnet.
The host IP 10.140.77.218/2110.140.77.218/21 belongs to the network 10.140.72.0/2110.140.72.0/21, covering IP range 10.140.72.010.140.72.0 to 10.140.79.25510.140.79.255.
A /21 mask uses a third-octet block size of 8 (256248=8256 - 248 = 8). Since 77÷8=977 \div 8 = 9 remainder 55, the network start is 9×8=729 \times 8 = 72.
2
Calculate the address space consumed by existing allocations.
Subnet 10.140.72.0/2310.140.72.0/23 consumes 10.140.72.010.140.72.0 - 10.140.73.25510.140.73.255, and subnet 10.140.74.0/2310.140.74.0/23 consumes 10.140.74.010.140.74.0 - 10.140.75.25510.140.75.255.
Each /23 subnet covers 2 third-octet values (22423=22^{24-23} = 2).
3
Identify the unallocated address space remaining in the /21 parent block.
The remaining contiguous range is 10.140.76.010.140.76.0 through 10.140.79.25510.140.79.255, which forms a single 10.140.76.0/2210.140.76.0/22 subnet.
The unallocated space spans 4 third-octet blocks (76,77,78,7976, 77, 78, 79), equivalent to a /22 prefix (256252=4256 - 252 = 4).
4
Calculate how many /26 subnets fit into the remaining /22 block.
Number of /26 subnets = 2(2622)=24=162^{(26 - 22)} = 2^4 = 16.
Subnet division is determined by exponentiation of the bit prefix difference between the parent block prefix and the target subnet prefix.

Anahtar Kavram

VLSM Subnet Allocation and CIDR Block Subdivisions
Soru 11Soru

A network administrator is provisioning a dedicated IPv4 subnet for a new VoIP deployment from an available 192.168.10.0/24192.168.10.0/24 network block. The subnet must accommodate 6060 IP phones, 22 router interfaces operating VRRP for default gateway redundancy, and 11 switch management interface. Which of the following CIDR prefixes represents the smallest subnet that fulfills all required host assignments while preserving the maximum amount of address space for future allocation?

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Cevap: /25/25

Cevap

The smallest subnet prefix that satisfies the host requirements while maximizing remaining address space is /25/25.
The scenario requires 6363 usable host IP addresses (6060 IP phones + 22 VRRP router addresses + 11 switch management address). The number of usable hosts in an IPv4 subnet is determined by 2h22^h - 2, where hh is the number of host bits. A /26/26 subnet has 66 host bits, yielding 262=622^6 - 2 = 62 usable addresses, which is 1 address short of the required total. A /25/25 subnet has 77 host bits, providing 272=1262^7 - 2 = 126 usable host IP addresses. Therefore, /25/25 is the smallest subnet mask (longest CIDR prefix) that meets the requirement while conserving IP space.

Adım Adım Çözüm

1
Calculate the total number of required usable host IP addresses.
Total usable host IPs required = 60 (phones)+2 (VRRP interfaces)+1 (switch management)=6360 \text{ (phones)} + 2 \text{ (VRRP interfaces)} + 1 \text{ (switch management)} = 63 addresses.
Every active host device and redundant gateway interface requires a distinct usable IP address within the subnet.
2
Apply the usable host capacity formula for IPv4 subnetting.
Usable host capacity = 2h22^h - 2, where h=32prefix lengthh = 32 - \text{prefix length}.
The network ID (all host bits 0) and broadcast address (all host bits 1) are reserved and cannot be assigned to hosts.
3
Evaluate candidate CIDR prefix lengths against the required 63 usable addresses.
For /26/26 (h=6h=6): 262=642=622^6 - 2 = 64 - 2 = 62 usable addresses (insufficient by 1 IP). For /25/25 (h=7h=7): 272=1282=1262^7 - 2 = 128 - 2 = 126 usable addresses (sufficient).
Selecting /26/26 by counting total addresses (6464) instead of usable addresses (6262) is a common subnetting calculation error.
4
Select the smallest subnet (longest prefix mask) that accommodates the minimum host requirement.
/25/25 is the smallest subnet mask that fulfills the 63 usable IP address requirement.
Although /24/24 also fits 63 hosts, it leaves 191 unused IP addresses in the subnet, wasting space compared to /25/25.

Anahtar Kavram

Subnet Usable Host Capacity Calculation (2h22^h - 2 Rule) and CIDR Prefix Selection
Tahmini Süre:2m 0s
Soru 12Soru

A network technician is deploying a new fleet of VoIP desktop phones across multiple subnets. To automate initial provisioning, each phone must automatically receive the IP address or hostname of its TFTP boot server from the central DHCP server upon starting up. Which DHCP option should the technician configure within the DHCP scope to supply this provisioning server information?

Cevabı ve açıklamayı göster

Cevap: Option 66

Cevap

Option 66 is used in DHCP scope configurations to specify the TFTP server hostname or IP address required by booting clients like VoIP phones and PXE-enabled computers.
DHCP Option 66 is explicitly designated in core networking standards to provide the TFTP server name or IP address to DHCP clients. This allows unconfigured network devices, such as VoIP phones or PXE boot workstations, to automatically contact their boot server and download operating configurations.

Adım Adım Çözüm

1
Identify the service requirement from the scenario.
The devices need automated TFTP provisioning server information delivered via DHCP.
VoIP phones require a boot server address during IP lease acquisition to download firmware and configuration files.
2
Map the requirement to standard DHCP option codes.
DHCP Option 66 is defined specifically for TFTP Server Name / IP Address.
Standardized DHCP options allow centralized addressing services to pass vendor-neutral and protocol-specific metadata to client endpoints.

Anahtar Kavram

DHCP Options for Automated IP Addressing Services
Tahmini Süre:1m 0s
Soru 13Soru

An enterprise border router maintains two paths to reach a remote corporate subnet (10.45.0.0/2410.45.0.0/24). Path 1 is learned via internal BGP (iBGP) with an Administrative Distance (AD) of 200, while Path 2 is learned via Open Shortest Path First (OSPF) with an AD of 110. Both routes report the exact same prefix length (/24/24). Network monitoring shows that traffic is actively taking Path 2 over the low-bandwidth OSPF backup link instead of the primary iBGP connection. Which configuration modification should the network administrator perform on the border router to ensure traffic prefers the primary iBGP link under normal operations?

Cevabı ve açıklamayı göster

Cevap: Decrease the Administrative Distance for the learned iBGP route to a value lower than 110.

Cevap

Decreasing the Administrative Distance of the iBGP route below 110 ensures the router prefers the primary iBGP link.
When a router receives multiple routes to the same network prefix with identical prefix lengths, it chooses the route with the lowest Administrative Distance (AD). Standard iBGP has a default AD of 200, whereas OSPF has a default AD of 110. Because 110 is lower than 200, the router prefers the OSPF path. Lowering the AD for the iBGP route below 110 ensures the router selects the iBGP path as primary while maintaining OSPF as the secondary backup path.

Adım Adım Çözüm

1
Evaluate the route selection hierarchy
Both candidate routes have identical prefix lengths (/24/24), so Longest Prefix Match is a tie.
When prefix lengths match, the router moves to the next evaluation criteria: Administrative Distance.
2
Compare Administrative Distance values
OSPF has an AD of 110, while iBGP has a default AD of 200.
Lower AD values indicate higher reliability/preference. The router currently installs the OSPF route (110 < 200).
3
Determine the necessary configuration change
Lower the iBGP AD below 110 (e.g., to 20 or 100).
This makes the iBGP route more believable to the routing table manager than the OSPF route.

Anahtar Kavram

Route Selection Hierarchy and Administrative Distance Preference
Soru 14Soru

A network technician needs to assign an IPv4 subnet mask to a network segment that must accommodate 2828 usable host devices. Which CIDR notation prefix fulfills this requirement while reserving the minimal amount of unused IP addresses?

Cevabı ve açıklamayı göster

Cevap: /27

Cevap

The /27 prefix is the smallest subnet mask that provides at least 28 usable host IP addresses.
A /27 prefix allocates 5 bits for host addressing (3227=532 - 27 = 5), resulting in 25=322^5 = 32 total IP addresses. Subtracting 22 for the network and broadcast addresses leaves 3030 usable IP addresses (322=3032 - 2 = 30). This is the smallest standard subnet mask that accommodates 2828 hosts.

Adım Adım Çözüm

1
Calculate the total required IP addresses by adding 2 reserved addresses (network ID and broadcast address) to the needed host count.
28 hosts+2 reserved=30 total addresses needed28\text{ hosts} + 2\text{ reserved} = 30\text{ total addresses needed}.
Every IPv4 subnet requires one network address and one broadcast address, which cannot be assigned to hosts.
2
Find the smallest block size (power of 2) that is greater than or equal to the total required addresses.
25=32 total addresses2^5 = 32\text{ total addresses}, requiring 55 host bits.
Subnet block sizes must be powers of two.
3
Subtract the host bits from 32 to determine the network prefix length.
325=/2732 - 5 = /27.
An IPv4 address contains 32 total bits.

Anahtar Kavram

IPv4 Host Capacity and Subnet Mask Selection
Tahmini Süre:45s
Soru 15Soru

A network host connected to interface FastEthernet 0/120/12 on a Layer 2 switch transmits an Address Resolution Protocol (ARP) request broadcast frame. Interface FastEthernet 0/120/12 is configured as an access port belonging to VLAN 5050. Which of the following actions will the switch perform upon receiving this frame?

Cevabı ve açıklamayı göster

Cevap: The switch forwards the frame out all operational access ports assigned to VLAN 5050, excluding interface FastEthernet 0/120/12.

Cevap

The switch forwards the broadcast frame out all active access ports configured for VLAN 50, excluding the receiving port FastEthernet 0/12.
A Virtual Local Area Network (VLAN) creates a logical Layer 2 broadcast domain. When a broadcast frame (such as an ARP request) is received on an access port, the switch floods that frame to every active port belonging to the same VLAN, while excluding the port on which the frame originally arrived to prevent loops.

Adım Adım Çözüm

1
Identify the frame type and ingress port configuration.
The frame is an ARP request broadcast (destination MAC ff:ff:ff:ff:ff:ff) arriving on access port FastEthernet 0/12 assigned to VLAN 50.
Understanding the PDU type and port membership determines how the switch handles forwarding.
2
Determine the Layer 2 broadcast domain boundary.
VLAN 50 forms a logical Layer 2 broadcast domain isolated from all other VLANs on the switch.
Broadcast frames are contained strictly within the originating VLAN.
3
Apply Layer 2 switch flooding behavior rules.
The switch floods the broadcast frame to all active interfaces assigned to VLAN 50, except the ingress port FastEthernet 0/12, without applying 802.1Q tags to access port output.
Standard switch logic dictates flooding broadcast traffic within the VLAN while preventing loopback to the source port.

Anahtar Kavram

VLAN Broadcast Domain Boundaries and Layer 2 Flooding Logic
Tahmini Süre:1m 15s
Soru 16Soru

A network engineer is implementing DHCP Relay Agent Information Option (Option 82) on Layer 2 access switches within an enterprise building. When a client sends a broadcast DHCPDISCOVERDHCPDISCOVER message, the relay agent appends sub-options to the request before forwarding it to a centralized DHCP server across a routed core. Which sub-option field specifically carries identifying information about the physical switch port and VLAN from which the client request originated?

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Cevap: Circuit ID (Sub-option 1)

Cevap

Circuit ID (Sub-option 1) is added by the DHCP relay agent to specify the client's physical port and VLAN information.
DHCP Option 82 allows a relay agent to attach location-specific metadata to client DHCP broadcasts. Sub-option 1 (Circuit ID) encodes the specific physical interface, module, and VLAN tag where the client broadcast was received.

Adım Adım Çözüm

1
Analyze the role of DHCP Option 82 (Relay Agent Information Option)
Option 82 allows relay agents to insert network topology details into client DHCP requests before forwarding them to the server.
This information enables the DHCP server to enforce security policies and assign IP addresses or parameters based on physical location.
2
Differentiate between Option 82 standard sub-options
Sub-option 1 (Circuit ID) identifies the incoming port/VLAN circuit, whereas Sub-option 2 (Remote ID) identifies the relay agent switch/router identity.
Circuit ID contains specific access port details (e.g., interface index, VLAN tag) necessary to identify where the host connects.

Anahtar Kavram

DHCP Option 82 and Relay Agent Information Sub-options
Soru 17Soru

A network administrator is configuring perimeter security rules to allow an internal Remote Authentication Dial-In User Service (RADIUS) server to authenticate wireless users against an external partner system. Which transport protocol and destination port pair must be permitted through the firewall for standard RADIUS authentication traffic?

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Cevap: UDP port 1812

Cevap

Standard RADIUS authentication traffic operates over UDP port 1812.
UDP port 1812 is the standard IANA-assigned port for RADIUS authentication and authorization traffic. RADIUS uses UDP for lightweight connectionless transmission.

Adım Adım Çözüm

1
Identify the protocol and required function in the scenario
The scenario requires RADIUS performing authentication.
RADIUS separates authentication/authorization (port 1812) from accounting functions (port 1813).
2
Determine the transport protocol used by RADIUS
RADIUS uses UDP rather than TCP at the Transport Layer.
RADIUS relies on connectionless UDP transport and handles retransmissions at the application layer.
3
Select the correct port and transport protocol pairing
UDP port 1812 is the official standard IANA port for RADIUS authentication.
UDP 1812 ensures RADIUS authentication requests reach the RADIUS daemon properly.

Anahtar Kavram

RADIUS Port Numbers and Protocols
Tahmini Süre:1m 0s
Soru 18Soru

A network engineer is configuring a core router that learns a path to the 10.50.0.0/1610.50.0.0/16 network through multiple routing sources simultaneously: an internal OSPF advertisement, a RIPv2 advertisement, and a manually configured static route. All sources advertise the exact same prefix length. Which path will the router install into its active routing table, and why?

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Cevap: The static route, because static routes have a lower Administrative Distance than OSPF and RIPv2.

Cevap

The static route is installed into the routing table because it has the lowest Administrative Distance (AD = 1) compared to OSPF (AD = 110) and RIPv2 (AD = 120) for the exact same destination prefix.
When a router receives multiple routes to the exact same destination network prefix from different routing sources, it uses Administrative Distance (AD) to select the best route. Default AD values are: Static route = 1, OSPF = 110, and RIPv2 = 120. Because lower AD indicates higher trustworthiness, the static route is selected.

Adım Adım Çözüm

1
Identify the advertised destination prefix lengths across all sources.
All three routes offer identical prefix lengths (10.50.0.0/1610.50.0.0/16), so Longest Prefix Match (LPM) ties.
When prefix lengths match, the router evaluates Administrative Distance (AD) to determine trustworthiness.
2
Compare the Administrative Distance values of the competing routing sources.
Static Route AD = 1, OSPF AD = 110, RIPv2 AD = 120.
Lower Administrative Distance values indicate higher administrative preference and trustworthiness.
3
Select the winning route to install in the IP routing table.
The static route with AD = 1 is selected.
The router selects the path with the lowest Administrative Distance.

Anahtar Kavram

Administrative Distance Hierarchy and Route Selection
Soru 19Soru

During a facility renovation, an IT team installs new workstations across multiple offices. A network technician needs to record the detailed mapping between individual wall outlet jacks, patch panel ports, and physical cable run paths back to the Intermediate Distribution Frame (IDF). Which type of documentation should the technician update to preserve this physical cabling location data?

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Cevap: Cable run schedule

Cevap

The cable run schedule is the appropriate document for recording physical wall jack locations, cabling paths, and patch panel port assignments.
A cable run schedule (or wiring documentation) provides an itemized index of physical network drops. It documents cable tag numbers, originating wall outlets, floor locations, and the corresponding patch panel and switch port assignments in an IDF or MDF cabinet.

Adım Adım Çözüm

1
Analyze the documentation requirements described in the scenario
The scenario requires tracking physical copper drops, wall jack numbers, patch panel ports, and cable routes between offices and the IDF.
Physical infrastructure details belong to Layer 1 documentation rather than logical configuration maps.
2
Compare network documentation types against the physical requirements
A cable run schedule provides a comprehensive matrix of cable IDs, floor locations, port terminations, and jack IDs.
It enables technicians to quickly trace physical connections during maintenance or troubleshooting.

Anahtar Kavram

Cable Run Schedule and Physical Network Documentation
Tahmini Süre:1m 0s
Soru 20Soru

A network administrator runs the `nslookup` command to check the mail server configuration for an enterprise domain and obtains the following output:

text
> set type=MX
> company.com
Server: dns1.internal.net
Address: 10.0.0.53

Non-authoritative answer:
company.com mail exchanger = 10 mail.company.com

Which DNS record type must be configured to map `mail.company.com` directly to its corresponding IPv4 address?

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Cevap: A record

Cevap

The host requires an A record to resolve the mail server hostname to an IPv4 address.
An A record specifically maps a domain name or hostname to a 32-bit IPv4 address, which is required for client systems to initiate direct IP connectivity to `mail.company.com` over IPv4.

Adım Adım Çözüm

1
Analyze the MX record output
The MX lookup returns the hostname `mail.company.com` as the mail exchanger for `company.com`.
MX records specify destination mail servers by hostname, requiring a secondary lookup to find their IP address.
2
Identify the target address family
The requirement specifies resolving `mail.company.com` to an IPv4 address.
Standard IPv4 host resolution is performed using Host Address (A) records.

Anahtar Kavram

DNS Record Functions and Resolution
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