Examkin

Network Fundamentals

398 soru

Tüm türler398Çoktan seçmeli205Çoklu seçim92Sayısal18Eşleştirme68Sıralama7Kısa cevap8
Soru 301Soru →

A Cisco router interface with a burned-in MAC address of `d46d.5012.abcd` is configured with the command `ipv6 address 2001:db8:100:1::/64 eui-64`. What is the complete 128-bit IPv6 global unicast address assigned to this interface?

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Cevap: 2001:db8:100:1:d66d:50ff:fe12:abcd; 2001:DB8:100:1:D66D:50FF:FE12:ABCD; 2001:db8:100:1:d66d:50ff:fe12:abcd/64; 2001:0db8:0100:0001:d66d:50ff:fe12:abcd

Cevap

2001:db8:100:1:d66d:50ff:fe12:abcd
To derive the IPv6 global unicast address using EUI-64, the router takes the assigned `/64` prefix `2001:db8:100:1::/64` and appends a 64-bit interface identifier generated from the interface's MAC address (`d46d.5012.abcd`). The MAC address is split into `d46d:50` and `12:abcd`, `FFFE` is inserted into the middle (`d46d:50ff:fe12:abcd`), and the 7th bit of the first byte is inverted (`d4` = `1101 0100` becomes `d6` = `1101 0110`). The final address is `2001:db8:100:1:d66d:50ff:fe12:abcd`.

Adım Adım Çözüm

1
Split the 48-bit MAC address into two 24-bit halves.
Left half: `d4:6d:50`, Right half: `12:ab:cd`.
EUI-64 process requires dividing the MAC address to insert the 16-bit midpoint padding.
2
Insert the 16-bit hexadecimal value `FFFE` between the two MAC halves.
Resulting 64-bit sequence: `d46d:50ff:fe12:abcd`.
EUI-64 expands a 48-bit MAC address into a 64-bit interface identifier using `FFFE`.
3
Invert the 7th bit (Universal/Local bit) of the first byte (`d4`).
Byte `d4` in binary is `1101 0100`. Inverting the 7th bit yields `1101 0110`, which is `d6` in hexadecimal. Modified interface identifier: `d66d:50ff:fe12:abcd`.
IEEE EUI-64 standard requires flipping the U/L bit (7th bit from left) to indicate global uniqueness.
4
Combine the configured `/64` IPv6 prefix with the modified 64-bit EUI-64 interface identifier.
Complete IPv6 address: `2001:db8:100:1:d66d:50ff:fe12:abcd`.
Appending the derived 64-bit Interface ID to the `2001:db8:100:1::/64` network prefix completes the 128-bit global unicast address.

Anahtar Kavram

Derivation of IPv6 EUI-64 Interface Identifier from a 48-bit MAC address
Tahmini Süre:2m 0s
Soru 302Soru →

A network administrator is evaluating transport layer requirements for a enterprise deployment. The network deployment includes a batch database synchronization service requiring guaranteed in-order delivery and a real-time voice streaming application sensitive to delay. Which TWO features distinguish TCP from UDP in this operational environment? (Select TWO.)

Geçerli olan tümünü seçin

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Cevap: TCP utilizes dynamic windowing mechanisms to adjust data transmission flow based on receiver buffer capacity.; TCP maintains session state tracking using sequence and acknowledgment fields within a 20-byte baseline header.

Cevap

TCP is distinguished from UDP by its use of dynamic windowing for flow control and session state tracking via sequence/acknowledgment numbers in a 20-byte baseline header.
TCP is a connection-oriented, stateful transport protocol that uses sequence and acknowledgment numbers within a baseline 20-byte header to guarantee ordered delivery. Additionally, TCP employs sliding dynamic windowing to adjust traffic flow based on receiver buffer readiness. UDP, in contrast, uses a fixed 8-byte header and lacks connection handshake, sequence numbers, or flow control mechanisms.

Adım Adım Çözüm

1
Analyze transport protocol mechanics required for guaranteed data delivery vs. low-overhead streaming.
Identified TCP as connection-oriented/stateful and UDP as connectionless/stateless.
TCP maintains session state and flow control to guarantee delivery, whereas UDP minimizes header size and latency by avoiding session management.
2
Evaluate the functional features of TCP.
TCP uses a minimum 20-byte header with sequence numbers, acknowledgment numbers, and dynamic window size fields for flow control.
Dynamic windowing allows the receiver to specify how much data can be transmitted before an acknowledgment is required.
3
Evaluate the functional features of UDP.
UDP uses a simple 8-byte header consisting of source port, destination port, length, and checksum.
UDP lacks sequence fields, error recovery retransmissions, and handshake mechanics, giving it low overhead suitable for real-time applications.

Anahtar Kavram

TCP vs UDP Transport Layer Features
Soru 303Soru →

A network administrator is reviewing interface addressing rules and EUI-64 auto-configuration for an enterprise IPv6 rollout. Which two statements regarding IPv6 addressing characteristics and EUI-64 generation are correct? (Choose two.)

Geçerli olan tümünü seçin

Cevabı ve açıklamayı göster

Cevap: The 7th bit of the MAC address's first octet (the Universal/Local bit) is inverted when constructing the 64-bit interface identifier.; The 16-bit hex value `FFFE` is inserted directly between the third and fourth octets (the middle) of the 48-bit MAC address.

Cevap

The two correct statements are: (1) The 7th bit of the MAC address's first octet (the Universal/Local bit) is inverted when constructing the 64-bit interface identifier, and (2) The 16-bit hex value `FFFE` is inserted directly between the third and fourth octets (the middle) of the 48-bit MAC address.
The EUI-64 process takes a 48-bit MAC address, splits it into two 24-bit sections, inserts `FFFE` in the middle between the third and fourth octets, and flips the 7th bit (Universal/Local bit) of the first byte. Thus, both statements describing these EUI-64 steps are correct.

Adım Adım Çözüm

1
Analyze the EUI-64 expansion process for a 48-bit MAC address.
Confirm that the 48-bit MAC address is split into two 24-bit halves, inserting `FFFE` in the middle (between octet 3 and octet 4).
EUI-64 requires creating a 64-bit interface ID from a 48-bit MAC address by inserting `FFFE` into the center.
2
Analyze the bit inversion rule for EUI-64 generation.
Confirm that the 7th bit (Universal/Local bit) of the first byte of the MAC address is flipped from 0 to 1 (or 1 to 0).
RFC 4291 mandates modifying the 7th bit to reflect whether the address is globally unique.
3
Evaluate the IPv6 host reservation mechanics versus IPv4.
Recognize that IPv6 does not utilize broadcast addresses and does not subtract 2 addresses for broadcast and network IDs like IPv4.
Multicast replaces broadcast in IPv6, so standard IPv4 subnet host subtraction rules do not apply.

Anahtar Kavram

Modified EUI-64 Interface Identifier Generation and IPv6 Address Structure
Soru 304Soru →

A network engineer is configuring a virtualized data center host to support multiple enterprise server workloads. The installation utilizes a bare-metal (Type 1) hypervisor to manage virtual machine (VM) compute and virtual switch networking features. Which statement accurately describes the operational role and architecture of a Type 1 hypervisor in this deployment?

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Cevap: It runs directly on the bare-metal physical hardware, managing hardware resources and virtual switch packet forwarding without an underlying host operating system.

Cevap

A Type 1 hypervisor runs directly on the bare-metal physical hardware, managing hardware resources and virtual switch packet forwarding without an underlying host operating system.
A Type 1 hypervisor (such as Cisco UCS ESXi or KVM) installs directly onto the bare-metal hardware of a host machine. It provides direct hardware control, lower latency, and embedded software vSwitching capabilities to switch network frames between guest virtual machines and uplink to physical network infrastructure.

Adım Adım Çözüm

1
Identify hypervisor architectural types
Recognize that Type 1 hypervisors are bare-metal systems installed directly on host server hardware, whereas Type 2 hypervisors run inside a conventional host operating system.
Understanding the installation layer determines resource efficiency and management responsibilities.
2
Evaluate internal networking and component roles
A Type 1 hypervisor abstracts physical Network Interface Cards (NICs) and creates software virtual switches (vSwitches) to forward frames directly between co-located virtual machines.
Virtual switches handle local Layer 2 communication directly within host memory.

Anahtar Kavram

Hypervisor Types and Virtualization Architecture
Soru 305Soru →

A network administrator assigns an IP address of 192.168.50.77/27192.168.50.77/27 to a server interface. What is the broadcast address for the subnet to which this server belongs?

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Cevap: 192.168.50.95192.168.50.95

Cevap

The broadcast address for the subnet containing host 192.168.50.77/27192.168.50.77/27 is 192.168.50.95192.168.50.95.
A /27/27 subnet prefix leaves 55 host bits, yielding a subnet block size of 25=322^5 = 32 addresses. Subnets increment by 3232 in the fourth octet (192.168.50.0192.168.50.0, 192.168.50.32192.168.50.32, 192.168.50.64192.168.50.64, 192.168.50.96192.168.50.96). The host address 192.168.50.77192.168.50.77 lies between 192.168.50.64192.168.50.64 and 192.168.50.95192.168.50.95. The final address in this range, 192.168.50.95192.168.50.95, is the broadcast address.

Adım Adım Çözüm

1
Determine the subnet block size from the prefix length.
A /27/27 prefix leaves 32−27=532 - 27 = 5 host bits. The subnet block size (increment) is 25=322^5 = 32.
Calculating the block size identifies the boundary increments in the fourth octet.
2
Identify the subnet network boundaries in the fourth octet.
Subnet network addresses increment by 3232: .0,.32,.64,.96,.128.0, .32, .64, .96, .128, and so forth.
Listing the multiples of 3232 establishes exact subnet boundary ranges.
3
Locate the specific subnet range containing 192.168.50.77192.168.50.77.
Since 7777 falls between 6464 and 9595, the network address is 192.168.50.64192.168.50.64 and the broadcast address is 192.168.50.95192.168.50.95.
The last address in a subnet block range is reserved as the broadcast address.

Anahtar Kavram

IPv4 Subnet Calculation and Broadcast Address Identification
Tahmini Süre:45s
Soru 306Soru →

A data center engineering team is selecting virtualization software to host enterprise services on physical hardware. The team chooses a bare-metal (Type 1) hypervisor architecture rather than a hosted (Type 2) hypervisor architecture. Which statement correctly identifies a key operational characteristic of the selected hypervisor type?

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Cevap: It runs directly on the physical hardware of the server, managing hardware resources without relying on an underlying host operating system.

Cevap

It runs directly on the physical hardware of the server, managing hardware resources without relying on an underlying host operating system.
A Type 1 (bare-metal) hypervisor is installed directly on the physical server hardware. It manages hardware resources natively and provides higher performance, efficiency, and security for enterprise virtualization environments because no intermediate host operating system is present.

Adım Adım Çözüm

1
Analyze the core architectural definition of a Type 1 (bare-metal) hypervisor.
Identify that Type 1 hypervisors install directly onto host hardware without requiring a separate host OS.
Direct hardware control maximizes performance, scalability, and security for enterprise workloads.
2
Compare Type 1 characteristics against Type 2 hypervisors.
Recognize that hosted hypervisors (Type 2) run on top of an OS, whereas Type 1 hypervisors act as the primary operating layer directly managing physical CPU, memory, and networking.
Differentiating these architecture types is essential for CCNA virtualization fundamentals.

Anahtar Kavram

Bare-Metal (Type 1) vs. Hosted (Type 2) Hypervisors
Tahmini Süre:1m 0s
Soru 307Soru →

A network technician is auditing host IP configurations across an enterprise network. Which of the following IPv4 addresses belong to RFC 1918 private address space? (Select TWO.)

Geçerli olan tümünü seçin

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Cevap: 10.50.100.1; 172.20.15.5

Cevap

The addresses 10.50.100.1 and 172.20.15.5 are private IPv4 addresses designated by RFC 1918.
The addresses 10.50.100.1 and 172.20.15.5 are correct because they reside inside RFC 1918 private IPv4 spaces: 10.0.0.0/8 (10.0.0.0 to 10.255.255.255) and 172.16.0.0/12 (172.16.0.0 to 172.31.255.255), respectively.

Adım Adım Çözüm

1
Recall the defined RFC 1918 private IPv4 address blocks
The three RFC 1918 private blocks are: 10.0.0.0/810.0.0.0/8 (10.0.0.010.0.0.0 – 10.255.255.25510.255.255.255), 172.16.0.0/12172.16.0.0/12 (172.16.0.0172.16.0.0 – 172.31.255.255172.31.255.255), and 192.168.0.0/16192.168.0.0/16 (192.168.0.0192.168.0.0 – 192.168.255.255192.168.255.255).
RFC 1918 reserves these specific ranges for unrouted, internal IP addressing.
2
Evaluate each choice against the reserved ranges
The address 10.50.100.110.50.100.1 falls inside 10.0.0.0/810.0.0.0/8 and 172.20.15.5172.20.15.5 falls inside 172.16.0.0/12172.16.0.0/12. However, 172.32.10.1172.32.10.1 exceeds the upper limit of 172.31.255.255172.31.255.255, and 192.169.1.10192.169.1.10 falls outside 192.168.0.0/16192.168.0.0/16.
Addresses outside the three designated blocks are globally routable public addresses.

Anahtar Kavram

RFC 1918 Private IPv4 Address Allocation
Soru 308Soru →

Match each network topology architecture to its corresponding operational characteristic and traffic flow optimization profile.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Spine-Leaf (Clos) Fabric
Traditional 3-Tier Campus
Collapsed Core (2-Tier)
WAN Hub-and-Spoke

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

Spine-Leaf (Clos) Fabric matches with East-West low-latency traffic using ECMP; Traditional 3-Tier Campus matches with separate functional tiers for structured North-South flow; Collapsed Core matches with combining core and distribution functions into a single switch pair; WAN Hub-and-Spoke matches with centralizing inspection and routing remote branch traffic through a central hub site.
Each topology architecture addresses specific scalability, redundancy, and traffic flow requirements: Spine-Leaf optimizes East-West data center flows using ECMP over a two-tier bipartite mesh; Traditional 3-Tier structures campus environments into distinct access, distribution, and core roles for policy control; Collapsed Core merges core and distribution layers for smaller networks to lower cost; WAN Hub-and-Spoke centralizes security and reduces link costs by routing branch-to-branch traffic through a central hub.

Adım Adım Çözüm

1
Analyze the structural traffic requirements of modern data center versus campus environments.
Identified that Spine-Leaf designs specifically target East-West (server-to-server) traffic via non-blocking ECMP routing across spine switches.
Data center workloads require low, deterministic latency across leaf switches regardless of physical placement.
2
Evaluate the functional division of campus topologies.
Identified that 3-Tier Campus separates core, distribution, and access, placing policy enforcement at the distribution layer, whereas Collapsed Core merges core and distribution into a single switch pair for smaller footprints.
Merging layers reduces cost in smaller networks, while dedicated tiers scale policy enforcement in larger enterprise networks.
3
Examine wide area network (WAN) deployment topologies.
Identified Hub-and-Spoke as the topology that centralizes policy enforcement and minimizes WAN link costs by routing all branch communications through a central hub.
Full mesh topologies require far more connections, making hub-and-spoke cost-effective when central security inspection is required.

Anahtar Kavram

Enterprise Network Topology Architectures and Traffic Flow Characteristics
Tahmini Süre:1m 30s
Soru 309Soru →

A network administrator is assigning an IP address to a router interface using a CIDR prefix length of /26/26. Which dotted-decimal subnet mask corresponds to this prefix?

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Cevap: 255.255.255.192

Cevap

255.255.255.192
A CIDR prefix of /26 indicates that 26 bits are set to 1 in the subnet mask. The first 24 bits populate the first three octets fully as 255.255.255. The remaining 2 network bits in the fourth octet correspond to binary 11000000, which equals 128+64=192128 + 64 = 192, producing the dotted-decimal subnet mask 255.255.255.192.

Adım Adım Çözüm

1
Determine the bit allocation across octets for a /26 prefix.
The first three octets use 24 bits (8 + 8 + 8 = 24), leaving 2 network bits for the fourth octet.
An IPv4 subnet mask has 32 total bits.
2
Convert the 2 network bits in the fourth octet to decimal.
Binary 11000000 equals 128+64=192128 + 64 = 192.
Subnet mask bits are filled from left to right.
3
Combine all four octets in dotted-decimal format.
255.255.255.192
The first three octets consist of 8 set bits each, yielding 255.255.255.

Anahtar Kavram

Converting CIDR notation to dotted-decimal subnet masks
Tahmini Süre:45s
Soru 310Soru →

An organization is deploying an in-house real-time voice communications protocol across their WAN links. The network team recommends using User Datagram Protocol (UDP) instead of Transmission Control Protocol (TCP) for the media transport path. Which fundamental Layer 4 header property supports this recommendation by minimizing protocol overhead for delay-sensitive traffic?

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Cevap: UDP utilizes a fixed header size of 8 bytes, whereas TCP requires a minimum header size of 20 bytes.

Cevap

UDP utilizes a fixed header size of 8 bytes, whereas TCP requires a minimum header size of 20 bytes.
The correct answer highlights that UDP maintains a fixed header size of only 8 bytes (comprising Source Port, Destination Port, Length, and Checksum). In contrast, a TCP header has a minimum size of 20 bytes and can expand up to 60 bytes with options. For small real-time voice payloads sent at high packet rates, reducing Layer 4 header overhead from 20 bytes to 8 bytes significantly decreases network bandwidth consumption and latency.

Adım Adım Çözüm

1
Analyze transport layer requirements for real-time voice traffic.
Real-time voice requires minimal delay and low protocol overhead; missing payload data is preferred over delayed retransmissions.
Late audio packets are discarded by voice codecs anyway, making TCP reliability mechanics counterproductive.
2
Compare Layer 4 header overhead between UDP and TCP.
UDP features a lean 8-byte fixed header (Source Port, Destination Port, Length, Checksum), while TCP has a 20-byte base header (plus options up to 60 bytes).
The smaller header footprint reduces bandwidth consumption and processing overhead per packet.

Anahtar Kavram

TCP vs UDP Header Overhead and Protocol Mechanics
Soru 311Soru →

A network engineer is provisioning subnets from the allocated summary block 172.25.192.0/20172.25.192.0/20 to expand an enterprise corporate network. The design specification requires Subnet A to accommodate at least 1,000 usable host endpoints and to be allocated starting from the very first available boundary of the 172.25.192.0/20172.25.192.0/20 block. Additionally, all assigned addresses must strictly comply with RFC 1918 private IPv4 space. What is the last usable IPv4 host address on Subnet A?

Cevabı ve açıklamayı göster

Cevap: 172.25.195.254

Cevap

172.25.195.254
To support at least 1,000 usable hosts, a subnet requires 10 host bits because 210−2=1,0222^{10} - 2 = 1,022 usable addresses (29−2=5102^9 - 2 = 510 is insufficient). Subtracting 10 host bits from 32 total bits yields a /22/22 subnet mask (255.255.252.0255.255.252.0). Allocating a /22/22 subnet from the start of the block 172.25.192.0/20172.25.192.0/20 results in a subnet spanning 172.25.192.0172.25.192.0 to 172.25.195.255172.25.195.255. The last usable IPv4 address is one less than the broadcast address, which is 172.25.195.254172.25.195.254. Furthermore, 172.25.192.0/20172.25.192.0/20 is fully within the RFC 1918 private IP range (172.16.0.0/12172.16.0.0/12).

Adım Adım Çözüm

1
Determine required host bits (hh) and prefix length
2h−2≥1,000  ⟹  210−2=1,022≥1,0002^h - 2 \ge 1,000 \implies 2^{10} - 2 = 1,022 \ge 1,000, so h=10h = 10 host bits. The prefix length is 32−10=/2232 - 10 = /22.
Subnet sizing requires finding the smallest power of 2 minus 2 for network and broadcast overhead that satisfies the host capacity requirement.
2
Verify RFC 1918 private addressing compliance
The block 172.25.192.0/20172.25.192.0/20 resides within the Class B private address space of 172.16.0.0/12172.16.0.0/12 (172.16.0.0172.16.0.0 to 172.31.255.255172.31.255.255).
RFC 1918 mandates that enterprise private networks use designated non-routable public IP blocks.
3
Calculate the subnet boundary and broadcast address
A /22/22 prefix has a block size of 44 in the third octet (210−8=42^{10-8} = 4). Starting at 172.25.192.0172.25.192.0, the network spans 172.25.192.0172.25.192.0 to 172.25.195.255172.25.195.255. The broadcast address is 172.25.195.255172.25.195.255.
Adding the block size minus 1 to the starting third octet gives the highest octet boundary for the subnet.
4
Identify the last usable host address
The last usable IP address is one position below the broadcast address: 172.25.195.254172.25.195.254.
The usable host range is bounded between the network ID (+1+1) and the broadcast ID (−1-1).

Anahtar Kavram

VLSM host calculation and RFC 1918 private IPv4 address boundaries
Soru 312Soru →

A network administrator inspects a Cisco router interface configured with dynamic IPv6 address generation and observes the assigned global unicast address `2001:db8:a1b2:c3d4:0211:22ff:fe33:4455/64`. Based on IEEE EUI-64 specification rules, which MAC address was assigned to the physical interface?

Cevabı ve açıklamayı göster

Cevap: 0011.2233.4455

Cevap

The MAC address assigned to the physical interface is 0011.2233.4455.
To derive the original MAC address from a Modified EUI-64 IPv6 interface ID (`0211:22ff:fe33:4455`), remove the inserted `FFFE` hex value from the center to yield six bytes (`02:11:22:33:44:55`) and invert the 7th bit of the first byte (`02` hex = `0000 0010` binary -> inverted to `0000 0000` binary = `00` hex). This reveals the original MAC address `0011.2233.4455`.

Adım Adım Çözüm

1
Identify the 64-bit Interface Identifier (Interface ID) from the IPv6 address.
The Interface ID is `0211:22ff:fe33:4455`.
In a /64 prefix, the last 64 bits represent the host portion (Interface ID).
2
Remove the 16-bit `ff:fe` sequence inserted into the middle of the MAC address during EUI-64 generation.
Removing `ff:fe` from `0211:22ff:fe33:4455` leaves 6 bytes: `02-11-22-33-44-55`.
EUI-64 creation inserts `FFFE` between the OUI (first 24 bits) and NIC specific bits (last 24 bits).
3
Invert the 7th bit (Universal/Local bit) of the first byte.
The first byte `02` in binary is `0000 0010`. Inverting the 7th bit yields `0000 0000`, which is `00` in hex. Combining all bytes yields `0011.2233.4455`.
The 7th bit is flipped from 0 to 1 when constructing Modified EUI-64, so it must be flipped back from 1 to 0 to restore the original MAC address.

Anahtar Kavram

Modified EUI-64 MAC-to-IPv6 Address Derivation
Soru 313Soru →

A network administrator is configuring virtual networking on a bare-metal hypervisor host equipped with redundant physical network interface cards (pNICs). The host accommodates multiple virtual machines (VMs) belonging to different department networks. Which two statements accurately describe the forwarding and operational characteristics of a virtual switch (vSwitch) in this environment? (Select two.)

Geçerli olan tümünü seçin

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Cevap: Intra-host traffic between virtual machines on the same VLAN is switched internally by the vSwitch without egressing to the physical network interface.; Port groups on the vSwitch allow administrators to apply network policies and VLAN tags to isolate traffic from distinct groups of virtual interfaces.

Cevap

The correct statements are that intra-host traffic between virtual machines on the same VLAN is switched internally by the vSwitch without egressing to the physical network interface, and port groups on the vSwitch allow administrators to apply network policies and VLAN tags to isolate traffic from distinct groups of virtual interfaces.
Virtual switches (vSwitches) operate inside the hypervisor to provide Layer 2 connectivity for virtual machines. When two VMs on the same physical host communicate within the same VLAN, the vSwitch forwards traffic locally in system memory, avoiding unnecessary transmission over physical NICs. Additionally, port groups on a vSwitch act as logical switchport configurations, allowing network administrators to define VLAN tags, traffic shaping, and security policies for connected vNICs.

Adım Adım Çözüm

1
Analyze how virtual switches handle local VM-to-VM communication on the same physical host.
Recognize that when two VMs are on the same host and VLAN, the vSwitch forwards frames directly between virtual ports inside memory without sending frames out through the physical network interface cards (pNICs).
This reduces physical link utilization and optimizes intra-host throughput.
2
Evaluate how logical segment isolation and policy enforcement are managed on a virtual switch.
Identify that vSwitches use port groups to group virtual machine network interfaces (vNICs) and assign configuration settings such as 802.1Q VLAN IDs and security settings.
Port groups act as logical access ports or trunks to enforce traffic isolation per tenant or department.
3
Evaluate incorrect assertions regarding Type 1 hypervisor architecture and Spanning Tree Protocol.
Dismiss the ideas that a Type 1 hypervisor relies on a standard host operating system network stack or that vSwitches participate in Spanning Tree Protocol (STP) BPDU exchanges.
Type 1 hypervisors run directly on hardware, and standard vSwitches do not run STP or form loops because they do not forward frames between physical uplinks.

Anahtar Kavram

Virtual Switch Architecture and Traffic Forwarding
Tahmini Süre:1m 30s
Soru 314Soru →

A network administrator provisions a small subnetwork for a server cluster using a subnet mask with a prefix length of /29/29. How many usable host IPv4 addresses are available on this subnet?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The subnet provides 6 usable host IPv4 addresses.
With a prefix length of /29/29, there are 32−29=332 - 29 = 3 host bits available. The formula to calculate usable host addresses is 2n−22^n - 2, where nn is the host bit count. Calculating 23−2=8−2=62^3 - 2 = 8 - 2 = 6 yields 6 usable host addresses.

Adım Adım Çözüm

1
Determine host bits from CIDR prefix
3 host bits
An IPv4 address consists of 32 bits, so 32−29=332 - 29 = 3 bits remain for host identification.
2
Calculate total IP address space
8 total addresses
Total address space equals 2n2^n where nn is the host bit count (23=82^3 = 8).
3
Subtract reserved addresses
6 usable host addresses
The network address (all host bits 0) and broadcast address (all host bits 1) cannot be assigned to end devices.

Anahtar Kavram

IPv4 Usable Host Calculation
Tahmini Süre:45s
Soru 315Soru →

A network administrator is evaluating Layer 4 protocol mechanisms for application deployment across an enterprise network. Which of the following features are characteristics unique to TCP when compared to UDP? (Select TWO.)

Geçerli olan tümünü seçin

Cevabı ve açıklamayı göster

Cevap: Explicit connection setup using a three-way handshake sequence before transmitting data; Flow control managed dynamically through receiver window size adjustments

Cevap

TCP uniquely provides stateful session establishment via a three-way handshake and flow control through dynamic sliding window adjustments, whereas UDP is stateless, connectionless, and lacks windowing flow control.
TCP is a connection-oriented protocol that establishes a stateful connection via a three-way handshake prior to data exchange and employs dynamic window size adjustments for flow control to match receiver processing capacity.

Adım Adım Çözüm

1
Analyze transport protocol connection management mechanisms.
TCP requires a three-way handshake (SYN, SYN-ACK, ACK) to establish a connection state between endpoints before payload exchange, whereas UDP is connectionless and sends datagrams without prior signaling.
Establishing connection state is essential for TCP reliability.
2
Analyze transport protocol traffic regulation capabilities.
TCP implements flow control using dynamic window size fields in its header to prevent sender buffer overrun at the receiver. UDP contains no window field or flow control mechanism.
Windowing allows TCP to adapt transmission rates to receiver processing limits.
3
Evaluate distractor properties against protocol specifications.
An 8-byte header and stateless delivery belong to UDP (TCP headers are at least 20 bytes). Multicast/broadcast support is exclusive to connectionless transport (UDP), as TCP is limited to unicast traffic.
Distractors represent characteristics inherent to UDP rather than TCP.

Anahtar Kavram

TCP vs UDP Operational Differences and Mechanisms
Soru 316Soru →

A network administrator is allocated the IPv4 address block 172.20.64.0/21172.20.64.0/21 for datacenter VLAN expansion. The architectural requirement mandates that the block be divided into the maximum possible number of equal-sized subnets such that each subnet can accommodate at least 5050 usable host addresses. If exactly 5050 hosts are assigned in every created subnet, how many usable host IP addresses across the entire /21/21 block remain unassigned?

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Cevap: 384

Cevap

384 usable host IP addresses remain unassigned across all created subnets.
To satisfy the requirement of at least 5050 usable host addresses per subnet while maximizing the number of subnets, 66 host bits are required (26−2=622^6 - 2 = 62 usable host addresses per subnet), resulting in a /26/26 prefix length. Dividing the /21/21 block into /26/26 subnets produces 226−21=322^{26-21} = 32 subnets. With 5050 host addresses assigned per subnet, each subnet has 62−50=1262 - 50 = 12 unassigned usable host addresses. Across all 3232 subnets, the total number of unassigned usable host IP addresses is 32×12=38432 \times 12 = 384.

Adım Adım Çözüm

1
Determine the minimum host bits required to support at least 50 usable host addresses per subnet.
6 host bits (h=6h = 6), which yields 26−2=622^6 - 2 = 62 usable host addresses per subnet and a prefix length of /26/26.
Using 5 host bits (25−2=302^5 - 2 = 30) is insufficient for 50 hosts. 6 host bits meet the minimum requirement while maximizing subnet creation.
2
Calculate the total number of /26/26 subnets created from the /21/21 summary block.
226−21=25=322^{26 - 21} = 2^5 = 32 subnets.
Borrowing 5 subnet bits from a /21/21 block creates 32 subnets of size /26/26.
3
Determine the unassigned usable host addresses per subnet when 50 hosts are assigned.
62−50=1262 - 50 = 12 unassigned usable host addresses per subnet.
Network and broadcast addresses are already excluded when calculating the 62 usable host addresses.
4
Multiply the unassigned usable host addresses per subnet by the total number of created subnets.
32 subnets×12 addresses=38432 \text{ subnets} \times 12 \text{ addresses} = 384 total unassigned usable host IP addresses.
Aggregating the unused host capacity across all equal-sized subnets within the allocated block.

Anahtar Kavram

IPv4 Subnet Sizing and Usable Host Capacity Calculation
Soru 317Soru →

A network engineer is designing a topology for a branch office with approximately 150 users across two floors. To minimize capital expenditure and management overhead while maintaining VLAN redundancy and routing capabilities between local subnets, the engineer selects a Cisco Collapsed Core topology instead of a traditional Cisco Three-Tier hierarchy. Which architectural modification defines this Collapsed Core implementation?

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Cevap: The core layer and distribution layer functions are consolidated into a single pair of switches.

Cevap

The core layer and distribution layer functions are consolidated into a single pair of switches.
The defining characteristic of a Cisco Collapsed Core (two-tier) architecture is merging the core and distribution layers into a single pair of switches. This maintains key Layer 3 routing, policy enforcement, and redundancy functions while reducing overall equipment cost and administration complexity for small to medium enterprise sites.

Adım Adım Çözüm

1
Analyze the requirements of the branch office network design
The scenario requires connecting 150 users efficiently while reducing costs and complexity compared to a full three-tier campus design.
Smaller sites do not have sufficient scale or port-density demands to justify separate Core and Distribution switch pairs.
2
Compare traditional Three-Tier architecture against Collapsed Core (Two-Tier) architecture
A traditional Three-Tier topology consists of distinct Access, Distribution, and Core layers. A Collapsed Core merges the Core (high-speed transport) and Distribution (routing, ACLs, policy enforcement) responsibilities into one tier.
Consolidating these two layers into a single switch pair (collapsed core switches) provides Layer 3 boundaries and inter-VLAN routing while decreasing hardware footprint.

Anahtar Kavram

Cisco Collapsed Core Topology Architecture
Soru 318Soru →

Host A with MAC address 0011.2233.4455 is connected to interface FastEthernet0/5 assigned to VLAN 20 on switch SW-Central. The MAC address table on SW-Central contains no existing entries for VLAN 20. When Host A transmits an ARP request frame to resolve its default gateway IP address, which operational sequence will SW-Central perform upon receiving the ingress frame?

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Cevap: Learn source MAC 0011.2233.4455 on interface FastEthernet0/5 in VLAN 20, then flood the frame out all active interfaces assigned to VLAN 20 except FastEthernet0/5.

Cevap

The switch records MAC 0011.2233.4455 on interface FastEthernet0/5 in VLAN 20 and floods the frame out all active interfaces in VLAN 20 except FastEthernet0/5.
When a Layer 2 switch receives an Ethernet frame, it first updates its MAC address table by associating the source MAC address (0011.2233.4455) with the receiving port (FastEthernet0/5) and VLAN (VLAN 20). Next, because an ARP request uses the Ethernet broadcast destination address (FFFF.FFFF.FFFF), the switch floods the frame out every port in VLAN 20 except the ingress interface.

Adım Adım Çözüm

1
Examine the source MAC address of the incoming Ethernet frame.
The source MAC address 0011.2233.4455 is identified on port FastEthernet0/5.
Layer 2 switches perform dynamic MAC address learning by inspecting the source MAC field of all ingress frames.
2
Update the switch MAC address table for VLAN 20.
An entry mapping MAC 0011.2233.4455 to port FastEthernet0/5 in VLAN 20 is created.
MAC table entries are stored per-VLAN to maintain Layer 2 traffic isolation.
3
Examine the destination MAC address of the frame.
The destination MAC address is FFFF.FFFF.FFFF (Ethernet broadcast).
ARP requests use the broadcast destination MAC address so that all devices in the broadcast domain receive the query.
4
Forward the broadcast frame according to Layer 2 switching rules.
The frame is flooded to all active ports belonging to VLAN 20, excluding ingress port FastEthernet0/5.
Broadcast frames must reach all hosts in the same VLAN domain, but split-horizon logic prevents sending the frame back out its ingress port.

Anahtar Kavram

L2 Switch MAC Table Operation and Broadcast Flooding Rules
Tahmini Süre:1m 15s
Soru 319Soru →

A network engineer is deploying IPv6 across an enterprise branch using the assigned global unicast prefix 2001:db8:85a3::/482001:\text{db8}:85\text{a}3::/48. Interfaces are set to use SLAAC with EUI-64 interface identifier generation. Which two statements correctly describe the prefix subnetting and EUI-64 address generation for this implementation? (Select TWO.)

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Cevap: Subnetting the 2001:db8:85a3::/482001:\text{db8}:85\text{a}3::/48 prefix into standard /64/64 subnets yields 65,53665,536 (2162^{16}) unique subnets.; A host with MAC address 70:6D:15:AB:CD:EF70:6\text{D}:15:\text{AB}:\text{CD}:\text{EF} using EUI-64 auto-configuration generates the interface identifier 726D:15FF:FEAB:CDEF726\text{D}:15\text{FF}:\text{FEAB}:\text{CDEF}.

Cevap

Subnetting a /48/48 prefix into /64/64 subnets produces 65,53665,536 unique subnets, and converting the MAC address 70:6D:15:AB:CD:EF70:6\text{D}:15:\text{AB}:\text{CD}:\text{EF} via EUI-64 produces 726D:15FF:FEAB:CDEF726\text{D}:15\text{FF}:\text{FEAB}:\text{CDEF}.
Subnetting a /48/48 prefix to /64/64 provides 1616 subnet bits (64−48=1664 - 48 = 16), giving 216=65,5362^{16} = 65,536 subnets without subtracting any reserved subnets. Additionally, converting the MAC address 70:6D:15:AB:CD:EF70:6\text{D}:15:\text{AB}:\text{CD}:\text{EF} to EUI-64 splits the address into 706D:15706\text{D}:15 and AB:CD:EF\text{AB}:\text{CD}:\text{EF}, inserts FFFE\text{FFFE} to form 706D:15FF:FEAB:CDEF706\text{D}:15\text{FF}:\text{FEAB}:\text{CDEF}, and flips the 7th bit of 0x700\text{x}70 (0111 000020111\,0000_2) to 0x720\text{x}72 (0111 001020111\,0010_2), giving 726D:15FF:FEAB:CDEF726\text{D}:15\text{FF}:\text{FEAB}:\text{CDEF}.

Adım Adım Çözüm

1
Calculate the available subnet bits and total subnets from /48/48 to /64/64.
Subnet bits = 64−48=1664 - 48 = 16. Total subnets = 216=65,5362^{16} = 65,536.
IPv6 does not deduct subnets for network or broadcast designations.
2
Perform EUI-64 interface ID conversion on MAC address 70:6D:15:AB:CD:EF70:6\text{D}:15:\text{AB}:\text{CD}:\text{EF}.
Insert FFFE\text{FFFE} between 70:6D:1570:6\text{D}:15 and AB:CD:EF\text{AB}:\text{CD}:\text{EF} to get 706D:15FF:FEAB:CDEF706\text{D}:15\text{FF}:\text{FEAB}:\text{CDEF}. Flip the 7th bit of byte 11 (0x70=0111 00002→0111 00102=0x720\text{x}70 = 0111\,0000_2 \rightarrow 0111\,0010_2 = 0\text{x}72), yielding 726D:15FF:FEAB:CDEF726\text{D}:15\text{FF}:\text{FEAB}:\text{CDEF}.
EUI-64 standard requires inserting FFFE\text{FFFE} in the center of the MAC address and setting the U/L bit to 1 for globally unique scope.

Anahtar Kavram

IPv6 Subnetting & EUI-64 Address Generation
Soru 320Soru →

A Cisco Catalyst switch receives an ingress Ethernet frame on interface GigabitEthernet0/1 assigned to VLAN 10. The frame contains a source MAC address that does not currently exist in the MAC address table and a destination MAC address that is already registered on interface GigabitEthernet0/2 in VLAN 10. Arrange the switch processing actions in the correct chronological sequence from initial frame reception to egress forwarding.

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Cevap

The correct sequence is: (1) Frame reception and header parsing on GigabitEthernet0/1, (2) Learning the source MAC address by adding an entry for GigabitEthernet0/1 in VLAN 10, (3) Looking up the destination MAC address in the MAC address table, and (4) Forwarding the unicast frame out GigabitEthernet0/2.
When a switch receives a frame, it follows a strict sequence of operations: first receiving and parsing the header, then learning the source MAC address (adding/updating the MAC table entry for the ingress port), looking up the destination MAC address in the table, and finally forwarding the frame out the designated destination interface.

Adım Adım Çözüm

1
Analyze frame ingress
The switch receives the frame on interface GigabitEthernet0/1, identifying VLAN 10, source MAC, and destination MAC.
Header parsing is the initial hardware step when a frame enters a switchport.
2
Process source MAC learning
The switch adds a dynamic entry mapping the unknown source MAC address to GigabitEthernet0/1 in VLAN 10.
Cisco switches perform Layer 2 source MAC learning before evaluating destination forwarding.
3
Perform destination lookup
The MAC address table search matches the destination MAC to interface GigabitEthernet0/2.
Destination address matching dictates whether a frame is forwarded to a single port or flooded.
4
Execute frame forwarding
The frame is forwarded directly out GigabitEthernet0/2.
Known unicast frames are forwarded solely out the egress port specified in the MAC table.

Anahtar Kavram

Layer 2 Switch Ingress Processing & MAC Address Table Operation
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