Network Fundamentals Alıştırma Soruları — Cisco CCNA — Sayfa 20

Soru 381Soru →

An enterprise network administrator is allocated the IPv4 address block $172.16.48.0/21$ to provision subnets for multiple VLANs in a regional branch office. One specific VLAN requires support for at least $100$ host devices while maximizing address efficiency. Which subnetwork address and CIDR prefix pair represents a valid, correctly sized subnet within the assigned parent block?

172.16.50.0/25

172.16.50.0/26

172.32.50.0/25

172.16.56.0/25

Cevabı ve açıklamayı göster

Cevap: $172.16.50.0/25$

Cevap

The subnet address

172.16.50.0/25

is correct because it falls within the

172.16.48.0/21

parent block, uses valid RFC 1918 private addressing, and provides 126 usable host addresses.

The subnet

172.16.50.0/25

is correct because a

/25

prefix provides

2^7 - 2 = 126

usable host addresses, which satisfies the 100 host requirement efficiently. Furthermore,

172.16.50.0

lies well within the assigned parent block of

172.16.48.0/21

(range

172.16.48.0

-

172.16.55.255

) and belongs to valid RFC 1918 private IP space.

Adım Adım Çözüm

1

Determine the address range of the allocated parent block

172.16.48.0/21

.

The block has 21 network bits, leaving 11 host bits. The 3rd octet ranges from 48 (

00011000_2

) to 55 (

00011111_2

). The valid IP range is

172.16.48.0

through

172.16.55.255

.

Any valid subnets must fall strictly within these boundaries.

2

Calculate the minimum prefix length needed for at least 100 host devices.

Using

2^h - 2 \ge 100

, for

h=6

,

2^6 - 2 = 62

(insufficient). For

h=7

,

2^7 - 2 = 126

(sufficient). The required prefix length is

32 - 7 = 25

(

/25

).

Selecting

/25

provides enough host IPs while conserving space compared to a larger mask like

/24

.

3

Evaluate candidate

/25

subnets against the parent block boundaries and RFC 1918 rules.

172.16.50.0/25

starts at a valid

/25

boundary in the 4th octet (

0

), resides within the

172.16.48.0 - 172.16.55.255

range, and uses Class B RFC 1918 private space (

172.16.0.0/12

).

This confirms all sizing, allocation, and RFC standards are met.

Anahtar Kavram

IPv4 Subnet Sizing and RFC 1918 Boundary Analysis

Tahmini Süre:1m 30s

Soru 382Soru →

A network engineer is selecting physical media and transceiver modules for an enterprise network expansion. Match each Ethernet interface standard or component on the left with its corresponding physical specification on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

10GBASE-SR matches the 850 nm multimode fiber specification supporting up to 400 meters over OM4. 100BASE-FX matches the Fast Ethernet multimode fiber specification supporting up to 2 km. 10GBASE-T matches the Category 6A UTP copper requirement for 100-meter 10 Gbps links. SFP+ matches the hot-swappable 10 Gbps transceiver form factor.

Each physical media standard uniquely pairs with its physical medium, speed rating, and distance capability. 10GBASE-SR operates on 850 nm multimode fiber up to 400m over OM4; 100BASE-FX operates on Fast Ethernet multimode fiber up to 2 km; 10GBASE-T requires Category 6A UTP for 10 Gbps over 100 meters; SFP+ is the 10 Gbps transceiver module form factor.

Adım Adım Çözüm

1

Identify the optical wavelength and distance limits for 10 Gbps multimode fiber.

10GBASE-SR specifies 850 nm optics over multimode fiber reaching up to 300m on OM3 or 400m on OM4.

The 'SR' designation stands for Short Reach optical transmission.

2

Determine the distance specification for Fast Ethernet optical links.

100BASE-FX provides 100 Mbps speed over multimode fiber up to 2 km.

100BASE-FX extends Fast Ethernet reach beyond the 100-meter copper limit using multimode fiber.

3

Evaluate UTP copper cabling categories for 10 Gbps Ethernet.

10GBASE-T requires Category 6A UTP to achieve the full 100-meter distance limit.

Standard Category 6 is limited to 37–55 meters for 10GBASE-T due to alien crosstalk; Cat6A is needed for 100m.

4

Match the modular interface module type to its bandwidth capacity.

SFP+ is the enhanced modular form factor designed for 10 Gbps throughput.

Standard SFP modules support up to 1 Gbps, while SFP+ modules support 10 Gbps.

Anahtar Kavram

Ethernet Physical Layer Standards, Fiber/Copper Cabling Media Specifications, and Transceiver Form Factors

Soru 383Soru →

An network engineer is configuring a Cisco router interface using IPv6 link-local autoconfiguration. The burned-in MAC address of the interface is `0800.279f.a1b2`. Which IPv6 link-local address will be automatically generated for this interface using the EUI-64 standard?

fe80::a00:27ff:fe9f:a1b2

fe80::800:27ff:fe9f:a1b2

fe80::a00:27ff:ffff:a1b2

fe80::a00:279f:a1b2:fffe

Cevabı ve açıklamayı göster

Cevap: fe80::a00:27ff:fe9f:a1b2

Cevap

fe80::a00:27ff:fe9f:a1b2

The address fe80::a00:27ff:fe9f:a1b2 is correct because the EUI-64 process takes the MAC address 0800.279f.a1b2, inserts FFFE between the 24th and 25th bits (0800:27FF:FE9F:A1B2), and flips the 7th bit of the first byte from 0 to 1 (converting 08 to 0A). Combining this 64-bit interface ID with the FE80::/64 link-local prefix yields fe80::a00:27ff:fe9f:a1b2.

Adım Adım Çözüm

1

Split the 48-bit MAC address into two 24-bit halves.

Left half: 0800.27, Right half: 9f.a1b2

EUI-64 requires inserting a 16-bit payload into the center of the 48-bit hardware address.

2

Insert the hex string FFFE between the two halves.

0800:27FF:FE9F:A1B2

FFFE is the standard 16-bit sequence specified by IEEE EUI-64 calculation.

3

Invert the 7th bit (Universal/Local bit) of the first byte.

First byte 08 (hex 0000 0000 -> 0000 1000) becomes 0A (hex 0000 1010). Resulting Interface ID: 0A00:27FF:FE9F:A1B2 (shortened to a00:27ff:fe9f:a1b2).

Inverting the U/L bit indicates that the address is globally unique / universally administered.

4

Prepend the IPv6 link-local prefix FE80::/64 to the interface ID.

fe80::a00:27ff:fe9f:a1b2

IPv6 link-local addresses use the prefix FE80::/64 combined with the 64-bit EUI-64 interface identifier.

Anahtar Kavram

EUI-64 Interface Identifier Generation

Soru 384Soru →

A network administrator is auditing interface configurations across an enterprise network. A host interface is assigned the IPv4 address $172.20.108.90$ with a subnet mask of $255.255.255.224$ . Which IPv4 address represents the broadcast address for this subnetwork?

172.20.108.95

172.20.108.94

172.20.108.96

172.20.108.127

Cevabı ve açıklamayı göster

Cevap: $172.20.108.95$

Cevap

The broadcast address for the subnetwork is

172.20.108.95

.

For a

/27

subnet mask (

255.255.255.224

), the subnet block size is

256 - 224 = 32

. The subnets increment by

32

in the fourth octet (

0, 32, 64, 96, 128

). The host address

172.20.108.90

falls into the subnet starting at

172.20.108.64

. The broadcast address is the highest IP address within this range, which is

172.20.108.95

(one less than the next subnet ID of

172.20.108.96

).

Adım Adım Çözüm

1

Determine the CIDR prefix length and block size from the subnet mask.

The mask

255.255.255.224

corresponds to a

/27

prefix (

24 + 3

bits). The block size in the fourth octet is

256 - 224 = 32

.

The subnet mask defines the boundary between the network and host portions of the IP address.

2

Identify the network ID for the given IP address.

Subnet multiples of

32

in the fourth octet are

0, 32, 64, 96, 128...

Since

90

lies between

64

and

95

, the network ID is

172.20.108.64

.

The network ID is the lowest address in the subnet block.

3

Calculate the broadcast address.

The next subnet starts at

172.20.108.96

. Subtracting

1

from the next network ID yields the broadcast address

172.20.108.95

.

The broadcast address is always the last address in the subnet range where all host bits are set to 1.

Anahtar Kavram

IPv4 Subnet Boundaries and Broadcast Address Calculation

Tahmini Süre:1m 0s

Soru 385Soru →

A Cisco router interface with MAC address 0011.4567.89ab is configured to dynamically derive its IPv6 interface identifier using EUI-64. What is the resulting 64-bit interface identifier in standard colon-separated hexadecimal IPv6 format?

Cevabı ve açıklamayı göster

Cevap: 0211:45ff:fe67:89ab; 211:45ff:fe67:89ab

Cevap

0211:45ff:fe67:89ab

To derive a 64-bit IPv6 interface identifier from the 48-bit MAC address 0011.4567.89ab using EUI-64: 1) Insert FFFE into the middle of the MAC address, yielding 0011:45FF:FE67:89AB. 2) Invert the 7th bit (Universal/Local bit) of the first octet. The first octet 0x00 (0000 0000 in binary) becomes 0x02 (0000 0010 in binary). Combining these steps yields the final interface identifier 0211:45ff:fe67:89ab (or 211:45ff:fe67:89ab without leading zero).

Adım Adım Çözüm

1

Split the 48-bit MAC address into two 24-bit halves.

First half: 00:11:45, Second half: 67:89:ab

EUI-64 calculation begins by separating the Organizationally Unique Identifier (OUI) from the vendor-assigned NIC portion.

2

Insert the 16-bit hexadecimal string FFFE between the two halves.

0011:45FF:FE67:89AB

IEEE EUI-64 conversion requires expanding the 48-bit MAC address into a 64-bit value by inserting FFFE at the midpoint.

3

Invert the 7th bit (Universal/Local bit) of the first byte.

First byte 0x00 (binary 00000000) becomes 0x02 (binary 00000010), producing 0211:45ff:fe67:89ab.

Inverting the U/L bit sets it to 1, indicating that the IEEE MAC-derived interface ID is globally unique.

Anahtar Kavram

IPv6 EUI-64 Interface Identifier Derivation

Soru 386Soru →

A network administrator needs to perform the initial configuration on a new Cisco Catalyst switch via an out-of-band management session. The administrator connects a laptop terminal emulator directly to the RJ-45 console port of the switch using an RS-232 serial cable adapter. Which cable pinout scheme is required to correctly map the laptop terminal adapter signals to the switch console interface?

A rollover cable pinout where pin 1 on one connector connects to pin 8 on the opposite connector.

A straight-through cable pinout adhering to T568B specifications on both connector ends.

A two-pair crossover cable pinout swapping pin 1 with pin 3 and pin 2 with pin 6.

A four-pair crossover cable pinout swapping pins 1/3, 2/6, 4/7, and 5/8.

Cevabı ve açıklamayı göster

Cevap: A rollover cable pinout where pin 1 on one connector connects to pin 8 on the opposite connector.

Cevap

A rollover cable pinout (where pin 1 connects to pin 8, pin 2 to pin 7, etc.) is required to connect a terminal serial interface to a Cisco switch RJ-45 console port.

Cisco RJ-45 console ports require a rollover cable (often called a console cable). In a rollover cable, the pinout on one end is completely inverted relative to the other end—pin 1 maps to pin 8, pin 2 maps to pin 7, pin 3 maps to pin 6, and pin 4 maps to pin 5. When combined with an RS-232 terminal adapter (DB-9 or USB-to-serial), this pin inversion routes the laptop serial transmitter pin to the console receiver pin and vice versa.

Adım Adım Çözüm

1

Identify the type of connection being established.

The connection is an out-of-band management terminal connection between a host PC/laptop terminal adapter and an RJ-45 Cisco console port, not an Ethernet network data link.

Console management uses asynchronous RS-232 serial signalling over an RJ-45 physical port interface rather than standard Ethernet framing.

2

Determine the physical pin mapping for Cisco RJ-45 console interfaces.

Cisco console ports use a rollover pinout convention where the wire sequence completely reverses from one RJ-45 connector to the other end (Pin 1 to Pin 8, Pin 2 to Pin 7, Pin 3 to Pin 6, Pin 4 to Pin 5).

This reversal correctly aligns the Transmit Data (TxD), Receive Data (RxD), and Ground signals between the RJ-45 console port and the DB-9 / USB serial adapter attached to the management PC.

Anahtar Kavram

Cisco Console Port and Rollover Cable Pinouts

Soru 387Soru →

A network engineer is selecting physical media and transceiver modules to establish a 1 Gbps backbone link between two switches located in different buildings separated by a distance of 400 meters. Which two transceiver and cabling combinations will successfully support a Gigabit Ethernet link across this distance? (Select two.)

Geçerli olan tümünü seçin

A 1000BASE-LX SFP transceiver connected using single-mode fiber-optic (SMF) cabling.

A 1000BASE-LX SFP transceiver connected using OM3 multimode fiber-optic (MMF) cabling with a mode-conditioning patch cord.

A 1000BASE-SX SFP transceiver connected using legacy OM1 (62.5/125 µm) multimode fiber-optic cabling.

A 1000BASE-T SFP transceiver connected using Category 6a unshielded twisted-pair (UTP) copper cabling.

A 1000BASE-ZX SFP transceiver connected using Category 7 shielded twisted-pair (STP) copper cabling.

Cevabı ve açıklamayı göster

Cevap: A 1000BASE-LX SFP transceiver connected using single-mode fiber-optic (SMF) cabling.; A 1000BASE-LX SFP transceiver connected using OM3 multimode fiber-optic (MMF) cabling with a mode-conditioning patch cord.

Cevap

The correct options are the 1000BASE-LX SFP transceiver with single-mode fiber (SMF) cabling, and the 1000BASE-LX SFP transceiver with OM3 multimode fiber (MMF) cabling using a mode-conditioning patch cord.

Both selected options meet or exceed the 400-meter distance constraint. 1000BASE-LX transceivers operating over single-mode fiber can transmit up to 5 kilometers. Additionally, 1000BASE-LX transceivers can achieve up to 550 meters over multimode fiber when paired with a mode-conditioning patch cord.

Adım Adım Çözüm

1

Identify the distance constraint and speed requirement from the scenario.

The link distance is 400 meters and the required bandwidth is 1 Gbps (Gigabit Ethernet).

Selection of transceivers and cable types depends directly on distance limits defined by IEEE 802.3 standards.

2

Evaluate copper cabling limitations.

Category 6a UTP (1000BASE-T) is limited to 100 meters total channel length.

Copper cabling cannot span 400 meters without active repeaters or switches in between.

3

Evaluate short-reach optical transceiver specifications (1000BASE-SX).

1000BASE-SX over OM1 (62.5/125 µm) MMF supports a maximum distance of 275 meters.

Short-wavelength optics (850 nm) attenuate faster over legacy multimode fiber cores.

4

Evaluate long-reach optical transceiver specifications (1000BASE-LX).

1000BASE-LX over SMF supports up to 5 km. 1000BASE-LX over laser-optimized MMF (using a mode-conditioning patch cord) supports up to 550 meters.

Long-wavelength optics (1310 nm) satisfy the 400-meter requirement across both SMF and suitably conditioned MMF media.

Anahtar Kavram

Gigabit Ethernet transceiver distance limits and fiber media specifications (1000BASE-LX vs 1000BASE-SX vs 1000BASE-T)

Soru 388Soru →

A network engineer is provisioning static IP addresses for servers assigned to the subnetwork block $10.45.192.0/21$ . Which TWO of the following IPv4 addresses are valid, usable host addresses within this subnetwork?

Geçerli olan tümünü seçin

10.45.192.0

10.45.192.1

10.45.199.254

10.45.199.255

Cevabı ve açıklamayı göster

Cevap: 10.45.192.1; 10.45.199.254

Cevap

The addresses 10.45.192.1 and 10.45.199.254 are valid usable host addresses.

For a /21 prefix (

255.255.248.0

), the subnetwork spans from network address

10.45.192.0

to broadcast address

10.45.199.255

. The usable host range includes all IP addresses between the network address and broadcast address, which are

10.45.192.1

through

10.45.199.254

. Therefore,

10.45.192.1

and

10.45.199.254

are both valid usable host addresses.

Adım Adım Çözüm

1

Determine the subnet mask and block size from the prefix length.

A /21 prefix corresponds to a subnet mask of 255.255.248.0. The block size in the third octet is 256 - 248 = 8.

Prefix length /21 leaves 11 bits for hosts (32 - 21 = 11).

2

Calculate the network address and broadcast address for 10.45.192.0/21.

Network Address: 10.45.192.0; Broadcast Address: 10.45.199.255.

The third octet increments in steps of 8 (192, 200, etc.). The current subnet spans from 10.45.192.0 to 10.45.199.255.

3

Identify the usable host address range.

Usable host range is 10.45.192.1 through 10.45.199.254.

Usable host addresses exclude the network address (first IP) and broadcast address (last IP).

Anahtar Kavram

Usable IPv4 Host Range Calculation

Soru 389Soru →

A network administrator is assigning an IP address to a newly installed interface on a database server. The server is configured with the IP address $172.28.140.197$ and a subnet mask of $255.255.252.0$ . Which IPv4 address represents the subnet broadcast address for this server's subnetwork?

172.28.143.255

172.28.140.255

172.28.143.254

172.32.143.255

Cevabı ve açıklamayı göster

Cevap: $172.28.143.255$

Cevap

172.28.143.255

For the IP address

172.28.140.197

with subnet mask

255.255.252.0

(/22), the third octet operates in increments of

4

(

256 - 252 = 4

). Since

140

is a multiple of

4

, the subnetwork begins at

172.28.140.0

and extends up to

172.28.143.255

. Therefore, the address

172.28.143.255

is the broadcast address for this subnetwork.

Adım Adım Çözüm

1

Convert the dotted-decimal subnet mask to CIDR notation and identify the interesting octet.

The subnet mask

255.255.252.0

corresponds to a

/22

prefix (

8 + 8 + 6 + 0 = 22

bits). The interesting octet is the third octet.

Determining the prefix length isolates which octet controls the subnetwork boundaries.

2

Calculate the subnetwork block size for the third octet.

Block size

= 256 - 252 = 4

.

The block size specifies the interval at which subnetwork addresses increment in the third octet.

3

Find the network address containing the host IP address

172.28.140.197

.

Dividing

140

by

4

gives exactly

35

with no remainder. Thus, the network ID is

172.28.140.0

.

The network address is the starting boundary of the subnet block.

4

Determine the broadcast address by adding (block size - 1) to the third octet and setting all host bits in the fourth octet to 1s (

255

).

Third octet upper bound

= 140 + 4 - 1 = 143

. Broadcast address

= 172.28.143.255

.

The broadcast address is the last IP address in the subnetwork range.

Anahtar Kavram

IPv4 Subnet Mask Evaluation and Broadcast Address Calculation

Soru 390Soru →

A network administrator needs to carve out a new subnetwork from the $172.16.64.0/22$ enterprise address block for a remote branch office. The branch requires static IPv4 addresses for $26$ workstation PCs, $2$ IP phones, and $2$ network printers. Which IPv4 network address and prefix length combination provides the most efficient assignment to accommodate all required host addresses without wasting address space?

172.16.64.0/28

172.16.64.0/27

172.16.64.0/26

172.32.64.0/27

Cevabı ve açıklamayı göster

Cevap: $172.16.64.0/27$

Cevap

The most efficient subnet assignment is

172.16.64.0/27

because it provides exactly 30 usable IPv4 addresses within the valid RFC 1918 private block.

The required endpoint count is 30 hosts (

26 + 2 + 2 = 30

). To accommodate 30 usable hosts, 5 host bits are needed (

2^5 - 2 = 30

). Subtracting 5 host bits from 32 IPv4 total bits leaves a /27 prefix. Subnetwork

172.16.64.0/27

resides within the allocated

172.16.64.0/22

block and RFC 1918 private space, fulfilling all requirements with zero wasted IP addresses.

Adım Adım Çözüm

1

Calculate the total host IP address requirement

Total endpoints =

26 \text{ PCs} + 2 \text{ phones} + 2 \text{ printers} = 30 \text{ usable IP addresses}

.

All active physical and virtual endpoints require individual host IPv4 addresses on the subnetwork.

2

Determine the host bit requirement (

h

)

Using

2^h - 2 \ge 30

, solving for

h

yields

h = 5

host bits (

2^5 - 2 = 30

).

Two IP addresses per subnet are unassignable to hosts: the network address (all host bits set to 0) and the subnet broadcast address (all host bits set to 1).

3

Calculate the subnet prefix length

Prefix length =

32 - 5 = /27

.

Subtracting the 5 host bits from the 32 total bits in an IPv4 address defines the 27 network prefix bits.

4

Verify address space validity and boundaries

Subnetwork

172.16.64.0/27

falls within the

172.16.64.0/22

parent block and belongs to the RFC 1918 private Class B range (

172.16.0.0 - 172.31.255.255

).

Internal enterprise subnets must utilize allocated private address space.

Anahtar Kavram

IPv4 Subnet Sizing and Usable Host Calculation

Soru 391Soru →

A network engineer is allocated the IPv4 address block $172.30.128.0/20$ for an enterprise campus expansion project. The engineer needs to divide this block into smaller, equal-sized subnetworks using a prefix mask of $/25$ . What is the total number of subnets that can be created from this original $/20$ block?

Cevabı ve açıklamayı göster

Cevap: 32

Cevap

The total number of subnets that can be created is 32.

Moving from a

/20

prefix to a

/25

prefix borrows 5 host bits (

25 - 20 = 5

). Calculating

2^5

gives 32 distinct subnets.

Adım Adım Çözüm

1

Calculate the number of borrowed subnet bits

5 bits borrowed (

25 - 20 = 5

)

The difference between the target prefix length and the starting prefix length represents the network bits borrowed from the host portion.

2

Calculate the total subnet count

32 subnets (

2^5 = 32

)

Each borrowed bit doubles the number of created subnets (

2^n

, where

n

is the number of borrowed bits).

Anahtar Kavram

Subnet counting using CIDR prefix borrowing

Soru 392Soru →

A network administrator configures a core switch interface with the IPv4 address $10.50.180.75/21$ . What is the broadcast address for the subnetwork to which this interface belongs?

10.50.183.255

10.50.180.255

10.50.183.254

10.50.191.255

Cevabı ve açıklamayı göster

Cevap: $10.50.183.255$

Cevap

The broadcast address for the subnetwork is

10.50.183.255

.

The prefix /21 designates 21 subnet bits, yielding a subnet mask of

255.255.248.0

. In the third octet, subnets increment in steps of 8. For the IP address

10.50.180.75

, the containing subnet boundary starts at

10.50.176.0

and spans up to

10.50.183.255

. Since the broadcast address requires all host bits set to 1,

10.50.183.255

is the correct subnet broadcast address.

Adım Adım Çözüm

1

Determine the subnet mask and block size from the CIDR prefix.

A /21 prefix provides 21 network bits and 11 host bits. In dotted-decimal form, this is

255.255.248.0

. The host portion spans into the third octet. The block size in the third octet is

256 - 248 = 8

.

Prefix length identifies where the network portion ends and the host portion begins.

2

Calculate the network address for the given IP address

10.50.180.75

.

Dividing the third octet value (180) by the block size (8) gives

180 / 8 = 22.5

. Multiplying

22 \times 8 = 176

. Thus, the network address is

10.50.176.0/21

.

The network address is the lowest boundary of the subnet block.

3

Calculate the broadcast address.

Adding the block size minus 1 (

8 - 1 = 7

) to the third octet network boundary gives

176 + 7 = 183

. Setting all host bits in the fourth octet to binary 1s gives 255. The broadcast address is

10.50.183.255

.

The broadcast address is the highest possible IP address within the subnet range.

Anahtar Kavram

IPv4 Subnet Boundary and Broadcast Address Derivation

Tahmini Süre:1m 30s

Soru 393Soru →

A network technician is provisioning host interfaces within the enterprise block assigned as $10.80.192.0/22$ . Which of the following IPv4 addresses are valid, usable host addresses within this specific subnetwork? (Select TWO.)

Geçerli olan tümünü seçin

10.80.192.1

10.80.195.250

10.80.192.0

10.80.195.255

10.80.196.1

Cevabı ve açıklamayı göster

Cevap: $10.80.192.1$ ; $10.80.195.250$

Cevap

The addresses

10.80.192.1

and

10.80.195.250

are the valid usable host addresses within the

10.80.192.0/22

subnetwork.

The prefix

/22

yields a subnet mask of

255.255.252.0

, providing

2^{10} - 2 = 1022

usable host addresses per subnet. The subnet boundaries run from network ID

10.80.192.0

to broadcast address

10.80.195.255

. Thus, the valid host range spans from

10.80.192.1

through

10.80.195.254

. Both

10.80.192.1

and

10.80.195.250

fall strictly inside this range.

Adım Adım Çözüm

1

Determine the subnet mask from prefix length

Prefix

/22

corresponds to a subnet mask of

255.255.252.0

.

Converting

/22

to binary gives 22 network bits and 10 host bits (

11111111.11111111.11111100.00000000

).

2

Calculate block size and network boundary

Block size in the 3rd octet is

256 - 252 = 4

. The subnet starts at

10.80.192.0

and ends right before

10.80.196.0

.

The 3rd octet increments in multiples of 4 (

0, 4, 8, \dots, 192, 196

). Network ID is

10.80.192.0

.

3

Identify broadcast address and usable host range

Broadcast address is

10.80.195.255

. Usable host range is

10.80.192.1

to

10.80.195.254

.

Usable host addresses are all IP values between the network ID and the broadcast ID.

4

Evaluate choices against the usable range

10.80.192.1

and

10.80.195.250

lie within

[10.80.192.1, 10.80.195.254]

.

10.80.192.0

is the network ID,

10.80.195.255

is the broadcast ID, and

10.80.196.1

is in the next subnet.

Only IP addresses inside the calculated host range can be assigned to devices.

Anahtar Kavram

IPv4 Subnet Boundaries and Host Address Ranges

Soru 394Soru →

A network engineer is assigned the IPv4 block $10.100.0.0/21$ to create subnetworks for regional offices. Each regional office network requires a $/26$ subnet mask. How many $/26$ subnetworks can be created from this $10.100.0.0/21$ block?

Cevabı ve açıklamayı göster

Cevap: 32

Cevap

32 subnets can be created from the

10.100.0.0/21

block.

Subtracting the original prefix length (

/21

) from the new subnet prefix length (

/26

) yields

5

subnet bits (

26 - 21 = 5

). Calculating

2^5

gives

32

subnets available within the

10.100.0.0/21

parent block.

Adım Adım Çözüm

1

Determine the number of subnet bits borrowed by subtracting the initial block prefix length from the target subnet prefix length.

Borrowed bits =

26 - 21 = 5

bits.

The difference between the new prefix length and the original prefix length represents the bits available to create subnets.

2

Calculate the total number of subnets using

2^n

, where

n

is the number of borrowed bits.

2^5 = 32

subnets.

Each additional bit assigned to the network prefix doubles the number of valid subnets.

Anahtar Kavram

Calculating Subnet Count from CIDR Prefix Differences

Tahmini Süre:1m 30s

Soru 395Soru →

A network administrator is assigning an IP address to a newly deployed web server. The default gateway interface for the server's subnet is configured with the IP address $172.20.108.1$ and a subnet mask of $255.255.252.0$ . Which of the following IPv4 addresses is a valid usable host address on the same subnet as the default gateway?

172.20.110.50

172.20.111.255

172.20.112.1

172.32.109.10

Cevabı ve açıklamayı göster

Cevap: $172.20.110.50$

Cevap

The IP address

172.20.110.50

is a valid usable host address on the

172.20.108.0/22

subnet.

The subnet mask

255.255.252.0

(/22) establishes a block size of

4

in the third octet (

256 - 252 = 4

). For the default gateway IP

172.20.108.1

, the subnet network address is

172.20.108.0/22

, giving a usable host range of

172.20.108.1

to

172.20.111.254

and a broadcast address of

172.20.111.255

. The IP address

172.20.110.50

falls directly within this valid host range.

Adım Adım Çözüm

1

Determine the subnet block size in the third octet

A subnet mask of

255.255.252.0

(/22 prefix length) leaves

2

host bits in the third octet. The block size is calculated as

256 - 252 = 4

.

Subnet boundaries in the third octet occur in increments of

4

.

2

Calculate the network address, broadcast address, and host range

Since

108

is a multiple of

4

(

4 \times 27 = 108

), the network address is

172.20.108.0/22

. The next subnet starts at

172.20.112.0

, making the broadcast address

172.20.111.255

.

The broadcast address is one integer below the next subnet's network address.

3

Identify the usable host range and evaluate candidate IP addresses

The usable host IP range is from

172.20.108.1

through

172.20.111.254

. The IP address

172.20.110.50

lies within this range.

Any IP strictly between the network ID and broadcast ID can be assigned to host interfaces.

Anahtar Kavram

IPv4 Subnet Masking and Usable Host Range Calculation

Soru 396Soru →

A network engineer is provisioning host IP addresses within the enterprise subnet block $192.168.48.0/20$ . Which of the following IP addresses are valid usable host addresses for endpoints on this subnetwork? (Select two.)

Geçerli olan tümünü seçin

192.168.50.100

192.168.63.254

192.168.48.0

192.168.63.255

Cevabı ve açıklamayı göster

Cevap: $192.168.50.100$ ; $192.168.63.254$

Cevap

The addresses

192.168.50.100

and

192.168.63.254

are valid usable host addresses within the

192.168.48.0/20

subnet.

For the prefix

/20

, the third octet increments in blocks of 16 (

256 - 240 = 16

). For the subnet starting at

192.168.48.0

, the subnet range spans from

192.168.48.0

to

192.168.63.255

. Subtracting the network ID (

192.168.48.0

) and broadcast ID (

192.168.63.255

) leaves a usable host range of

192.168.48.1

through

192.168.63.254

. Therefore,

192.168.50.100

and

192.168.63.254

are valid host assignments.

Adım Adım Çözüm

1

Determine the subnet mask block size for the /20 prefix length.

A

/20

prefix corresponds to a subnet mask of

255.255.240.0

. The block size in the third octet is

256 - 240 = 16

.

Prefix length

/20

uses 4 subnet bits in the third octet (

128 + 64 + 32 + 16 = 240

).

2

Identify the network address and broadcast address boundaries.

Subnet Network ID:

192.168.48.0

. Next Subnet ID:

192.168.64.0

. Broadcast Address:

192.168.63.255

.

Adding the block size of 16 to the third octet start (

48 + 16

) gives the next subnet boundary at

192.168.64.0

.

3

Calculate the usable host address range.

Usable host range:

192.168.48.1

to

192.168.63.254

.

Usable hosts exclude the network ID (

192.168.48.0

) and broadcast address (

192.168.63.255

).

4

Evaluate the choices against the usable host address range.

192.168.50.100

and

192.168.63.254

fall within the usable range, whereas

192.168.48.0

(network ID) and

192.168.63.255

(broadcast ID) are reserved.

Only IP addresses between the first host and last host can be assigned to endpoint interfaces.

Anahtar Kavram

IPv4 Subnet Boundaries and Usable Host Range Calculation

Tahmini Süre:1m 30s

Soru 397Soru →

A network administrator is assigning an IP address to a core switch interface with the IPv4 address $172.16.138.200$ and a subnet mask of $255.255.248.0$ . What is the broadcast address of this subnetwork?

172.16.143.255

172.16.143.254

172.16.139.255

172.32.143.255

Cevabı ve açıklamayı göster

Cevap: $172.16.143.255$

Cevap

The broadcast address for the subnetwork containing

172.16.138.200/21

is

172.16.143.255

.

For a subnet mask of

255.255.248.0

(/21), the third octet block size is

256 - 248 = 8

. Subnet boundaries in the third octet occur at multiples of 8 (

0, 8, 16, \dots, 136, 144

). Since the host IP is

172.16.138.200

, it falls in the subnet starting at

172.16.136.0

. The broadcast address is one less than the next subnet boundary (

172.16.144.0

), resulting in

172.16.143.255

.

Adım Adım Çözüm

1

Determine the prefix length and block size from the subnet mask

255.255.248.0

.

The subnet mask

255.255.248.0

corresponds to a /21 prefix. The interesting octet is the third octet, giving a block size of

256 - 248 = 8

.

Identifying the block size determines the increment between consecutive subnet boundaries.

2

Calculate the network ID containing the host address

172.16.138.200

.

Dividing the third octet value

138

by

8

yields

17.25

. Multiplying

17 \times 8 = 136

establishes the network boundary at

172.16.136.0/21

.

Finds the starting network address of the subnet.

3

Calculate the broadcast address by subtracting 1 from the next subnet boundary.

The next subnet boundary is

172.16.136.0 + 8 = 172.16.144.0

. Subtracting 1 gives

172.16.143.255

.

The broadcast address is the highest address in the subnet with all host bits set to 1.

Anahtar Kavram

Subnet Block Size and Broadcast Address Derivation

Soru 398Soru →

An enterprise network administrator needs to design an IPv4 subnet for a new IP security camera deployment that must support 500 host devices. What is the maximum number of usable host IPv4 addresses provided by the smallest standard CIDR subnet block that can accommodate this requirement?

Cevabı ve açıklamayı göster

Cevap: 510

Cevap

510 usable host IPv4 addresses

To support 500 host devices, the number of host bits

h

must satisfy

2^h - 2 \ge 500

. A

/24

subnet (

h = 8

) yields only

2^8 - 2 = 254

usable hosts, which is insufficient. The smallest subnet size that meets the requirement is a

/23

subnet (

h = 9

), which provides

2^9 - 2 = 510

usable host addresses.

Adım Adım Çözüm

1

Determine the host bit requirement formula

Use

2^h - 2 \ge 500

, where

h

represents host bits.

In every standard IPv4 subnet, two addresses are reserved: the network ID (all host bits 0) and the broadcast address (all host bits 1).

2

Calculate the smallest number of host bits that accommodates 500 hosts

2^8 - 2 = 254

usable addresses (too small for 500 hosts);

2^9 - 2 = 510

usable addresses (meets the requirement).

Selecting 9 host bits leaves

32 - 9 = 23

network bits, resulting in a

/23

prefix length.

3

Calculate the usable host capacity

2^9 - 2 = 512 - 2 = 510

usable host addresses.

Subtracting 2 reserved addresses from the 512 total addresses gives 510 valid host assignments.

Anahtar Kavram

IPv4 Subnet Sizing and Usable Host Calculation

Network Fundamentals

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