Networking Concepts Practice Questions — CompTIA Network+ — Page 21

Question 401Question →

Match each dynamic routing protocol to its corresponding architectural classification and primary metric selection criteria.

Click a left item, then click its matching right item

Show answer & explanation

Answer

OSPF matches Link-state IGP using cost derived from bandwidth; BGP matches Path-vector EGP using Autonomous System paths; EIGRP matches Advanced distance-vector IGP using bandwidth and delay; RIPv2 matches Distance-vector IGP using hop count with a 15-hop limit.

Each routing protocol is defined by its operational algorithm and metrics: OSPF is a link-state IGP calculating path cost from bandwidth; BGP is a path-vector EGP routing between Autonomous Systems; EIGRP is an advanced distance-vector protocol using a composite metric of bandwidth and delay; and RIPv2 is a traditional distance-vector protocol using hop count up to a maximum limit of 15 hops.

Step-by-Step Solution

1

Identify protocol scopes (IGP vs EGP)

BGP is the only Exterior Gateway Protocol (EGP) in the list used for inter-domain routing across Autonomous Systems.

Categorizing by protocol scope isolates BGP to the path-vector EGP description.

2

Classify remaining Interior Gateway Protocols by algorithm type

OSPF uses link-state (Shortest Path First), RIPv2 uses distance-vector (Bellman-Ford), and EIGRP uses advanced distance-vector (DUAL).

Protocol algorithms dictate link state database propagation versus route vector updates.

3

Evaluate metric calculation methods for each IGP

OSPF calculates cost based on bandwidth; EIGRP uses a composite metric (bandwidth and delay); RIPv2 uses simple hop count (max 15).

Each routing protocol utilizes specific quantitative metrics to determine the optimal route to a destination.

Key Concept

Dynamic Routing Protocol Classification and Metric Evaluation

Estimated Time:1m 30s

Question 402Question →

A network administrator is sub-dividing an allocated IPv4 address block of $172.28.16.0/20$ into contiguous subnets with a $/23$ prefix length for datacenter switches. What is the broadcast address of the second $/23$ subnet?

172.28.19.255

172.28.18.255

172.28.17.255

172.28.19.254

Show answer & explanation

Answer: $172.28.19.255$

Answer

The broadcast address of the second

/23

subnet is

172.28.19.255

.

Dividing

172.28.16.0/20

into

/23

subnets yields block increments of

2

in the third octet. The first subnet spans

172.28.16.0

to

172.28.17.255

. The second subnet starts at

172.28.18.0

and spans through

172.28.19.255

. Therefore,

172.28.19.255

is the broadcast address of the second subnet.

Step-by-Step Solution

1

Determine the size and block increment of a

/23

subnet in the third octet.

A

/23

subnet has

32 - 23 = 9

host bits (

2^9 = 512

total addresses). In the third octet, each subnet increments by

512 / 256 = 2

.

Calculating the block size allows identification of subnet boundaries.

2

Calculate the network address boundaries for the first and second subnets starting from

172.28.16.0/20

.

Subnet 1 starts at

172.28.16.0/23

. Subnet 2 starts at

172.28.18.0/23

. Subnet 3 starts at

172.28.20.0/23

.

Adding the third octet increment of

2

identifies each subnet's network ID.

3

Identify the broadcast address for the second subnet (

172.28.18.0/23

).

The broadcast address is one address below the start of Subnet 3 (

172.28.20.0

), which is

172.28.19.255

.

The highest numerical address in a subnet range is reserved as the broadcast address.

Key Concept

IPv4 Subnet Boundary and Broadcast Address Calculation

Question 403Question →

During a network architecture review, an engineer explains how distance-vector routing protocols prevent routing loops on a local subnetwork. Specifically, the protocol prevents a router from advertising a learned route back out of the exact interface from which it received that update. Which mechanism is the engineer describing?

Split horizon

Time to Live (TTL) decrementing

Default gateway redirection

Command line port assignment

Show answer & explanation

Answer: Split horizon

Answer

Split horizon is the loop-prevention rule that prevents a router from advertising a network route back through the same interface from which it was learned.

Split horizon directly prevents distance-vector routing loops by prohibiting a router from sending routing information about a network back out of the interface through which the router originally learned about that network.

Step-by-Step Solution

1

Analyze the operational constraint described in the scenario.

The scenario highlights a rule prohibiting route propagation back toward the source interface.

Distance-vector routing protocols rely on neighbor-reported distance updates and require loop-mitigation techniques.

2

Differentiate between packet-level and protocol-level loop mitigation.

IP header TTL operates on individual packets, whereas split horizon governs route advertisement announcements between routers.

Preventing invalid route announcements prevents routing loops from forming in the routing table initially.

Key Concept

Distance-Vector Loop Prevention Mechanics (Split Horizon)

Question 404Question →

A network administrator partitions the IPv4 block $192.168.50.0/24$ into subnets with a $/27$ prefix length. Which of the following IP addresses represent valid usable host addresses within the third subnet block? (Select TWO.)

Select all that apply

192.168.50.64

192.168.50.67

192.168.50.90

192.168.50.95

Show answer & explanation

Answer: $192.168.50.67$ ; $192.168.50.90$

Answer

The valid usable host addresses for the third

/27

subnet are

192.168.50.67

and

192.168.50.90

.

For the allocated block

192.168.50.0/24

, partitioning with a

/27

subnet mask creates blocks of 32 IP addresses. The third subnet spans from

192.168.50.64

to

192.168.50.95

. Subtracting the network ID (

192.168.50.64

) and broadcast address (

192.168.50.95

) yields the valid host IP range

192.168.50.65

through

192.168.50.94

. Therefore,

192.168.50.67

and

192.168.50.90

are correct valid host IP assignments.

Step-by-Step Solution

1

Determine the block size for a

/27

subnet mask.

A

/27

mask gives

32 - 27 = 5

host bits. Block size

= 2^5 = 32

addresses.

Calculating block size establishes the boundary boundaries for consecutive subnets.

2

Calculate network ranges for the first three subnets starting from

192.168.50.0

.

Subnet 1:

192.168.50.0 - 192.168.50.31

, Subnet 2:

192.168.50.32 - 192.168.50.63

, Subnet 3:

192.168.50.64 - 192.168.50.95

.

Identifying the third subnet boundaries isolates the specific address range being evaluated.

3

Determine the usable host IP address range for the third subnet.

Network ID:

192.168.50.64

, Broadcast Address:

192.168.50.95

, Usable Range:

192.168.50.65

to

192.168.50.94

.

Usable host addresses exclude the first address (Network ID) and last address (Broadcast) of the block.

4

Compare given choices against the usable host IP address range.

192.168.50.67

and

192.168.50.90

fall within

192.168.50.65 - 192.168.50.94

, whereas

192.168.50.64

and

192.168.50.95

are network/broadcast addresses.

Confirms which addresses can actually be statically assigned to network host interfaces.

Key Concept

IPv4 Subnet Boundaries & Usable Host Ranges

Question 405Question →

A network administrator is creating an Access Control List (ACL) on an enterprise router to allow branch office workstations to obtain initial authentication tickets from an internal Active Directory Domain Controller. Which of the following port and transport protocol combinations must be permitted through the firewall to enable this Kerberos ticket-granting service?

TCP and UDP port 88

TCP port 389

UDP port 445

TCP port 636

Show answer & explanation

Answer: TCP and UDP port 88

Answer

TCP and UDP port 88 is the correct combination required for Kerberos authentication services.

Kerberos is the primary authentication mechanism for Active Directory domains. It operates over TCP and UDP port 88 to communicate with the Key Distribution Center (KDC) to request and issue authentication tickets.

Step-by-Step Solution

1

Identify the protocol required for ticket-granting domain authentication.

Active Directory domain ticket-granting services rely on the Kerberos authentication protocol.

Kerberos handles user identification and initial ticket issuing within Active Directory.

2

Determine the standard port and transport protocol binding for Kerberos.

Kerberos uses port 88 over both UDP (for typical ticket requests) and TCP (for larger payload responses).

Firewalls must allow port 88 for both TCP and UDP traffic to maintain proper authentication functionality.

Key Concept

Standard Well-Known Ports and Authentication Protocols

Question 406Question →

A network engineer is provisioning a dedicated VLAN for an isolated building management network that requires static IP address assignments for exactly 27 environmental control sensors. To conserve IPv4 address space while accommodating all devices, which subnet mask should be configured on the network interface?

255.255.255.224

255.255.255.240

255.255.255.192

255.255.255.248

Show answer & explanation

Answer: 255.255.255.224

Answer

The subnet mask 255.255.255.224 (/27) is the correct choice as it provides 30 usable IP addresses, meeting the 27-host requirement with minimal address waste.

To support 27 host devices, the smallest block size needed must yield at least 27 usable IPs. Using 5 host bits gives 2^5 - 2 = 30 usable host IP addresses, which corresponds to a /27 prefix length or 255.255.255.224. This satisfies the requirement while minimizing address waste.

Step-by-Step Solution

1

Determine the required number of usable host IP addresses.

The network requires at least 27 usable host IP addresses.

Each sensor requires a unique static IP address on the subnet.

2

Calculate the host bit requirement using the formula 2^h - 2 >= required_hosts.

For h = 5, 2^5 - 2 = 30 usable hosts (since 2^4 - 2 = 14 is too small).

Two addresses in every IPv4 subnet are reserved for the network ID and broadcast address.

3

Convert the host bits to a CIDR prefix and dotted-decimal subnet mask.

32 total bits - 5 host bits = /27 prefix. In dotted-decimal notation, /27 corresponds to 255.255.255.224.

The fourth octet has 3 network bits enabled (128 + 64 + 32 = 224).

Key Concept

Subnet Host Capacity and Mask Calculation

Estimated Time:1m 15s

Question 407Question →

A network administrator is assigned the IPv4 address block $10.20.0.0/22$ for a new branch location. The administrator needs to partition this block into equal-sized subnets, where each subnet must support at least $50$ usable host IP addresses. What is the maximum number of such subnets that can be created from this address block?

Show answer & explanation

Answer: 16

Answer

The maximum number of subnets that can be created is 16.

To accommodate at least 50 hosts, each subnet requires 6 host bits because

2^6 - 2 = 62

usable addresses (5 host bits only provides 30 usable addresses). A subnet with 6 host bits uses a

/26

prefix (

32 - 6 = 26

). Subnetting a

/22

prefix into

/26

subnets borrows 4 bits (

26 - 22 = 4

), resulting in

2^4 = 16

total subnets.

Step-by-Step Solution

1

Determine the required host bits for each subnet.

6 host bits are required.

The formula for usable hosts per subnet is

2^h - 2

. Setting

h = 6

yields

2^6 - 2 = 62

usable host IP addresses, satisfying the requirement of at least 50 hosts.

2

Determine the subnet CIDR prefix length.

The prefix length is

/26

.

Subtracting 6 host bits from the total 32 IPv4 bits gives

32 - 6 = 26

network bits.

3

Calculate the total number of

/26

subnets within the original

/22

block.

16 subnets.

The difference between prefix lengths is

26 - 22 = 4

borrowed subnet bits. Calculating

2^4

gives 16 subnets.

Key Concept

IPv4 Subnetting and Host Capacity Calculation

Estimated Time:1m 30s

Question 408Question →

An engineer is allocating static IP addresses for client workstations on a local network segment provisioned with the CIDR block $172.16.50.160/27$ . Which of the following IP addresses represent valid usable host addresses within this subnet? (Select TWO.)

Select all that apply

172.16.50.165

172.16.50.160

172.16.50.188

172.16.50.191

Show answer & explanation

Answer: $172.16.50.165$ ; $172.16.50.188$

Answer

The IP addresses

172.16.50.165

and

172.16.50.188

are valid usable host addresses for the

172.16.50.160/27

subnet.

For the CIDR block

172.16.50.160/27

, the prefix length leaves 5 host bits, yielding

2^5 = 32

total addresses per subnet. The block starts at network ID

172.16.50.160

and ends at broadcast address

172.16.50.191

. Subtracting these two reserved addresses gives a usable host IP range of

172.16.50.161

through

172.16.50.190

. Both

172.16.50.165

and

172.16.50.188

fall squarely within this range.

Step-by-Step Solution

1

Calculate the block size and subnet boundaries for a

/27

prefix length.

A

/27

mask reserves

32 - 27 = 5

host bits. The total address block size is

2^5 = 32

addresses.

Knowing the block size allows determination of the starting network ID and ending broadcast address.

2

Determine the network address, broadcast address, and usable host IP range.

Network ID =

172.16.50.160

. Broadcast address =

172.16.50.160 + 31 = 172.16.50.191

. Usable host range =

172.16.50.161

to

172.16.50.190

.

The network address (all host bits 0) and broadcast address (all host bits 1) are reserved and must be excluded from host assignment.

3

Evaluate the candidate options against the usable host IP range.

172.16.50.165

and

172.16.50.188

lie strictly within the range of

172.16.50.161

to

172.16.50.190

.

172.16.50.160

is the network ID and

172.16.50.191

is the broadcast address.

Only IP addresses inside the usable host boundary are assignable to workstation interfaces.

Key Concept

Subnet host range calculation and exclusion of network ID and broadcast address

Question 409Question →

A network engineer is configuring static host IP addresses for a newly created VLAN assigned the subnet block $10.50.160.0/20$ . Which of the following IPv4 addresses are valid, usable host addresses within this subnet? (Select TWO.)

Select all that apply

10.50.160.1

10.50.172.45

10.50.160.0

10.50.175.255

Show answer & explanation

Answer: $10.50.160.1$ ; $10.50.172.45$

Answer

The valid usable host addresses in the subnet are

10.50.160.1

and

10.50.172.45

.

For a

/20

subnet prefix (

255.255.240.0

), the block size in the third octet is

16

(

256 - 240 = 16

). Starting at

10.50.160.0

, the subnet encompasses all addresses up through

10.50.175.255

. Reserving the network address (

10.50.160.0

) and the broadcast address (

10.50.175.255

) yields a usable host range of

10.50.160.1

through

10.50.175.254

. Both

10.50.160.1

and

10.50.172.45

reside within this usable host range.

Step-by-Step Solution

1

Determine the subnet mask and block size

A

/20

prefix corresponds to a subnet mask of

255.255.240.0

. The block size in the third octet is

256 - 240 = 16

.

Calculating the block size identifies the boundary increments for subnets.

2

Calculate the network address and broadcast address

The network address is

10.50.160.0

. The next subnet starts at

10.50.176.0

, making the broadcast address for this subnet

10.50.175.255

.

The broadcast address is one address lower than the start of the next subnet.

3

Identify the usable host IP address range

The usable host range spans from

10.50.160.1

(first usable) to

10.50.175.254

(last usable).

Subtracting 2 reserved addresses (network ID and broadcast) defines the assignable IP range.

4

Evaluate the choices against the usable range

10.50.160.1

and

10.50.172.45

fall inside the usable range, whereas

10.50.160.0

(network ID) and

10.50.175.255

(broadcast) do not.

Host assignments must strictly exclude network and broadcast IP addresses.

Key Concept

Determining usable IPv4 host address boundaries for a given CIDR prefix.

Question 410Question →

A network administrator is configuring a subnet dedicated to security IP cameras in a warehouse facility. The subnet must support at least $58$ host devices, with each camera assigned a static IPv4 address. To minimize wasted IP address space, what is the minimum CIDR prefix length (represented as an integer $N$ for $/N$ ) that should be assigned to this subnet?

Show answer & explanation

Answer: 26

Answer

The minimum CIDR prefix length required to support at least 58 usable host addresses is 26.

To support 58 static host devices, the subnet must contain at least 58 usable host IP addresses. The formula for usable host addresses is

2^h - 2

, where

h

is the number of host bits. Setting

h = 5

gives

2^5 - 2 = 30

usable addresses (insufficient). Setting

h = 6

gives

2^6 - 2 = 62

usable addresses (sufficient). Subtracting 6 host bits from the 32 total bits in an IPv4 address yields

32 - 6 = 26

. Thus, the minimum CIDR prefix length is 26.

Step-by-Step Solution

1

Determine the minimum host bits required

6 host bits (

h = 6

)

Using 5 host bits yields

2^5 - 2 = 30

usable IP addresses, which cannot accommodate 58 devices. Using 6 host bits yields

2^6 - 2 = 62

usable addresses, which satisfies the requirement.

2

Calculate the CIDR prefix length

Prefix length of 26 (

32 - 6 = 26

)

IPv4 addresses consist of 32 total bits. Subtracting the 6 host bits leaves 26 bits dedicated to the network prefix.

Key Concept

Calculating minimum CIDR prefix length from host capacity requirements

Question 411Question →

A network administrator is provisioning a dedicated subnet for a newly established server cluster that requires static IP assignments for exactly $31$ usable host devices. To conserve address space, which of the following is the most efficient subnet mask that satisfies this requirement?

255.255.255.224

255.255.255.192

255.255.255.240

255.255.255.128

Show answer & explanation

Answer: 255.255.255.192

Answer

The subnet mask 255.255.255.192 is the most efficient subnet mask because a /26 network provides 62 usable host IP addresses, satisfying the 31 host requirement with minimal wasted IP addresses.

The option specifying '255.255.255.192' corresponds to a prefix length of

/26

. A

/26

subnet contains

6

host bits, yielding

2^6 = 64

total IP addresses. Subtracting

2

for the reserved network and broadcast addresses yields

62

usable host IPs, which is the smallest standard subnet capable of housing

31

hosts.

Step-by-Step Solution

1

Determine the required number of usable host IP addresses.

The scenario specifies a requirement of

31

usable host IP addresses.

Subnet capacity must accommodate all host interfaces plus non-usable network and broadcast addresses.

2

Calculate required host bits using the formula

2^h - 2 \ge 31

.

For

h = 5

bits:

2^5 - 2 = 30

usable IPs (insufficient). For

h = 6

bits:

2^6 - 2 = 62

usable IPs (sufficient).

Two IP addresses per subnet are reserved for the network ID and broadcast address.

3

Convert the required host bits to a subnet mask.

The prefix length is

32 - 6 = /26

. Converting

/26

to dotted-decimal yields

255.255.255.192

.

A

/26

prefix uses

26

network bits (

11111111.11111111.11111111.11000000_2 = 255.255.255.192

).

Key Concept

IPv4 Subnetting & Usable Host Calculation

Question 412Question →

A network administrator is configuring a management server in an infrastructure VLAN. The default gateway interface for this subnet is assigned the IPv4 address $192.168.4.129$ with a subnet mask of $255.255.255.192$ . Which of the following IPv4 addresses is a valid usable host address that can be assigned to the new server on this subnet?

192.168.4.150

192.168.4.128

192.168.4.191

192.168.4.192

Show answer & explanation

Answer: 192.168.4.150

Answer

192.168.4.150 is the correct choice as it resides within the valid usable host range of the subnet.

The subnet mask

255.255.255.192

(/26) creates subnets of 64 addresses each. The gateway IP

192.168.4.129

belongs to the

192.168.4.128/26

subnet, which has a network address of

192.168.4.128

, a broadcast address of

192.168.4.191

, and a usable host address range of

192.168.4.129

through

192.168.4.190

. The address

192.168.4.150

falls squarely within this usable host range.

Step-by-Step Solution

1

Determine the block size from the subnet mask.

A mask of

255.255.255.192

corresponds to a

/26

prefix length. The block size is

256 - 192 = 64

.

Knowing the block size allows identification of subnet boundaries.

2

Identify the network address containing the default gateway IP

192.168.4.129

.

Subnet blocks increment by

64

:

.0

,

.64

,

.128

,

.192

. The gateway address

192.168.4.129

falls into the

192.168.4.128/26

subnet.

The network address defines the lower bound of the subnet boundary.

3

Calculate the broadcast address and usable host range.

The broadcast address is

192.168.4.191

(one less than the next subnet

192.168.4.192

). The usable host range spans from

192.168.4.129

to

192.168.4.190

.

Usable host addresses exclude the network ID and broadcast address.

4

Evaluate the options against the usable host range.

192.168.4.150

is the only option that falls within

192.168.4.129

–

192.168.4.190

.

Host IP addresses must be strictly within the calculated usable range.

Key Concept

IPv4 Subnet Boundaries and Usable Host Allocation

Question 413Question →

A network administrator is designing a subnetwork scheme for a branch facility assigned the IPv4 block $192.168.16.0/22$ . The design requirement specifies creating subnets that can each support a minimum of $50$ usable host IP addresses for individual department segments while maximizing the total number of subnets created. Which CIDR prefix length meets these requirements, and what is the exact number of usable host IP addresses provided per subnet?

/26

prefix length providing

62

usable host addresses

/26

prefix length providing

64

usable host addresses

/25

prefix length providing

126

usable host addresses

/27

prefix length providing

30

usable host addresses

Show answer & explanation

Answer: $/26$ prefix length providing $62$ usable host addresses

Answer

The

/26

prefix length providing

62

usable host addresses

To accommodate at least

50

usable host IP addresses per subnet, the minimum number of host bits

h

needed is

6

, because

2^6 - 2 = 62 \ge 50

. Subtracting

6

host bits from

32

gives a

/26

prefix. The usable host count is

62

after subtracting the network and broadcast addresses.

Step-by-Step Solution

1

Determine the required number of host bits (

h

).

h = 6

bits, since

2^6 - 2 = 62 \ge 50

.

Five host bits (

2^5 - 2 = 30

) would not provide enough IP addresses, while six bits provides sufficient capacity.

2

Calculate the CIDR prefix length.

Prefix length is

32 - 6 = /26

.

Subtracting the host bits from the 32-bit total IPv4 address length yields the subnet prefix length.

3

Calculate the usable host count per subnet.

2^6 - 2 = 62

usable host IP addresses.

The network address (all host bits 0) and broadcast address (all host bits 1) cannot be assigned to hosts.

Key Concept

IPv4 Subnetting and Usable Host Calculation

Estimated Time:1m 30s

Question 414Question →

A network administrator needs to provision an IPv4 subnet for a new database cluster requiring a maximum of 28 usable host IP addresses. Which CIDR prefix length represents the smallest subnet that meets this requirement?

/27

/28

/26

/29

Show answer & explanation

Answer: /27

Answer

/27 is the smallest CIDR prefix length that provides at least 28 usable host IP addresses.

The prefix length /27 allocates 5 host bits (

32 - 27 = 5

). Calculating

2^5 - 2 = 30

usable host addresses confirms that a /27 subnet is the smallest subnet capacity that safely supports 28 host devices without wasting excess address space.

Step-by-Step Solution

1

Determine the required number of usable IP addresses

The network segment requires at least 28 usable host IP addresses.

Host requirement stated in the scenario.

2

Calculate total IP addresses needed including reserved addresses

Add 2 reserved addresses (network ID and broadcast address):

28 + 2 = 30

total IP addresses.

Every IPv4 subnet reserves the first address for the network ID and the last address for the broadcast address.

3

Find the smallest power of 2 that is greater than or equal to the total required addresses

2^5 = 32

total IP addresses (

32 \ge 30

). Thus, 5 host bits are required.

Subnet block sizes must be powers of 2.

4

Subtract the host bits from 32 to find the CIDR prefix length

32 - 5 = 27

, resulting in a prefix length of /27.

An IPv4 address consists of 32 bits divided into network bits and host bits.

Key Concept

IPv4 CIDR prefix calculation and usable host address capacity

Estimated Time:1m 0s

Question 415Question →

A network administrator is assigned the IPv4 address block $172.28.40.0/23$ to provision subnets for new departmental VLANs. Each VLAN requires a minimum of 31 usable host IP addresses. What is the maximum number of subnets meeting this requirement that can be created from the assigned block?

8

16

4

2

Show answer & explanation

Answer: 8

Answer

The maximum number of valid subnets meeting the requirement is 8.

To support a minimum of 31 usable host IPs, the usable host capacity formula

2^h - 2 \ge 31

must be evaluated. With 5 host bits,

2^5 - 2 = 30

addresses are available, which is insufficient. Therefore, 6 host bits are required, providing

2^6 - 2 = 62

usable host IPs. Subtracting 6 host bits from 32 gives a prefix length of

/26

. Dividing the parent

/23

block into

/26

subnets borrows

26 - 23 = 3

bits, yielding

2^3 = 8

maximum subnets.

Step-by-Step Solution

1

Determine the required host bits for each subnet

6 host bits are needed (

2^6 - 2 = 62

usable addresses). 5 host bits only yield 30 usable addresses (

2^5 - 2 = 30

), which fails the 31 host requirement.

Every IPv4 subnet reserves the first address for network identification and the last address for directed broadcast, requiring 2 addresses to be subtracted from the total host capacity.

2

Calculate the subnet prefix length

Subnet mask prefix length is

/26

(

32 - 6 = 26

).

An IPv4 address contains 32 total bits; subtracting 6 host bits leaves 26 network bits.

3

Calculate the total number of subnets from the parent block

2^{26 - 23} = 2^3 = 8

subnets.

Subdividing a

/23

block into

/26

subnets borrows 3 additional subnet bits (

26 - 23 = 3

).

Key Concept

Subnet Host Capacity Formula and Prefix Allocation

Estimated Time:1m 30s

Question 416Question →

A network engineer is configuring static host IP addresses for infrastructure servers within the allocated CIDR block $10.240.16.0/21$ . Which of the following IP addresses are valid usable host addresses within this subnet? (Select TWO.)

Select all that apply

10.240.16.0

10.240.17.100

10.240.20.255

10.240.23.255

Show answer & explanation

Answer: $10.240.17.100$ ; $10.240.20.255$

Answer

The addresses

10.240.17.100

and

10.240.20.255

are valid usable host IP addresses within the

10.240.16.0/21

subnet.

For the subnet

10.240.16.0/21

, the usable host range spans from

10.240.16.1

through

10.240.23.254

. Both

10.240.17.100

and

10.240.20.255

fall within this valid range. Note that

10.240.20.255

is valid because the broadcast address for the entire

/21

block is

10.240.23.255

.

Step-by-Step Solution

1

Determine the subnet mask and block size from CIDR notation.

A

/21

mask corresponds to

255.255.248.0

. The block size in the third octet is

256 - 248 = 8

.

Calculating block size establishes the boundary of the subnetwork.

2

Identify the network ID and broadcast address for the

10.240.16.0/21

subnet.

Network ID:

10.240.16.0

. Next subnet ID:

10.240.24.0

. Broadcast Address:

10.240.23.255

.

The network address is the first address in the block, and the broadcast address is the last address before the next subnet boundary.

3

Determine the range of usable host IP addresses.

Usable host range is from

10.240.16.1

to

10.240.23.254

.

The first usable IP is Network ID

+ 1

, and the last usable IP is Broadcast ID

- 1

.

4

Evaluate the candidate options against the valid host range.

10.240.17.100

and

10.240.20.255

fall strictly between

10.240.16.1

and

10.240.23.254

.

10.240.16.0

is the Network ID and

10.240.23.255

is the Broadcast ID.

Only IP addresses between the network address and broadcast address can be assigned to host interfaces.

Key Concept

Subnet Host Range Boundaries & Intermediate Octet .255 Host Addresses

Question 417Question →

A network administrator is allocating IP addresses for a high-availability firewall cluster. The segment is assigned a subnet mask of $255.255.255.248$ . How many usable host IP addresses are available for assignment in this subnetwork?

Show answer & explanation

Answer: 6

Answer

There are 6 usable host IP addresses available in a subnetwork assigned a

255.255.255.248

subnet mask.

A subnet mask of

255.255.255.248

corresponds to a

/29

CIDR prefix length. This leaves 3 host bits (

32 - 29 = 3

), providing

2^3 = 8

total IP addresses per subnet block. Subtracting 2 for the network ID (all host bits 0) and the directed broadcast address (all host bits 1) leaves exactly 6 usable host IP addresses.

Step-by-Step Solution

1

Calculate the total block size of IP addresses in the subnet

256 - 248 = 8

total IP addresses

The subnet mask octet of 248 indicates a block size of 8 total addresses (

2^3 = 8

host bits).

2

Subtract reserved network and broadcast addresses

8 - 2 = 6

usable host IP addresses

The network ID (first address) and broadcast address (last address) cannot be assigned to hosts.

Key Concept

Calculating Usable Host IP Addresses from Subnet Masks

Estimated Time:1m 0s

Question 418Question →

A network engineer is configuring static host IP addresses for infrastructure devices within the newly assigned network block $10.88.16.0/20$ . Which of the following IPv4 addresses represents the last valid, usable host address in this subnet?

10.88.31.254

10.88.31.255

10.88.32.254

10.88.30.254

Show answer & explanation

Answer: $10.88.31.254$

Answer

10.88.31.254

For the subnet

10.88.16.0/20

, the subnet mask is

255.255.240.0

, giving a block size of 16 in the third octet. The network span is from

10.88.16.0

to

10.88.31.255

. Reserving

10.88.16.0

for the network ID and

10.88.31.255

for the broadcast address leaves

10.88.16.1

through

10.88.31.254

as usable host addresses. Therefore,

10.88.31.254

is the last valid usable host IP.

Step-by-Step Solution

1

Calculate the subnet mask and third-octet block size for a /20 prefix.

A /20 prefix corresponds to the mask

255.255.240.0

. The block size in the third octet is

256 - 240 = 16

.

The third octet contains 4 network bits and 4 host bits (

2^4 = 16

).

2

Determine the network boundary and broadcast address.

Starting at network ID

10.88.16.0

, adding 16 to the third octet gives the next subnet starting at

10.88.32.0

. Thus, the broadcast address is

10.88.31.255

.

The broadcast address is one address before the start of the next subnet.

3

Identify the last usable host IP address.

Subtracting 1 from the broadcast address (

10.88.31.255 - 1

) yields

10.88.31.254

.

The last usable host address is always the IP immediately preceding the subnet broadcast address.

Key Concept

IPv4 Subnet Host Boundaries and Allocation

Estimated Time:1m 15s

Question 419Question →

A network administrator is assigning static IP addresses to server interfaces within the network block $192.168.100.0/25$ . Which of the following IPv4 addresses are valid, usable host addresses within this subnet? (Select TWO.)

Select all that apply

192.168.100.1

192.168.100.126

192.168.100.0

192.168.100.127

Show answer & explanation

Answer: $192.168.100.1$ ; $192.168.100.126$

Answer

The addresses

192.168.100.1

and

192.168.100.126

are the correct usable host IP addresses.

For a CIDR block of

192.168.100.0/25

, the address range spans from

192.168.100.0

through

192.168.100.127

. The network address is

192.168.100.0

and the broadcast address is

192.168.100.127

. Excluding these two reserved addresses leaves the usable host range as

192.168.100.1

through

192.168.100.126

. Both

192.168.100.1

and

192.168.100.126

fall strictly within this usable range.

Step-by-Step Solution

1

Determine the total block size and network boundary for CIDR prefix /25.

A /25 prefix provides

2^{32-25} = 2^7 = 128

total IP addresses (

192.168.100.0

to

192.168.100.127

).

The CIDR prefix defines the number of host bits (

32 - 25 = 7

bits).

2

Identify the reserved Network ID and Broadcast address.

The Network ID is

192.168.100.0

(first address) and the Broadcast address is

192.168.100.127

(last address).

The first IP address identifies the network itself and the last IP address is reserved for broadcast traffic within the subnet.

3

Calculate the range of usable host IP addresses.

Usable host IP range is from

192.168.100.1

to

192.168.100.126

, yielding

128 - 2 = 126

usable host addresses.

Usable host addresses consist of all IP addresses between the Network ID and the Broadcast address.

Key Concept

Usable IPv4 Host Range Calculation for Subnets

Estimated Time:1m 30s

Question 420Question →

A network security analyst auditing firewall logs observes traffic originating from an internal host with IP address $10.50.84.142/21$ . To configure a targeted network access rule, the analyst must determine the exact boundaries of the local subnet. Which of the following identifies the correct network address and broadcast address for the subnet containing this host?

Network address

10.50.80.0

and broadcast address

10.50.87.255

Network address

10.50.84.0

and broadcast address

10.50.84.255

Network address

10.50.80.0

and broadcast address

10.50.88.255

Network address

10.50.80.1

and broadcast address

10.50.87.254

Show answer & explanation

Answer: Network address $10.50.80.0$ and broadcast address $10.50.87.255$

Answer

Network address

10.50.80.0

and broadcast address

10.50.87.255

For the host IP address

10.50.84.142/21

, the

/21

mask corresponds to

255.255.248.0

. The third octet operates in block increments of

8

. The network boundary starts at

80

(

10 \times 8

), making the network address

10.50.80.0

. The subnet spans from

10.50.80.0

through

10.50.87.255

, establishing

10.50.87.255

as the broadcast address.

Step-by-Step Solution

1

Determine the subnet mask from the CIDR prefix length

/21

.

A

/21

mask has

21

network bits:

255.255.248.0

.

Converting CIDR notation to dotted-decimal format identifies which octet contains the subnet boundary.

2

Calculate the block size (increment) in the third octet.

Block size =

256 - 248 = 8

.

Subnet boundaries in the third octet recur in multiples of

8

.

3

Identify the network address by finding the highest multiple of

8

less than or equal to the host's third octet (

84

).

Multiples of

8

:

0, 8, 16, \dots, 72, 80, 88

. Since

80 \leq 84 < 88

, the network third octet is

80

. Network address =

10.50.80.0

.

The network address uses all zeros in the host portion.

4

Calculate the broadcast address for this subnet.

The next subnet begins at

10.50.88.0

, so the broadcast address is

10.50.87.255

.

The broadcast address is one IP less than the next subnet's network address (all host bits set to 1).

Key Concept

IPv4 Subnet Boundaries and Broadcast Address Calculation

Estimated Time:1m 30s

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