IPv4 Addressing and Subnetting Practice Questions — Cisco CCNA

Question 1Question →

A network engineer is configuring IP address assignments for client workstations within a corporate VLAN allocated the $10.150.32.0/21$ IPv4 block. Which two IP addresses are valid host addresses that can be assigned to devices on this subnet?

Select all that apply

10.150.35.255

10.150.39.254

10.150.39.255

10.150.40.0

Show answer & explanation

Answer: $10.150.35.255$ ; $10.150.39.254$

Answer

The valid host addresses are

10.150.35.255

and

10.150.39.254

.

For the prefix

/21

, the subnet mask is

255.255.248.0

, giving a block size of

8

in the third octet. The subnet

10.150.32.0/21

ranges from network address

10.150.32.0

to broadcast address

10.150.39.255

. The usable host range is

10.150.32.1

through

10.150.39.254

. Both

10.150.35.255

and

10.150.39.254

fall within this usable range.

Step-by-Step Solution

1

Determine the subnet mask and block size for a /21 prefix

Prefix /21 corresponds to subnet mask 255.255.248.0. The third octet block size is 256 - 248 = 8.

Calculating block size in the interesting octet establishes subnet boundaries.

2

Calculate the network and broadcast addresses for the 10.150.32.0/21 subnet

Network address is 10.150.32.0. The next subnet starts at 10.150.40.0. Therefore, the broadcast address is 10.150.39.255.

Subnet boundaries define the start and end of the address block.

3

Determine the usable host IP address range

Usable host range spans from 10.150.32.1 to 10.150.39.254 inclusive.

Usable host IPs exclude the network ID (all host bits 0) and broadcast ID (all host bits 1).

4

Evaluate the given options against the usable host range

10.150.35.255 and 10.150.39.254 lie between 10.150.32.1 and 10.150.39.254. Address 10.150.39.255 is the broadcast IP and 10.150.40.0 belongs to the next subnet.

Verifies which IP addresses fall within the valid host envelope.

Key Concept

Usable host address range determination for IPv4 subnets with non-byte-aligned prefixes

Question 2Question →

A network administrator is configuring a interface on a switch assigned to the $192.168.4.128/26$ IPv4 network segment. Which IPv4 address represents a valid usable host address for this interface?

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

The address

192.168.4.150

is a valid usable host address in the

192.168.4.128/26

subnet.

For a

/26

prefix, 6 bits are available for hosts, providing a total of

2^6 = 64

addresses. The

192.168.4.128/26

subnet spans from

192.168.4.128

(Network ID) through

192.168.4.191

(Broadcast Address). The usable host range is

192.168.4.129

to

192.168.4.190

. The address

192.168.4.150

falls within this valid usable range.

Step-by-Step Solution

1

Determine the total block size from the prefix length.

A

/26

prefix leaves

32 - 26 = 6

host bits. The block size is

2^6 = 64

addresses.

Calculating block size establishes the boundary for subnet ranges.

2

Identify the network and broadcast addresses of the subnet.

The network address is

192.168.4.128

. The broadcast address is

192.168.4.128 + 64 - 1 = 192.168.4.191

.

Network and broadcast addresses mark the unassignable outer boundaries of the IPv4 block.

3

Calculate the range of usable host addresses.

First usable host =

192.168.4.129

; Last usable host =

192.168.4.190

.

Usable host addresses consist of all IP addresses strictly between the network ID and broadcast address.

4

Evaluate the given choices against the usable host range.

The address

192.168.4.150

falls within the range

[192.168.4.129, 192.168.4.190]

.

Only IP addresses in the usable range can be statically assigned to network interfaces.

Key Concept

IPv4 Subnet Boundaries and Usable Host Identification

Estimated Time:1m 0s

Question 3Question →

A network engineer is configuring a static IPv4 address for a server residing on a network segment assigned the network ID $172.24.16.0/22$ . Which IPv4 address represents a valid usable host address that can be assigned to this server?

172.24.19.254

172.24.19.255

172.24.20.1

172.32.16.5

Show answer & explanation

Answer: $172.24.19.254$

Answer

172.24.19.254

is a valid usable host address within the

172.24.16.0/22

subnet.

The IP address

172.24.19.254

is the last valid usable host IP address in the

172.24.16.0/22

subnet range, which spans from

172.24.16.1

to

172.24.19.254

.

Step-by-Step Solution

1

Determine the subnet mask and block size

A prefix of

/22

corresponds to a subnet mask of

255.255.252.0

. The block size in the third octet is

256 - 252 = 4

.

Calculating the block size identifies how the subnets increment across the third octet.

2

Identify the network boundary and next subnet

The network address is

172.24.16.0

. Adding the block size of

4

to the third octet gives the next subnet network address of

172.24.20.0

.

The next subnet boundary defines the upper limit of the current subnet.

3

Calculate the broadcast address and usable host range

The broadcast address is

172.24.19.255

(one IP before

172.24.20.0

). The usable host range spans from

172.24.16.1

through

172.24.19.254

.

Usable host addresses lie strictly between the network ID and the broadcast ID.

Key Concept

IPv4 Subnet Boundaries and Usable Host Identification

Question 4Question →

A system engineer is configuring a static IPv4 address for a database server located on a segment designated with a network ID of $172.30.100.0$ and a subnet mask of $255.255.254.0$ . Company security policy requires that the configured address must be a valid usable host address within the assigned subnet and must also belong to the RFC 1918 private IPv4 address space. Which IPv4 address is a valid host address for this server?

172.30.101.250

172.30.101.255

172.32.100.50

172.30.102.1

Show answer & explanation

Answer: $172.30.101.250$

Answer

172.30.101.250

The subnet mask

255.255.254.0

creates subnets with a block size of

2

in the third octet. For the network ID

172.30.100.0

, the subnet encompasses

172.30.100.0

through

172.30.101.255

, giving a usable host range of

172.30.100.1

to

172.30.101.254

. The IP address

172.30.101.250

falls cleanly inside this usable host range and is also a valid RFC 1918 private address (within

172.16.0.0/12

).

Step-by-Step Solution

1

Determine the CIDR prefix and block size for subnet mask 255.255.254.0.

The subnet mask 255.255.254.0 corresponds to a /23 prefix length. The block size in the third octet is calculated as 256 - 254 = 2.

Calculating the block size identifies how the subnets increment across octets.

2

Calculate the network ID, broadcast address, and usable host range for the 172.30.100.0/23 subnet.

The network address is 172.30.100.0, the broadcast address is 172.30.101.255, and the range of valid usable host IPs is 172.30.100.1 through 172.30.101.254.

Host IP addresses cannot use the subnet network ID (first IP) or broadcast address (last IP).

3

Verify compliance with RFC 1918 private IPv4 address boundaries.

RFC 1918 specifies the Class B private IP range as 172.16.0.0 to 172.31.255.255 (172.16.0.0/12 block). The address 172.30.101.250 falls inside this range.

Addresses starting with 172.32.x.x fall outside RFC 1918 limits and are public addresses.

Key Concept

IPv4 Subnet Mask Calculation and RFC 1918 Private Address Boundaries

Question 5Question →

A network technician is configuring static IPv4 addresses for new network monitoring probes on a subnetwork assigned the address block $192.168.50.144/28$ . Which of the following IP addresses are valid usable host addresses on this subnet? (Select two.)

Select all that apply

192.168.50.146

192.168.50.157

192.168.50.144

192.168.50.159

Show answer & explanation

Answer: $192.168.50.146$ ; $192.168.50.157$

Answer

The addresses

192.168.50.146

and

192.168.50.157

are valid usable host addresses on the

192.168.50.144/28

subnet.

For the network block

192.168.50.144/28

, the

/28

prefix yields a block size of 16 (

256 - 240

). The subnet boundaries are defined by the network ID

192.168.50.144

and the broadcast address

192.168.50.159

. Therefore, valid host IP addresses must fall strictly between

192.168.50.145

and

192.168.50.158

. The addresses

192.168.50.146

and

192.168.50.157

fall within this range.

Step-by-Step Solution

1

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

A

/28

prefix corresponds to a subnet mask of

255.255.255.240

. The block size (increment) per subnet is

256 - 240 = 16

.

Calculating the block size identifies the boundary and increment for the subnet range.

2

Identify the network ID and broadcast address for the subnet.

The network ID is

192.168.50.144

. The next subnet starts at

144 + 16 = 160

, making the broadcast address

192.168.50.159

.

Network ID and broadcast address delineate the extreme boundaries of the subnet.

3

Calculate the usable host address range.

The first usable host address is

192.168.50.145

and the last usable host address is

192.168.50.158

.

Usable host addresses exclude the network ID and broadcast address (

2^n - 2

).

4

Evaluate the candidate options against the usable host range.

192.168.50.146

and

192.168.50.157

lie strictly within the range

192.168.50.145

–

192.168.50.158

, while

192.168.50.144

is the network ID and

192.168.50.159

is the broadcast address.

Only IP addresses within the usable host range can be statically configured on end-device interfaces.

Key Concept

IPv4 Usable Host Address Range Calculation

Question 6Question →

A network administrator is provisioning a new VLAN for an enterprise VoIP deployment that requires support for up to $500$ usable IP phones. The network block allocated for this subnetwork starts at $172.25.100.0$ . To accommodate the required number of host devices while minimizing unallocated address space, what is the broadcast address of this subnetwork?

172.25.101.255

172.25.100.255

172.25.101.254

172.32.101.255

Show answer & explanation

Answer: $172.25.101.255$

Answer

The broadcast address of the subnetwork is

172.25.101.255

.

To support 500 usable host devices, 9 host bits are required (

2^9 - 2 = 510

hosts). This yields a prefix length of

/23

(

32 - 9 = 23

). Starting at network address

172.25.100.0

, a

/23

subnet encompasses the range

172.25.100.0

through

172.25.101.255

, making

172.25.101.255

the correct broadcast address.

Step-by-Step Solution

1

Determine the minimum number of host bits required.

n = 9

host bits because

2^9 - 2 = 510 \ge 500

.

A host bit count of 8 (

2^8 - 2 = 254

) is insufficient for 500 hosts.

2

Calculate the prefix length and subnet mask.

Prefix length is

/23

(

32 - 9 = 23

), corresponding to subnet mask

255.255.252.0

.

Subtracting 9 host bits from 32 total bits yields a 23-bit network prefix.

3

Determine the subnet boundaries for

172.25.100.0/23

.

Network address is

172.25.100.0

, usable host range is

172.25.100.1

to

172.25.101.254

, and the broadcast address is

172.25.101.255

.

The block size in the third octet for a /23 subnet is

256 - 252 = 4

or

2^1 = 2

, spanning

172.25.100.0

through

172.25.101.255

.

Key Concept

Subnet Mask Derivation and Broadcast Address Calculation

Estimated Time:1m 30s

Question 7Question →

A network administrator is selecting an IPv4 subnetwork block for a new internal department segment that requires support for up to $60$ usable host interfaces. The network design policy requires using an RFC 1918 private IPv4 address block and selecting the most efficient prefix length to avoid wasting addresses. Which subnetwork address and CIDR prefix satisfies all requirements?

10.40.16.0/26

172.32.16.0/26

10.40.16.0/27

192.168.10.0/25

Show answer & explanation

Answer: $10.40.16.0/26$

Answer

The subnetwork

10.40.16.0/26

meets all design requirements by using an RFC 1918 private IPv4 address and providing

62

usable host IP addresses, which is the most efficient subnet size for

60

hosts.

The subnetwork

10.40.16.0/26

utilizes an address within the RFC 1918 Class A private block (

10.0.0.0/8

). A

/26

prefix leaves

6

host bits (

32 - 26 = 6

), providing

2^6 - 2 = 62

usable host IP addresses. This accommodates the required

60

host interfaces with the minimum possible address waste.

Step-by-Step Solution

1

Determine the required number of host bits (

h

) for

60

usable hosts

Using the formula

2^h - 2 \ge 60

, we test

h = 5 \rightarrow 30

usable hosts (too small) and

h = 6 \rightarrow 62

usable hosts (sufficient). Thus,

6

host bits are needed.

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

2

Calculate the CIDR prefix length

Prefix length

= 32 - 6 = /26

.

An IPv4 address consists of 32 total bits.

3

Verify RFC 1918 private IPv4 address range compliance

10.40.16.0

is inside the

10.0.0.0/8

private address block (

10.0.0.0

to

10.255.255.255

).

RFC 1918 defines private address spaces:

10.0.0.0/8

,

172.16.0.0/12

, and

192.168.0.0/16

.

Key Concept

IPv4 Subnet Sizing and RFC 1918 Private Addressing

Question 8Question →

A network engineer is assigned the IPv4 subnetwork block $192.168.100.0/23$ for a high-density enterprise server rack. What is the maximum number of usable host IPv4 addresses that can be assigned to active network interfaces within this subnet?

Show answer & explanation

Answer: 510

Answer

510 usable host addresses

A

/23

prefix leaves

32 - 23 = 9

bits for host addresses. Total addresses equal

2^9 = 512

. Subtracting 2 for the network address (

192.168.100.0

) and broadcast address (

192.168.101.255

) leaves exactly 510 usable host IP addresses.

Step-by-Step Solution

1

Calculate the number of host bits from the prefix length.

Host bits =

32 - 23 = 9

bits

An IPv4 address consists of 32 bits. Subtracting the prefix length yields the remaining host portion bits.

2

Calculate the total number of IPv4 addresses in the subnetwork block.

2^9 = 512

total addresses

The formula

2^n

gives the complete block size in total IP addresses.

3

Subtract reserved subnetwork and broadcast addresses to determine usable host capacity.

512 - 2 = 510

usable hosts

The lowest address in the subnet acts as the subnet ID, and the highest address acts as the directed broadcast address; neither can be assigned to host interfaces.

Key Concept

IPv4 Usable Host Calculation

Question 9Question →

A network engineer is configuring static IPv4 addresses for new host interfaces within a server cluster assigned the subnetwork block $172.24.16.0/21$ . Which TWO of the following IPv4 addresses are valid, usable host addresses within this subnetwork?

Select all that apply

172.24.22.255

172.24.23.255

172.24.16.1

172.32.20.5

Show answer & explanation

Answer: $172.24.22.255$ ; $172.24.16.1$

Answer

The valid usable host addresses for the

172.24.16.0/21

subnetwork are

172.24.22.255

and

172.24.16.1

.

For a /21 prefix (

255.255.248.0

), the third octet increments by 8. The network block starting at

172.24.16.0

covers

172.24.16.0

to

172.24.23.255

. The network address is

172.24.16.0

and the broadcast address is

172.24.23.255

. Therefore, the range of usable host addresses is

172.24.16.1

through

172.24.23.254

. Both

172.24.16.1

and

172.24.22.255

fall within this valid usable host range.

Step-by-Step Solution

1

Determine the subnet mask and block size for CIDR prefix /21.

A /21 prefix corresponds to the dotted-decimal subnet mask

255.255.248.0

. The block size in the third octet is

256 - 248 = 8

.

Calculating block size establishes the boundary interval for networks incrementing in the third octet.

2

Calculate the network address, broadcast address, and usable host range.

Network ID:

172.24.16.0

; Broadcast address:

172.24.23.255

; Usable host range:

172.24.16.1

through

172.24.23.254

.

The third octet spans from

16

up to

16 + 8 - 1 = 23

.

3

Evaluate candidate addresses against the usable host range.

172.24.16.1

is the first usable host address.

172.24.22.255

falls strictly between

172.24.16.1

and

172.24.23.254

.

172.24.23.255

is the reserved broadcast address.

172.32.20.5

is outside the network and not in the private RFC 1918 Class B space.

Identify which candidates lie within the inclusive host boundary without matching the network or broadcast ID.

Key Concept

Subnet Boundary and Usable Host Range Calculation for Supernets/Subnets wider than /24

Question 10Question →

A network engineer is designing a Variable Length Subnet Mask (VLSM) IP addressing scheme for a newly provisioned branch office using the summary block $192.168.10.0/24$ . Subnets must be allocated sequentially without wasting address space, starting from the lowest available IP address in descending order of host requirements:

- Subnet A (Staff): Requires $55$ usable host IPv4 addresses
- Subnet B (Sales): Requires $28$ usable host IPv4 addresses
- Subnet C (VoIP): Requires $12$ usable host IPv4 addresses
- Subnet D (WAN Link): Requires $2$ usable host IPv4 addresses

What is the broadcast address of Subnet C?

192.168.10.110

192.168.10.111

192.168.10.127

172.32.10.111

Show answer & explanation

Answer: 192.168.10.111

Answer

The broadcast address of Subnet C is 192.168.10.111.

Subnet C requires 12 host addresses. Satisfying

2^n - 2 \ge 12

requires

n=4

host bits (

14

usable IPs), which yields a

/28

prefix with a block size of 16. Allocating Subnet A (

55

hosts

\rightarrow /26

, range .0–.63) and Subnet B (

28

hosts

\rightarrow /27

, range .64–.95) places Subnet C at network address 192.168.10.96. Adding the block size of 16 gives the subnet boundary 192.168.10.96 to 192.168.10.111, making 192.168.10.111 the broadcast address.

Step-by-Step Solution

1

Determine the prefix and range for Subnet A (55 hosts)

Subnet A requires a /26 prefix (block size 64). Subnet A spans 192.168.10.0 to 192.168.10.63.

Formula

2^n - 2 \ge 55

yields

n=6

host bits (

62

usable IPs). The prefix is

/26

(

32-6

).

2

Determine the prefix and range for Subnet B (28 hosts)

Subnet B starts at 192.168.10.64 with a /27 prefix (block size 32). Subnet B spans 192.168.10.64 to 192.168.10.95.

Formula

2^n - 2 \ge 28

yields

n=5

host bits (

30

usable IPs). The prefix is

/27

(

32-5

).

3

Calculate the network address, block size, and broadcast address for Subnet C (12 hosts)

Subnet C starts at 192.168.10.96 with a /28 prefix (block size 16). The network range is 192.168.10.96/28, giving a broadcast address of 192.168.10.111.

Formula

2^n - 2 \ge 12

yields

n=4

host bits (

14

usable IPs). The prefix is

/28

(

32-4

). Adding the block size of 16 to the network start 192.168.10.96 yields a range of 192.168.10.96 to 192.168.10.111.

Key Concept

VLSM block allocation order and subnet broadcast address determination

Estimated Time:2m 0s

Question 11Question →

A network administrator is designing a Variable Length Subnet Mask (VLSM) scheme starting from the $172.16.128.0/18$ network block. Department A requires a subnet that supports at least $500$ usable host addresses. If the administrator allocates the first available subnet of the minimum required size to Department A, which TWO statements regarding this allocation are true?

Select all that apply

The subnet mask required for Department A is

255.255.254.0

.

The broadcast address for Department A's allocated subnet is

172.16.129.255

.

The allocated subnet provides a maximum of

512

usable host addresses for Department A.

The allocated subnet address block falls outside the RFC 1918 private IPv4 address space.

Show answer & explanation

Answer: The subnet mask required for Department A is $255.255.254.0$ .; The broadcast address for Department A's allocated subnet is $172.16.129.255$ .

Answer

The correct statements are that the subnet mask required for Department A is

255.255.254.0

and the broadcast address for Department A's allocated subnet is

172.16.129.255

.

To support 500 usable host addresses, 9 host bits are necessary because

2^9 - 2 = 510

usable addresses. This yields a /

23

prefix, which translates to a subnet mask of

255.255.254.0

. Allocating the first available /

23

block from

172.16.128.0/18

results in the network address

172.16.128.0

and broadcast address

172.16.129.255

.

Step-by-Step Solution

1

Determine the required host bits and prefix length for Department A.

Department A requires

\ge 500

usable hosts. Using the formula

2^h - 2 \ge 500

,

h = 9

host bits (

2^9 - 2 = 510

usable host addresses). The prefix length is

32 - 9 = /23

.

Host bits must satisfy the required host capacity while preserving network bits.

2

Convert prefix length /

23

to dotted-decimal subnet mask format.

A /

23

prefix corresponds to mask

255.255.254.0

.

The first 23 bits are set to 1 (

11111111.11111111.11111110.00000000

).

3

Calculate the subnet boundaries for the first /

23

block within

172.16.128.0/18

.

Network address:

172.16.128.0/23

. Usable range:

172.16.128.1

–

172.16.129.254

. Broadcast address:

172.16.129.255

.

The block size in the 3rd octet is

256 - 254 = 2

, covering 3rd octet values 128 and 129.

4

Verify RFC 1918 private address classification.

The range

172.16.128.0

–

172.16.129.255

is inside the

172.16.0.0/12

private block.

RFC 1918 designates

172.16.0.0

to

172.31.255.255

as private IPv4 space.

Key Concept

VLSM Subnet Sizing and Boundary Calculation

Estimated Time:2m 0s

Question 12Question →

A network administrator needs to assign a single usable IPv4 host address to a server connected to the 192.168.1.16/28 subnet. Which of the following IPv4 addresses is a valid usable host address for this subnet?

192.168.1.16

192.168.1.25

192.168.1.31

192.168.1.32

Show answer & explanation

Answer: 192.168.1.25

Answer

The address 192.168.1.25 is a valid usable host address.

For the network block 192.168.1.16/28, the subnet range begins at the network address 192.168.1.16 and ends at the broadcast address 192.168.1.31. Consequently, the assignable usable host address range spans from 192.168.1.17 through 192.168.1.30. The address 192.168.1.25 falls directly within this valid range.

Step-by-Step Solution

1

Determine the block size from the prefix length.

A /28 prefix uses a subnet mask of 255.255.255.240. The block size is 256 - 240 = 16.

Calculating the block size identifies the increments between adjacent subnets.

2

Identify the network address and broadcast address for the subnet.

Network address is 192.168.1.16. Broadcast address is 192.168.1.31 (16 + 16 - 1).

The network address is the start of the subnet block, and the broadcast address is the last address in the block.

3

Determine the range of usable host addresses.

Usable host addresses range from 192.168.1.17 to 192.168.1.30.

Usable host IPs exclude both the network address and the broadcast address.

Key Concept

IPv4 Subnet Boundaries and Usable Host Identification

Question 13Question →

A network administrator is evaluating the IPv4 configuration on a core router interface assigned the IP address $172.31.223.254/22$ . Which of the following statements regarding this IPv4 addressing assignment are correct? (Select TWO.)

Select all that apply

The configured interface IP address represents the last usable host IP address within the subnet.

The IP address falls within the RFC 1918 private IPv4 address space.

The subnet provides a maximum capacity of

1,024

usable host IP addresses for connected endpoints.

The IP address is a globally routable public address because the Class B private space ends at

172.30.255.255

.

Show answer & explanation

Answer: The configured interface IP address represents the last usable host IP address within the subnet.; The IP address falls within the RFC 1918 private IPv4 address space.

Answer

The statements confirming that the configured interface IP address is the last usable host within the subnet and that the IP address belongs to the RFC 1918 private IPv4 space are correct.

The subnet mask 255.255.252.0 (/22) creates subnets in increments of 4 in the third octet. For 172.31.223.254, the network address is 172.31.220.0 and the broadcast address is 172.31.223.255. Therefore, 172.31.223.254 is the last assignable host address in the subnet. Furthermore, RFC 1918 specifies the Class B private range as 172.16.0.0 to 172.31.255.255, so this address is a valid private IPv4 address.

Step-by-Step Solution

1

Determine host bits and total/usable address capacity for a /22 prefix.

A /22 prefix leaves

32 - 22 = 10

host bits. Total addresses =

2^{10} = 1,024

. Usable host capacity =

2^{10} - 2 = 1,022

host addresses.

Network ID and broadcast ID cannot be assigned to individual endpoints.

2

Calculate subnet network boundaries for

172.31.223.254/22

.

The subnet mask is

255.255.252.0

. The third octet block size is

256 - 252 = 4

. Multiplying by 55 yields

55 \times 4 = 220

. The network ID is

172.31.220.0/22

, and the broadcast address is

172.31.223.255

.

Finding the block size in the interesting octet identifies the network boundary and broadcast address.

3

Determine the usable host IP address range.

The usable range is

172.31.220.1

through

172.31.223.254

. Thus,

172.31.223.254

is the last assignable host address.

Usable host addresses lie strictly between the network ID and the broadcast ID.

4

Verify RFC 1918 private address range compliance.

RFC 1918 designates

172.16.0.0/12

(

172.16.0.0 - 172.31.255.255

) as private space.

172.31.223.254

falls directly within this block.

Checking official RFC 1918 prefix allocations determines public versus private status.

Key Concept

IPv4 subnet boundary calculation, usable host calculation (

2^n - 2

), and RFC 1918 private IP address range boundaries.

Question 14Question →

An enterprise network architect is designing a Variable Length Subnet Mask (VLSM) address allocation plan starting at the base private IPv4 block $10.200.16.0/20$ . Three internal subnets must be allocated sequentially in contiguous order from the lowest available boundary:

- Subnet A (WLAN Clients): Requires at least $1,000$ usable host IP addresses
- Subnet B (VoIP Phones): Requires at least $450$ usable host IP addresses
- Subnet C (Management): Requires at least $60$ usable host IP addresses

If Subnet A is allocated first, followed immediately by Subnet B, and then Subnet C, what is the valid usable host IPv4 address range for Subnet C?

10.200.22.1 to 10.200.22.62

10.200.22.0 to 10.200.22.63

10.200.20.1 to 10.200.20.62

10.200.21.1 to 10.200.21.62

Show answer & explanation

Answer: 10.200.22.1 to 10.200.22.62

Answer

The valid usable host IPv4 address range for Subnet C is 10.200.22.1 to 10.200.22.62.

To calculate VLSM subnets sequentially from the base block 10.200.16.0/20:
1. Subnet A requires 1,000 usable host IPs. The smallest power of 2 minus 2 that satisfies this is 2^10 - 2 = 1,022 (10 host bits, /22 prefix). Allocated block: 10.200.16.0 to 10.200.19.255.
2. Subnet B starts at 10.200.20.0 and requires 450 usable host IPs. 2^9 - 2 = 510 (9 host bits, /23 prefix). Allocated block: 10.200.20.0 to 10.200.21.255.
3. Subnet C starts at 10.200.22.0 and requires 60 usable host IPs. 2^6 - 2 = 62 (6 host bits, /26 prefix). Network ID is 10.200.22.0 and broadcast is 10.200.22.63.
Thus, the usable host range for Subnet C is 10.200.22.1 through 10.200.22.62.

Step-by-Step Solution

1

Calculate host bits and prefix length for Subnet A (1,000 hosts).

Host bits needed:

h = 10

(

2^{10} - 2 = 1,022 \ge 1,000

). Prefix length:

/22

(

32 - 10

). Address block:

10.200.16.0/22

, spanning

10.200.16.0

through

10.200.19.255

.

Subnetting formula

2^h - 2

requires 10 host bits to fit 1,000 usable host addresses.

2

Determine starting address and size for Subnet B (450 hosts).

Start address:

10.200.20.0

. Host bits needed:

h = 9

(

2^9 - 2 = 510 \ge 450

). Prefix length:

/23

(

32 - 9

). Address block:

10.200.20.0/23

, spanning

10.200.20.0

through

10.200.21.255

.

Sequential allocation places Subnet B immediately after Subnet A's broadcast boundary.

3

Determine starting address, size, and usable range for Subnet C (60 hosts).

Start address:

10.200.22.0

. Host bits needed:

h = 6

(

2^6 - 2 = 62 \ge 60

). Prefix length:

/26

(

32 - 6

). Network ID:

10.200.22.0/26

. Broadcast address:

10.200.22.63

. Usable range:

10.200.22.1

to

10.200.22.62

.

Subnet C starts at

10.200.22.0/26

. Usable hosts range from Network ID

+ 1

to Broadcast

- 1

.

Key Concept

Variable Length Subnet Masking (VLSM) host requirement sizing and sequential subnet boundary calculation.

Question 15Question →

A network engineer is configuring a point-to-point connection and internal VLANs within an enterprise infrastructure. An interface is assigned the IPv4 address $10.144.67.100/21$ . What is the decimal value of the third octet of the broadcast address for the subnet to which this host belongs?

Show answer & explanation

Answer: 71

Answer

The third octet of the broadcast address is 71 (broadcast address is 10.144.71.255).

For the prefix /21 (subnet mask 255.255.248.0), the third octet operates with a block size of 8 (256 - 248 = 8). The third octet value 67 falls into the subnet starting at 64 (10.144.64.0). The next subnet starts at 72 (10.144.72.0), making the broadcast address for the current subnet 10.144.71.255. The third octet of this broadcast address is 71.

Step-by-Step Solution

1

Convert CIDR prefix /21 to dotted-decimal mask.

Mask is 255.255.248.0 (21 network bits: 8 + 8 + 5).

The third octet contains 5 subnet bits and 3 host bits.

2

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

256 - 248 = 8.

Subnets increment in steps of 8 in the third octet.

3

Find the third octet network boundary for IP 10.144.67.100.

Network address is 10.144.64.0.

The greatest multiple of 8 less than or equal to 67 is 64.

4

Determine the broadcast address of the subnet.

Broadcast address is 10.144.71.255.

The subnet spans from 10.144.64.0 to 10.144.71.255, one value prior to the next subnet 10.144.72.0.

Key Concept

Subnet Boundary and Broadcast Address Calculation

Question 16Question →

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

172.24.135.254

172.24.135.255

172.24.131.254

172.32.135.254

Show answer & explanation

Answer: $172.24.135.254$

Answer

The IPv4 address

172.24.135.254

must be assigned to the gateway interface.

For a

/21

network (

255.255.248.0

), the third octet increments by 8. Given the network ID

172.24.128.0

, the subnet range encompasses

172.24.128.0

through

172.24.135.255

. The broadcast address is

172.24.135.255

, making

172.24.135.254

the highest usable host IPv4 address.

Step-by-Step Solution

1

Determine the subnet mask and block size for a

/21

prefix

A

/21

prefix corresponds to a subnet mask of

255.255.248.0

. The block size in the third octet is

256 - 248 = 8

.

Calculating the block size identifies the boundary for the network increment.

2

Calculate the network ID and broadcast address

Starting at network ID

172.24.128.0

, adding the block size of

8

gives the next subnet starting at

172.24.136.0

. Therefore, the broadcast address for this subnet is

172.24.135.255

.

The broadcast address is the last address in the subnet before the start of the next subnet.

3

Identify the highest usable host IP address

Subtracting 1 from the broadcast address

172.24.135.255

yields

172.24.135.254

as the highest usable host address.

The network ID (

172.24.128.0

) and broadcast address (

172.24.135.255

) are reserved and cannot be assigned to network interfaces.

Key Concept

IPv4 Subnet Boundaries and Usable Host Range Calculation

Question 17Question →

An infrastructure team is provisioning a new subnet from the IPv4 address block $172.20.96.0/23$ for a server farm that requires support for up to $250$ usable host interfaces. If the team applies the most efficient prefix length that satisfies this requirement, which IPv4 address is the last usable host address in the first subnet created?

172.20.96.254

172.20.96.255

172.20.97.254

172.32.96.254

Show answer & explanation

Answer: 172.20.96.254

Answer

172.20.96.254

To support

250

usable host interfaces efficiently,

8

host bits are required (

2^8 - 2 = 254

usable hosts), corresponding to a

/24

subnet prefix. The first

/24

subnet carved from the parent block

172.20.96.0/23

has a network address of

172.20.96.0

and a broadcast address of

172.20.96.255

. The last usable host IP address is one position below the broadcast address, which is

172.20.96.254

.

Step-by-Step Solution

1

Determine the required host bits and prefix length for 250 hosts.

Using

2^n - 2 \ge 250

,

n = 8

host bits are needed (

2^8 - 2 = 254

usable addresses). The corresponding prefix length is

32 - 8 = /24

.

Selecting 8 host bits fulfills the requirement of at least 250 usable hosts with minimum wasted IP space.

2

Calculate the boundaries for the first /24 subnet from 172.20.96.0/23.

The first subnet is 172.20.96.0/24, spanning network address 172.20.96.0 to broadcast address 172.20.96.255.

A /24 subnet increments in the third octet by 1, starting at the base network address 172.20.96.0.

3

Identify the last usable host address of the 172.20.96.0/24 subnet.

Subtracting 1 from the broadcast address (172.20.96.255 - 1) gives 172.20.96.254.

Usable host IP addresses lie strictly between the network address and the broadcast address.

Key Concept

IPv4 Subnetting, Prefix Derivation, and Host Range Boundaries

Question 18Question →

A network engineer is configuring an interface on a Cisco router servicing an internal LAN segment. The interface is assigned the IPv4 address $192.168.1.37$ with a subnet mask of $255.255.255.240$ ( $/28$ ). Which IPv4 address represents the broadcast address for this subnetwork?

192.168.1.32

192.168.1.47

192.168.1.48

192.168.1.63

Show answer & explanation

Answer: $192.168.1.47$

Answer

192.168.1.47

The subnet mask

255.255.255.240

(

/28

) provides a block size of 16 addresses per subnet. Multiples of 16 in the 4th octet yield network addresses of

192.168.1.0

,

192.168.1.16

,

192.168.1.32

, and

192.168.1.48

. The host

192.168.1.37

resides in the

192.168.1.32/28

subnet, making

192.168.1.47

the broadcast address.

Step-by-Step Solution

1

Determine the block size from the subnet mask.

A subnet mask of

255.255.255.240

corresponds to a

/28

prefix, leaving

32 - 28 = 4

host bits. The block size is

2^4 = 16

.

Calculating the block size defines the increment between subnetwork boundaries.

2

Identify the network boundary for IP address

192.168.1.37

.

Subnet multiples of 16 in the 4th octet are 0, 16, 32, 48. Since 37 falls between 32 and 47, the subnetwork ID is

192.168.1.32

.

The network address is the lowest boundary address for the subnet.

3

Calculate the broadcast address for the subnet.

The broadcast address is one less than the next network address (

192.168.1.48 - 1 = 192.168.1.47

).

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

Key Concept

IPv4 Subnet Masking and Broadcast Address Calculation

Estimated Time:1m 0s

Question 19Question →

A network administrator is assigning a private IPv4 host address within the $172.16.0.0/12$ range for a branch office subnet defined as $172.30.50.16/28$ . Which IPv4 address is a valid usable host address for a device in this subnet?

172.30.50.25

172.30.50.16

172.30.50.31

172.32.50.20

Show answer & explanation

Answer: $172.30.50.25$

Answer

The address

172.30.50.25

is a valid usable host IPv4 address within the

172.30.50.16/28

subnet.

The subnet

172.30.50.16/28

has a block size of

16

addresses (

2^4

), covering

172.30.50.16

through

172.30.50.31

. Subtracting the network address (

172.30.50.16

) and broadcast address (

172.30.50.31

) leaves a usable host range of

172.30.50.17

to

172.30.50.30

. The address

172.30.50.25

lies within this valid host range and is within the RFC 1918 private block (

172.16.0.0/12

).

Step-by-Step Solution

1

Determine the network properties for

172.30.50.16/28

.

A

/28

subnet mask has

32 - 28 = 4

host bits, yielding a block size of

2^4 = 16

.

The block size dictates the increment between subnets.

2

Identify the network address, broadcast address, and usable host range.

Network Address:

172.30.50.16

; Broadcast Address:

172.30.50.31

; Usable Host Range:

172.30.50.17

to

172.30.50.30

.

The network address is the first address in the block, and the broadcast address is the last address in the block.

3

Verify RFC 1918 private address range compliance and match against options.

The range

172.16.0.0/12

spans

172.16.0.0

to

172.31.255.255

. The address

172.30.50.25

is both within the RFC 1918 block and within the usable host range

172.30.50.17

-

172.30.50.30

.

Only valid usable host addresses within the private range can be assigned to host interfaces.

Key Concept

IPv4 Subnet Boundaries and RFC 1918 Private Addressing

Question 20Question →

A network administrator is designing an IPv4 addressing plan for a new office department. The department requires static host IP assignments for 61 client workstations, 1 default gateway router interface, and 1 network printer, requiring a total of 63 active host IP addresses. Which subnet mask provides the most efficient address space utilization while satisfying all required host addresses?

255.255.255.128

255.255.255.192

255.255.255.224

255.255.255.0

Show answer & explanation

Answer: $255.255.255.128$

Answer

The subnet mask

255.255.255.128

(corresponding to a

/25

prefix) is the most efficient subnet mask that accommodates 63 host IP addresses.

To support 63 active host interfaces, the subnet must provide at least 63 usable IPv4 addresses. Using 6 host bits yields

2^6 - 2 = 62

usable addresses, which falls 1 short. Therefore, 7 host bits are required (

2^7 - 2 = 126

usable addresses). A subnet with 7 host bits leaves 25 prefix bits (

32 - 7 = 25

), corresponding to the dotted-decimal subnet mask

255.255.255.128

. This is the smallest subnet that satisfies the requirement without wasting excessive IP addresses.

Step-by-Step Solution

1

Determine total required host IP addresses

61 \text{ workstations} + 1 \text{ gateway} + 1 \text{ printer} = 63 \text{ usable host IP addresses}

Every device and gateway interface requiring an IP address counts toward the total host capacity.

2

Calculate required host bits using the usable host capacity formula

2^H - 2 \ge 63

For

H = 6

:

2^6 - 2 = 62

usable addresses (insufficient). For

H = 7

:

2^7 - 2 = 126

usable addresses (sufficient).

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

3

Derive prefix length and dotted-decimal subnet mask

Prefix length

= 32 - 7 = /25

, which converts to

255.255.255.128

Subtracting 7 host bits from 32 total IPv4 bits leaves 25 network bits.

Key Concept

Usable IPv4 Host Address Calculation (

2^H - 2

) and Subnet Mask Derivation

IPv4 Addressing and Subnetting

Answer

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