All practice questions

Question 1Question →

A systems deployment team is provisioning a dedicated IPv4 subnet for a smart office floor. The design plan requires static IP assignments for $42$ IoT sensors, $4$ wireless access points, and $2$ router interfaces configured for high-availability default gateway redundancy. What is the total number of usable host IPv4 addresses provided by the smallest standard subnet mask that accommodates all required devices?

Show answer & explanation

Answer: 62

Answer

The smallest subnet that satisfies the requirement for

48

IP addresses is a

/26

subnet, which provides

62

usable host IPv4 addresses.

To determine the total usable IPv4 host capacity of the smallest suitable subnet, first sum the required IP assignments:

42\text{ (sensors)} + 4\text{ (access points)} + 2\text{ (gateways)} = 48\text{ addresses}

. The formula for usable host addresses is

2^h - 2

, where

h

is the number of host bits. A

/27

block (

h=5

) yields

2^5 - 2 = 30

usable addresses, which is too small. A

/26

block (

h=6

) yields

2^6 - 2 = 62

usable host addresses, which fully supports the

48

required addresses.

Step-by-Step Solution

1

Calculate total host IP addresses required

Total host requirement =

42 + 4 + 2 = 48

usable IP addresses

Every connected device and default gateway interface requires a distinct usable IPv4 address.

2

Determine the required host bit count

h = 6

host bits (CIDR prefix

/26

)

2^5 - 2 = 30

hosts is insufficient for

48

addresses, whereas

2^6 - 2 = 62

hosts satisfies the requirement.

3

Calculate the usable host capacity of the subnet

62

usable host addresses

Subtracting

2

reserved addresses (network ID and broadcast address) from total

2^6 = 64

block addresses yields

62

usable host IPs.

Key Concept

Subnet Host Capacity Calculation (

2^h - 2

Rule)

Question 2Question →

A network engineer is provisioning subnets from the assigned IPv4 block $10.150.12.0/22$ for an industrial automation network. Each production segment requires a subnet capable of accommodating at least $110$ usable IP addresses for automated controllers and monitoring sensors. What is the maximum number of subnets meeting this requirement that can be created from the allocated block?

Show answer & explanation

Answer: 8

Answer

The maximum number of valid subnets that can be created is 8.

To host at least

110

usable IP addresses per segment,

7

host bits are required because

2^7 - 2 = 126

usable host IPs (

6

bits only provide

2^6 - 2 = 62

usable IPs). A

7

-bit host field corresponds to a

/25

prefix length (

32 - 7 = 25

). Subnetting a

/22

block into

/25

subnets yields

2^{25 - 22} = 2^3 = 8

subnets.

Step-by-Step Solution

1

Calculate the host bits needed for at least 110 usable IP addresses.

7 host bits are required because

2^7 - 2 = 126

usable hosts (

2^6 - 2 = 62

is insufficient).

Each subnet must accommodate 110 hosts, requiring 7 host bits.

2

Determine the required prefix length for each subnet.

Subnet prefix length is

/25

(

32 - 7 = 25

).

Subtracting host bits from 32 gives the network prefix length.

3

Determine the number of

/25

subnets created from a

/22

block.

2^{(25 - 22)} = 2^3 = 8

subnets.

The difference between the new prefix length (/25) and original prefix length (/22) provides 3 subnet bits.

Key Concept

Subnet sizing and host capacity calculation using IPv4 CIDR notation

Question 3Question →

A network administrator is assigning subnets for small remote branch offices. Each office requires an IPv4 subnet that can support at least $30$ usable host IP addresses. What is the minimum CIDR prefix length (enter the integer prefix number, such as 24 for a $/24$ subnet) that satisfies this requirement?

Show answer & explanation

Answer: 27

Answer

The minimum CIDR prefix length required to support 30 usable hosts is 27.

To host 30 usable devices, a subnet must provide at least 32 total IP addresses (

30

usable

+ 2

reserved for network and broadcast). Using the formula

2^h \ge 32

, we find

h = 5

host bits are needed. Subtracting

5

host bits from the total

32

bits in an IPv4 address yields a prefix length of

27

(a

/27

subnet).

Step-by-Step Solution

1

Calculate the total number of IP addresses required per subnet including overhead.

30\text{ usable hosts} + 1\text{ network ID} + 1\text{ broadcast ID} = 32\text{ total IP addresses}

.

Every IPv4 subnet requires two reserved IP addresses: one for the network ID and one for the broadcast address.

2

Determine the required host bits (

h

).

2^h = 32 \implies h = 5

host bits.

Five host bits provide

2^5 = 32

total IP addresses.

3

Calculate the prefix length by subtracting host bits from the total bits in an IPv4 address.

32 - 5 = 27

.

The prefix length represents the network portion of the address, calculated as

32 - h

.

Key Concept

Calculating CIDR Subnet Prefix Length for Host Requirements

Question 4Question →

A network architect is assigned the IPv4 address block $172.24.160.0/19$ to provision subnets for a new regional data center. To optimize routing and broadcast domain sizes, the architect must partition the entire block into equal-sized subnets such that each subnet can accommodate a minimum of $120$ usable host IP addresses. Using the longest possible CIDR prefix mask that satisfies this host requirement, what is the total maximum number of usable host IP addresses available across all subnets created?

Show answer & explanation

Answer: 8064

Answer

8064 usable host IP addresses

To support at least 120 usable host IP addresses per subnet while utilizing the longest possible CIDR prefix, a /25 subnet mask (7 host bits) is required, providing

2^7 - 2 = 126

usable host addresses per subnet. Partitioning the parent

172.24.160.0/19

block into /25 subnets yields

2^{25-19} = 2^6 = 64

subnets. Multiplying 64 subnets by 126 usable host IP addresses per subnet results in 8064 total usable host IP addresses.

Step-by-Step Solution

1

Determine the host bit requirement (

h

) for at least 120 usable hosts per subnet.

7 host bits are required because

2^7 - 2 = 126 \ge 120

usable host addresses.

6 host bits (

2^6 - 2 = 62

) are insufficient to meet the 120 host requirement.

2

Calculate the subnet prefix length (CIDR notation).

Prefix length is

/25

(

32 - 7 = 25

).

Subtracting 7 host bits from the 32 total IPv4 bits leaves a 25-bit network prefix.

3

Calculate total

/25

subnets created from the parent

/19

block.

64 subnets (

2^{25 - 19} = 2^6 = 64

).

The difference between the new prefix (/25) and parent prefix (/19) is 6 subnet bits.

4

Calculate total usable host IP addresses across all 64 subnets.

8064 usable host IP addresses (

64 \times 126 = 8064

).

Each of the 64 subnets reserves 2 IP addresses (network ID and broadcast address) out of 128 total addresses.

Key Concept

Partitioning IPv4 CIDR blocks using VLSM/subnetting and calculating total aggregate usable host space.

Question 5Question →

An enterprise operations team is allocated the network block $172.28.64.0/21$ to deploy microservices within a private cloud environment. To satisfy security policies, the administrator must divide this entire block into equal-sized subnets such that each subnet can accommodate at least $100$ usable host IP addresses while reserving as much space as possible for maximum subnetting density (using the longest possible subnet mask that meets the requirement). What is the total number of usable host IP addresses available across all created subnets combined?

Show answer & explanation

Answer: 2016

Answer

2016 usable host IP addresses

To host at least

100

usable IP addresses, each subnet requires

7

host bits (

2^7 - 2 = 126

usable hosts), which corresponds to a

/25

prefix (

32 - 7 = 25

). Subnetting a

/21

network block into

/25

subnets yields

2^{25-21} = 2^4 = 16

distinct subnets. Because each

/25

subnet reserves

1

network address and

1

broadcast address, each subnet has

126

usable host IPs. Across all

16

subnets, the total usable host capacity is

16 \times 126 = 2016

.

Step-by-Step Solution

1

Determine the required subnet mask for each subnet to support at least 100 hosts

A

/25

prefix (

255.255.255.128

) providing 126 usable host IP addresses per subnet

The formula for usable hosts is

2^h - 2 \ge 100

. For

h = 6

,

2^6 - 2 = 62

(insufficient). For

h = 7

,

2^7 - 2 = 126

(sufficient). The subnet prefix length is

32 - 7 = 25

.

2

Calculate how many

/25

subnets fit into the parent

/21

block

16 subnets

The difference between the new subnet prefix (

/25

) and the parent prefix (

/21

) is

25 - 21 = 4

bits. The number of subnets is

2^4 = 16

.

3

Calculate the total aggregate usable host capacity

2016 aggregate usable host addresses

Multiplying the total subnets (

16

) by the usable hosts per subnet (

126

) yields

16 \times 126 = 2016

.

Key Concept

Subnet partitioning and usable host capacity calculation under CIDR constraints

Question 6Question →

An enterprise network architect is assigned the IPv4 block $10.180.64.0/19$ . Using Variable Length Subnet Masking (VLSM), the architect sequentially allocates subnets starting from the lowest available IP address to satisfy the following host requirements:

- Subnet A (Data Center): $1,200$ usable hosts
- Subnet B (Corporate HQ): $500$ usable hosts
- Subnet C (Voice Gateway): $250$ usable hosts
- Subnet D (Branch Office): $120$ usable hosts

Each subnet is provisioned using the smallest viable CIDR block. After allocating these four subnets back-to-back without leaving gaps between them, what is the maximum number of usable host IP addresses that can be supported by the single largest contiguous subnet that can be formed within the remaining unallocated space of the $10.180.64.0/19$ prefix?

Show answer & explanation

Answer: 4094

Answer

The maximum number of usable host IP addresses in the single largest contiguous subnet remaining is 4094.

The sequentially allocated subnets occupy IP addresses up to 10.180.75.127. The unallocated space spans 10.180.75.128 to 10.180.95.255. Because IPv4 subnets must align on binary power-of-two boundaries matching their size, the largest valid contiguous subnet that can be formed in this space is 10.180.80.0/20 (size 4,096 IPs). Subtracting 2 for the network and broadcast addresses yields 4,094 usable hosts.

Step-by-Step Solution

1

Determine the minimum prefix size for each required subnet

Subnet A requires /21 (2048 IPs), Subnet B requires /23 (512 IPs), Subnet C requires /24 (256 IPs), Subnet D requires /25 (128 IPs)

Host requirement formula is 2^n - 2. Subnet A: 2^11 - 2 = 2046 >= 1200; Subnet B: 2^9 - 2 = 510 >= 500; Subnet C: 2^8 - 2 = 254 >= 250; Subnet D: 2^7 - 2 = 126 >= 120.

2

Map sequential subnet allocations across address space

Allocated address range spans 10.180.64.0 through 10.180.75.127

Subnet A: 10.180.64.0/21 (10.180.64.0 - 10.180.71.255). Subnet B: 10.180.72.0/23 (10.180.72.0 - 10.180.73.255). Subnet C: 10.180.74.0/24 (10.180.74.0 - 10.180.74.255). Subnet D: 10.180.75.0/25 (10.180.75.0 - 10.180.75.127).

3

Identify remaining contiguous IP space and evaluate boundary alignments

Unallocated space consists of 10.180.75.128/25, 10.180.76.0/22, and 10.180.80.0/20

Address 10.180.80.0 is divisible by 16 in the 3rd octet, aligning perfectly on a /20 boundary (4096 IPs) covering 10.180.80.0 through 10.180.95.255.

4

Calculate usable hosts for the largest valid contiguous subnet block (/20)

4,094 usable host IP addresses

Formula: 2^(32 - 20) - 2 = 2^12 - 2 = 4,096 - 2 = 4,094.

Key Concept

Variable Length Subnet Masking (VLSM) and Bit Boundary Alignment

Estimated Time:3m 0s

Question 7Question →

A network engineer is configuring an internal routing domain using the assigned IP address block $10.88.0.0/18$ . The network architecture requires dividing this entire $/18$ block into equal-sized subnets such that each subnet can accommodate at least $180$ usable host IP addresses while maximizing the number of subnets created. How many total subnets can be created from the original block using this optimal subnet mask?

Show answer & explanation

Answer: 64

Answer

64 subnets can be created from the original block.

To support at least 180 usable hosts per subnet, each subnet requires 8 host bits because

2^8 - 2 = 254

usable IP addresses (

2^7 - 2 = 126

is insufficient). An 8-host-bit subnet corresponds to a

/24

prefix (

32 - 8 = 24

). Dividing a

/18

network block into

/24

subnets yields

2^{24 - 18} = 2^6 = 64

equal-sized subnets.

Step-by-Step Solution

1

Calculate the minimum number of host bits required per subnet

8 host bits (

h = 8

)

Usable hosts formula is

2^h - 2

. For

180

hosts,

2^h - 2 \ge 180 \implies 2^h \ge 182

. Since

2^7 = 128

is insufficient,

2^8 = 256

is required, providing

254

usable host IP addresses.

2

Determine the optimal CIDR subnet mask prefix length

/24 prefix length (

255.255.255.0

)

Subtracting the

8

required host bits from

32

total IPv4 address bits yields

32 - 8 = 24

network/subnet bits.

3

Calculate the total number of subnets derived from the original /18 block

64 subnets

The number of borrowed subnet bits is

24 - 18 = 6

bits. The total number of available subnets is

2^6 = 64

.

Key Concept

Equal-length IPv4 Subnetting and Host Requirement Calculation

Question 8Question →

A network engineer is allocated the IPv4 address block $10.240.16.0/20$ to create subnets for distinct department segments. Each segment requires a minimum of $120$ usable host IP addresses. What is the maximum number of subnets of this minimum required size that can be created from the given block?

Show answer & explanation

Answer: 32

Answer

The maximum number of subnets of the minimum required size that can be created is 32.

To support a minimum of 120 hosts, 7 host bits are required (

2^7 - 2 = 126 \ge 120

). A subnet with 7 host bits has a prefix length of

/25

(

32 - 7 = 25

). Partitioning a

/20

block into

/25

subnets borrows 5 bits (

25 - 20 = 5

), resulting in

2^5 = 32

subnets.

Step-by-Step Solution

1

Calculate the host bits required for the minimum host capacity requirement

7 host bits are required because

2^7 - 2 = 126 \ge 120

.

Using 6 host bits only provides

2^6 - 2 = 62

usable addresses, which fails to meet the 120 host requirement.

2

Calculate the required subnet mask prefix length

The prefix length is

/25

, calculated as

32 - 7 = 25

.

Subtracting host bits from the 32 total IPv4 address bits determines the network prefix.

3

Calculate the total subnets formed by splitting a

/20

block into

/25

subnets

2^{25 - 20} = 2^5 = 32

subnets.

The number of subnets created is 2 raised to the power of the borrowed subnet bits.

Key Concept

IPv4 Subnetting and Host Capacity Calculation

Question 9Question →

An enterprise network administrator is assigned the IPv4 network block $192.168.16.0/22$ to provision subnets for a new regional office. Using Variable Length Subnet Masking (VLSM) to minimize address waste, subnets are created for the following minimum usable host requirements:

- Department A: $250$ usable host IPs
- Department B: $120$ usable host IPs
- Department C: $60$ usable host IPs
- Department D: $25$ usable host IPs

What is the total number of unallocated IP addresses (including network and broadcast addresses) remaining in the original $192.168.16.0/22$ block after these subnets are assigned?

Show answer & explanation

Answer: 544

Answer

544 unallocated IP addresses remain in the original /22 block.

A /22 block provides

1024

total addresses. By sizing each subnet to the smallest power of 2 that accommodates host requirements plus network and broadcast overhead, the allocations consume

256

(/24),

128

(/25),

64

(/26), and

32

(/27) addresses, totaling

480

addresses. Subtracting

480

from

1024

leaves

544

unallocated addresses.

Step-by-Step Solution

1

Determine total capacity of the assigned prefix

A /22 subnet contains

2^{10} = 1024

total IP addresses.

The prefix length of /22 leaves

32 - 22 = 10

host bits.

2

Calculate required CIDR subnet block size for each department using VLSM rules

Dept A needs 256 IPs (/24); Dept B needs 128 IPs (/25); Dept C needs 64 IPs (/26); Dept D needs 32 IPs (/27).

Each subnet must accommodate the usable hosts plus 2 overhead addresses (network ID and broadcast address), rounded up to the smallest power of 2.

3

Sum total allocated IP addresses

256 + 128 + 64 + 32 = 480

IP addresses allocated.

Combining block sizes yields the total addresses consumed across all four departmental subnets.

4

Calculate remaining unallocated IP addresses

1024 - 480 = 544

unallocated IP addresses.

Subtracting allocated addresses from the total block size leaves the unassigned address space.

Key Concept

VLSM Subnet Block Size Calculation & Host Address Allocation

Question 10Question →

A network administrator is assigned the IPv4 address block $172.28.64.0/21$ to provision subnets for a multi-department enterprise deployment. Using Variable Length Subnet Masking (VLSM), subnets must be allocated sequentially starting from the lowest available IP address in order of capacity requirements (from largest to smallest):

- Subnet A: Requires capacity for up to $400$ usable host IP addresses.
- Subnet B: Requires capacity for up to $180$ usable host IP addresses.
- Subnet C: Requires capacity for up to $60$ usable host IP addresses.
- Subnet D: Requires capacity for exactly $2$ usable host IP addresses.

Each subnet must be allocated using the smallest valid CIDR prefix length that accommodates its requirement. What is the total number of unallocated IP addresses remaining in the $172.28.64.0/21$ block after all four subnets are allocated?

Show answer & explanation

Answer: 1212

Answer

The total number of unallocated IP addresses remaining in the

172.28.64.0/21

parent block after provisioning all four subnets is 1212.

The parent block

172.28.64.0/21

contains

2048

total IP addresses. Allocating the subnets sequentially requires: Subnet A (

400

hosts)

\rightarrow /23

(

512

IPs), Subnet B (

180

hosts)

\rightarrow /24

(

256

IPs), Subnet C (

60

hosts)

\rightarrow /26

(

64

IPs), and Subnet D (

2

hosts)

\rightarrow /30

(

4

IPs). Total allocated IP space is

512 + 256 + 64 + 4 = 836

IP addresses. Subtracting

836

from

2048

leaves

1212

total unallocated IP addresses.

Step-by-Step Solution

1

Calculate total IP address capacity of the parent block

172.28.64.0/21

contains

2^{32-21} = 2^{11} = 2048

total IP addresses

A

/21

prefix uses 21 network bits, leaving 11 host bits for addressing.

2

Determine prefix length and block size for Subnet A

Prefix:

/23

, Block size:

512

total IPs (

172.28.64.0/23

)

To support

400

hosts, 9 host bits are needed (

2^9 - 2 = 510 \ge 400

), giving a

/23

prefix (

512

IPs).

3

Determine prefix length and block size for Subnet B

Prefix:

/24

, Block size:

256

total IPs (

172.28.66.0/24

)

To support

180

hosts, 8 host bits are needed (

2^8 - 2 = 254 \ge 180

), giving a

/24

prefix (

256

IPs).

4

Determine prefix length and block size for Subnet C

Prefix:

/26

, Block size:

64

total IPs (

172.28.67.0/26

)

To support

60

hosts, 6 host bits are needed (

2^6 - 2 = 62 \ge 60

), giving a

/26

prefix (

64

IPs).

5

Determine prefix length and block size for Subnet D

Prefix:

/30

, Block size:

4

total IPs (

172.28.67.64/30

)

To support

2

hosts (point-to-point link), 2 host bits are needed (

2^2 - 2 = 2 \ge 2

), giving a

/30

prefix (

4

IPs).

6

Calculate remaining unallocated IP space

2048 - (512 + 256 + 64 + 4) = 2048 - 836 = 1212

IP addresses

Subtracting the sum of all allocated CIDR block sizes from the total parent address space yields the remaining unallocated IP addresses.

Key Concept

Variable Length Subnet Masking (VLSM) allocation and host requirement calculations

Question 11Question →

A network technician is configuring a new subnet to accommodate a cluster of $40$ usable host interfaces. What is the minimum IPv4 prefix length (in CIDR notation, e.g., $24$ for $/24$ ) required to support this number of hosts?

Show answer & explanation

Answer: 26

Answer

The minimum IPv4 prefix length required is 26.

To support

40

usable hosts, a subnet requires

6

host bits because

2^6 - 2 = 62

usable IP addresses (

2^5 - 2 = 30

is too small). Subtracting

6

host bits from

32

total IPv4 bits results in a prefix length of

26

(or

/26

).

Step-by-Step Solution

1

Determine host capacity requirement

Need a subnet that provides at least

40

usable host IP addresses.

Each subnet reserves two IP addresses: one for the network ID and one for the broadcast address.

2

Find host bits needed

6

host bits are required (

2^6 - 2 = 62

usable hosts).

5

host bits only yield

30

usable addresses, which fails the requirement of

40

hosts.

3

Compute CIDR prefix length

32 - 6 = 26

.

An IPv4 address consists of

32

total bits; subtracting the host bits gives the network prefix bits.

Key Concept

Host capacity formula and CIDR prefix length calculation

Question 12Question →

An enterprise network operations team is allocating IPv4 subnets from the parent block $172.31.192.0/19$ to provision Point-to-Point WAN links between datacenter locations. Each link requires a dedicated subnet configured with a $/30$ mask to connect two router interfaces. If the team provisions $45$ operational Point-to-Point link subnets sequentially starting from the lowest IP address of the parent block, what is the maximum number of additional $/30$ link subnets that can be created from the remaining unallocated space within this $/19$ block?

Show answer & explanation

Answer: 2003

Answer

The maximum number of additional /30 link subnets that can be created from the remaining unallocated space is 2003.

The parent block has a prefix length of /19 and each link uses a /30 subnet mask. Subtracting the prefix lengths (

30 - 19 = 11

) gives

11

subnet bits, which equals

2^{11} = 2048

total possible /30 subnets within the /19 block. Subtracting the 45 already provisioned subnets leaves

2048 - 45 = 2003

unallocated /30 subnets.

Step-by-Step Solution

1

Calculate the total number of /30 subnets available within a /19 parent block

2048 subnets

The prefix length difference between the child subnet (/30) and parent block (/19) is

30 - 19 = 11

bits. The total number of /30 subnets created is

2^{11} = 2048

.

2

Subtract the number of already allocated /30 subnets from the total available subnet capacity

2003 subnets

Subtracting the

45

allocated subnets from the total capacity of

2048

(

2048 - 45 = 2003

) yields the remaining unallocated /30 subnets.

Key Concept

Subnet allocation and capacity calculations using CIDR prefix length differences

Question 13Question →

A network technician is configuring a local subnetwork for a department that requires a $/27$ prefix length. How many usable host IPv4 addresses can be assigned to devices within this subnetwork?

Show answer & explanation

Answer: 30

Answer

There are 30 usable host IPv4 addresses in a /27 subnet.

A /27 prefix leaves 5 host bits (

32 - 27 = 5

). Calculating

2^5

gives

32

total IP addresses. Subtracting

2

reserved addresses for the network ID and broadcast address results in exactly

30

usable host IPv4 addresses.

Step-by-Step Solution

1

Determine the number of host bits remaining in a /27 CIDR prefix.

32 total bits - 27 network bits = 5 host bits.

An IPv4 address contains 32 bits divided between network bits and host bits.

2

Calculate the total capacity of IP addresses for 5 host bits.

2^5 = 32 total IP addresses.

The number of total IP addresses in a subnet is calculated using 2 raised to the power of the host bits.

3

Subtract the reserved network and broadcast addresses.

32 - 2 = 30 usable host addresses.

The lowest IP address represents the network ID and the highest IP address represents the broadcast address, neither of which can be assigned to host devices.

Key Concept

IPv4 Usable Host Address Calculation

Estimated Time:45s

Question 14Question →

A network administrator is designing an IPv4 addressing scheme for a enterprise WLAN segment that must support at least $500$ usable host devices per subnet. What is the maximum CIDR prefix length (integer value of the mask bits) that can be assigned to meet this requirement without wasting IP address space?

Show answer & explanation

Answer: 23

Answer

The maximum CIDR prefix length that supports at least 500 usable host addresses is 23.

To host at least

500

usable devices, a subnet requires 9 host bits because

2^9 - 2 = 510

usable IP addresses. Subtracting 9 host bits from the 32 total bits in an IPv4 address results in a prefix length of 23.

Step-by-Step Solution

1

Calculate the required number of host bits

9 host bits are required.

Using the formula

2^h - 2

, 8 host bits yield only 254 usable addresses, whereas 9 host bits yield

510

usable addresses, which satisfies the

500

host minimum requirement.

2

Calculate the network prefix length in CIDR notation

The prefix length is 23.

An IPv4 address consists of 32 bits in total. Subtracting the 9 host bits (

32 - 9

) leaves 23 bits for the network and subnet identifier.

Key Concept

IPv4 Host Capacity and CIDR Prefix Calculation

Question 15Question →

A network administrator is configuring a new subnet assigned a $/26$ CIDR prefix. What is the maximum number of usable host IP addresses that can be assigned to devices on this subnet?

Show answer & explanation

Answer: 62

Answer

The maximum number of usable host IP addresses on a

/26

subnet is 62.

A

/26

subnet prefix leaves

32 - 26 = 6

bits for host assignment. The total address capacity is

2^6 = 64

IP addresses. Subtracting the network identifier and broadcast address yields

64 - 2 = 62

usable host addresses.

Step-by-Step Solution

1

Determine host portion bit length

6 bits (

32 - 26 = 6

)

IPv4 addresses consist of 32 bits total. Subtracting the prefix length yields the number of bits allocated for host addresses.

2

Calculate total address space

64 IP addresses (

2^6 = 64

)

The total number of unique address combinations is

2^h

, where

h

is the number of host bits.

3

Subtract reserved addresses

62 usable host addresses (

64 - 2 = 62

)

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

Key Concept

Calculating usable host count from IPv4 CIDR prefix length

Question 16Question →

A network technician is configuring a dedicated point-to-point wireless link between two enterprise buildings using the IPv4 subnet $192.168.10.16/28$ . What is the maximum number of usable host IPv4 addresses available for assignment to physical network interfaces in this subnet?

Show answer & explanation

Answer: 14

Answer

The maximum number of usable host IPv4 addresses in a

/28

subnet is 14.

For a

/28

subnet, there are 4 host bits (

32 - 28 = 4

). Calculating total IP addresses yields

2^4 = 16

. Subtracting 2 reserved addresses (one network address and one broadcast address) leaves 14 usable host IPv4 addresses.

Step-by-Step Solution

1

Calculate the number of host bits available from the CIDR prefix

4 host bits (

32 - 28 = 4

)

An IPv4 address contains 32 bits in total, so subtracting the prefix length gives the remaining bits allocated for host addresses.

2

Calculate the total number of IP addresses in the subnet

16 total IP addresses (

2^4 = 16

)

The number of total IP combinations for

n

host bits is

2^n

.

3

Calculate the usable host address count by excluding reserved addresses

14 usable host IPv4 addresses (

16 - 2 = 14

)

The first IP address in the range is reserved as the network address, and the final IP address is reserved as the directed broadcast address.

Key Concept

Calculating Usable Host Addresses in an IPv4 Subnet

Question 17Question →

A network engineer is configuring a newly provisioned server rack in a data center using the assigned IPv4 subnet block $10.50.16.0/27$ . What is the maximum number of usable host IPv4 addresses that can be assigned to active devices within this subnet block?

Show answer & explanation

Answer: 30

Answer

The maximum number of usable host IPv4 addresses in a /27 subnet is 30.

A CIDR prefix of /27 leaves 5 host bits (

32 - 27 = 5

). Raising 2 to the power of 5 gives 32 total addresses. Subtracting the network address and broadcast address (

32 - 2

) results in 30 usable host IP addresses.

Step-by-Step Solution

1

Determine the number of available host bits.

Host bits =

32 - 27 = 5

bits.

An IPv4 address has 32 bits total. Subtracting the prefix length yields the bits remaining for host addressing.

2

Calculate the total size of the address block.

Total IP addresses =

2^5 = 32

addresses.

Five host bits allow

2^5

distinct binary combinations.

3

Subtract reserved addresses to find the usable host count.

Usable host addresses =

32 - 2 = 30

addresses.

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

Key Concept

Usable Host Calculation in Subnetting

Question 18Question →

A network engineer is configuring Variable Length Subnet Masking (VLSM) within the assigned block $10.200.16.0/20$ . The first subnet (Subnet A) must be allocated from the start of the block to support at least $600$ usable host interfaces. The second subnet (Subnet B) must be provisioned immediately following the address space allocated to Subnet A to accommodate at least $250$ usable host interfaces. What is the decimal value of the third octet in the network ID of Subnet B?

Show answer & explanation

Answer: 20

Answer

The decimal value of the third octet in the network ID of Subnet B is 20.

To support at least 600 usable hosts in Subnet A, 10 host bits are required (

2^{10} - 2 = 1022

usable addresses), establishing a /22 subnet mask. With a /22 prefix starting at

10.200.16.0

, the block spans 4 values in the third octet (

16, 17, 18, 19

), covering

10.200.16.0

through

10.200.19.255

. The next available network address for Subnet B starts immediately at

10.200.20.0

. Thus, the third octet decimal value is 20.

Step-by-Step Solution

1

Determine the required prefix length for Subnet A.

Subnet A requires a /22 prefix length.

Subnet A needs at least 600 usable host IPs. Since

2^9 - 2 = 510

is insufficient, 10 host bits are required (

2^{10} - 2 = 1022

). The prefix length is

32 - 10 = 22

.

2

Calculate the block size and address range for Subnet A.

Subnet A spans

10.200.16.0

to

10.200.19.255

.

A /22 prefix has a block size of

2^{(32-22)} = 1024

total addresses (

4

block increments in the third octet). Starting at

10.200.16.0

, the subnet ends at

10.200.19.255

.

3

Find the starting network ID for Subnet B.

Subnet B begins at

10.200.20.0

.

Subnet B must be allocated immediately after the broadcast address of Subnet A (

10.200.19.255

).

4

Extract the third octet value.

20

In the IPv4 address

10.200.20.0

, the third octet is 20.

Key Concept

Variable Length Subnet Masking (VLSM) block size calculation and sequential subnet allocation.

Question 19Question →

A network engineer is configuring subnets within the $10.50.0.0/16$ address space for a new office building. Each subnet must support a minimum of $150$ usable host IP addresses. What is the maximum number of equal-sized subnets that can be created from this block while meeting the host requirement?

Show answer & explanation

Answer: 256

Answer

The maximum number of equal-sized subnets that can be created is 256.

To support at least 150 host devices, 8 host bits are necessary (

2^8 - 2 = 254

usable hosts). Subtracting 8 host bits from 32 total address bits yields a /24 subnet mask. Borrowing 8 bits from the original /16 network prefix produces

2^8 = 256

subnets.

Step-by-Step Solution

1

Determine the number of host bits (

h

) required for each subnet to accommodate at least 150 usable host IP addresses.

8 host bits are required because

2^8 - 2 = 254

usable host addresses (7 host bits only yield

2^7 - 2 = 126

usable host addresses).

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

2

Calculate the prefix length required for the new subnets.

Prefix length

/24

(

32 \text{ total bits} - 8 \text{ host bits} = 24

).

IPv4 addresses consist of 32 bits divided between network/subnet and host portions.

3

Determine the number of subnet bits borrowed from the original /16 network mask.

8 bits borrowed (

24 - 16 = 8

).

Subtracting the original network prefix length from the new subnet prefix length gives the borrowed bits.

4

Calculate the total number of equal-sized subnets that can be created.

256 subnets (

2^8 = 256

).

The number of subnets generated is

2^n

, where

n

is the number of borrowed subnet bits.

Key Concept

IPv4 Subnet Masking and Host Capacity Calculations

Question 20Question →

A network administrator is designing an IPv4 subnet for a new regional warehouse facility. The facility requires static and dynamic IP address assignments for $115$ endpoint devices, including inventory scanners, workstation PCs, and network printers. What is the maximum number of usable host IP addresses available in the smallest CIDR subnet mask that satisfies this host requirement?

Show answer & explanation

Answer: 126

Answer

The smallest CIDR block accommodating 115 hosts is a /25 prefix (255.255.255.128), which provides 126 usable host IP addresses.

To accommodate 115 hosts, a subnet requires at least

115 + 2 = 117

total IP addresses. The smallest power of 2 covering 117 is

2^7 = 128

, corresponding to a

/25

subnet prefix. Subtracting the Network ID and Broadcast address (

128 - 2

) results in 126 usable host IP addresses.

Step-by-Step Solution

1

Calculate the total required IP space including reserved network and broadcast addresses.

115 \text{ required host IPs} + 2 \text{ reserved addresses} = 117 \text{ total IP addresses}

.

Every IPv4 subnet reserves the first address for the Network ID and the final address for the Broadcast address.

2

Determine the smallest block size (

2^n

) that accommodates the total address requirement.

2^7 = 128 \text{ total addresses}

, corresponding to host bits

n=7

and prefix length

32 - 7 = /25

.

2^6 = 64

total addresses (

62

usable) is insufficient for

115

devices.

3

Subtract the 2 reserved addresses to find total usable host capacity.

128 - 2 = 126 \text{ usable host IP addresses}

.

Usable host capacity formula is

2^{(32 - \text{prefix})} - 2

.

Key Concept

IPv4 Subnet Mask Host Capacity Calculation

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept