All practice questions

Question 1Question →

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 2Question →

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 3Question →

A network architect is implementing Variable Length Subnet Masking (VLSM) for an enterprise branch using the assigned summary prefix $172.20.0.0/19$ . Four subnets must be allocated sequentially from the lowest available IP address of the block to support the following minimum host requirements:

- Location A: 1,000 usable hosts
- Location B: 480 usable hosts
- Location C: 250 usable hosts
- Location D: 120 usable hosts

Each subnet must be sized using the smallest possible prefix length to satisfy its host requirement. After completing these four allocations in order (Location A, then B, then C, then D), what is the maximum number of usable host IPv4 addresses that can be accommodated in the single largest contiguous unallocated subnet block remaining in the $172.20.0.0/19$ summary range?

Show answer & explanation

Answer: 4094

Answer

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

The base block

172.20.0.0/19

contains

2^{13} = 8192

total IP addresses (ranging from

172.20.0.0

to

172.20.31.255

). Allocating for Locations A (

/22

, 1024 IPs), B (

/23

, 512 IPs), C (

/24

, 256 IPs), and D (

/25

, 128 IPs) uses addresses

172.20.0.0

through

172.20.7.127

. The remaining space (

172.20.7.128

to

172.20.31.255

) decomposes into three valid CIDR subnets:

172.20.7.128/25

(128 IPs),

172.20.8.0/21

(2048 IPs), and

172.20.16.0/20

(4096 IPs). The largest single contiguous subnet is

172.20.16.0/20

, which accommodates

2^{12} - 2 = 4094

usable host addresses.

Step-by-Step Solution

1

Calculate required prefix lengths and allocated address ranges for each location sequentially starting at 172.20.0.0

Location A (1000 hosts) requires /22 (1024 total IPs: 172.20.0.0 - 172.20.3.255). Location B (480 hosts) requires /23 (512 total IPs: 172.20.4.0 - 172.20.5.255). Location C (250 hosts) requires /24 (256 total IPs: 172.20.6.0 - 172.20.6.255). Location D (120 hosts) requires /25 (128 total IPs: 172.20.7.0 - 172.20.7.127).

Each subnet host capacity is calculated using 2^h - 2, where h is the host bits. Minimum prefix blocks must align on natural network boundaries.

2

Analyze the remaining unallocated address range within the 172.20.0.0/19 block

The total /19 block spans 172.20.0.0 to 172.20.31.255 (8192 total IPs). Address space used so far is 172.20.0.0 through 172.20.7.127 (1920 IPs). Unallocated range is 172.20.7.128 to 172.20.31.255.

Subtracting allocated IP space from the base summary block isolates the remaining contiguous unassigned addresses.

3

Partition remaining unallocated space into standard summary subnets and find the largest block

Remaining space partitions into 172.20.7.128/25 (128 IPs), 172.20.8.0/21 (2048 IPs), and 172.20.16.0/20 (4096 IPs). The largest block is 172.20.16.0/20.

Subnets must align on binary bit boundaries corresponding to their prefix lengths.

4

Calculate usable hosts for the largest unallocated block (172.20.16.0/20)

2^12 - 2 = 4096 - 2 = 4094 usable host IPv4 addresses.

Subtracting the network ID and broadcast address from total block addresses gives usable host count.

Key Concept

VLSM block allocation, binary boundary alignment, and unallocated subnet summarization

Estimated Time:3m 0s

Question 4Question →

A network administrator assigns the IPv4 subnet mask $255.255.255.224$ to a VLAN interface on a Cisco switch. What is the maximum number of usable host IPv4 addresses that can be assigned to end devices in this subnet?

Show answer & explanation

Answer: 30

Answer

The maximum number of usable host IPv4 addresses is 30.

A subnet mask of

255.255.255.224

corresponds to a CIDR prefix of /27, leaving 5 bits for host addressing. Using the formula

2^n - 2

(where

n = 5

), the calculation yields

2^5 - 2 = 32 - 2 = 30

usable host IP addresses.

Step-by-Step Solution

1

Determine the number of host bits (

n

) available in the

255.255.255.224

subnet mask.

The mask has 27 network bits and 5 host bits (

32 - 27 = 5

).

The last octet value

224

in binary is

11100000_2

, meaning 3 bits belong to the network/subnet portion and 5 bits belong to the host portion.

2

Calculate the number of usable host IP addresses using the formula

2^n - 2

.

2^5 - 2 = 32 - 2 = 30

usable host addresses.

The formula

2^n - 2

accounts for the total IP addresses (

2^n

) minus the reserved network identifier (all host bits set to 0) and the broadcast address (all host bits set to 1).

Key Concept

Usable Host Address Calculation

Question 5Question →

A network administrator is designing an IPv4 addressing scheme for a local subnet that must support $60$ usable host IP addresses. The administrator assigns the smallest subnet prefix length (largest mask) that accommodates this host requirement. How many total usable host IPv4 addresses are provided by this assigned subnet?

Show answer & explanation

Answer: 62

Answer

The assigned subnet provides 62 total usable host IPv4 addresses.

To support 60 host addresses, a subnet must allocate 6 host bits because

2^6 - 2 = 62

usable addresses. 5 host bits only provide 30 usable addresses (

2^5 - 2

), which is insufficient.

Step-by-Step Solution

1

Determine the required number of host bits.

6 host bits are required (

2^6 - 2 = 62

).

5 host bits only yield

2^5 - 2 = 30

usable IP addresses, which is not enough for 60 hosts.

2

Calculate the usable host capacity of the subnet.

62 usable host addresses.

With 6 host bits, the total IP space is

2^6 = 64

. Subtracting the network address and the broadcast address leaves 62 usable host IP addresses.

Key Concept

IPv4 Usable Host Calculation

Estimated Time:45s

Question 6Question →

A network technician is configuring a new subnet with an IPv4 prefix length of $/28$ . How many usable host IPv4 addresses are available on this subnet?

Show answer & explanation

Answer: 14

Answer

14 usable IPv4 host addresses are available on a /28 subnet.

For a

/28

prefix, the host portion contains

32 - 28 = 4

bits. The formula to calculate usable hosts is

2^h - 2

, where

h

is the number of host bits. Therefore,

2^4 - 2 = 16 - 2 = 14

usable host IPv4 addresses.

Step-by-Step Solution

1

Determine the number of host bits available.

Host bits =

32 - 28 = 4

bits.

An IPv4 address consists of 32 bits. Subtracting the prefix length (/28) leaves the host portion of the address.

2

Calculate total IP address space for the subnet.

Total addresses =

2^4 = 16

addresses.

The formula

2^h

determines total IP address combinations, where

h

is the number of host bits.

3

Subtract reserved addresses to get usable host addresses.

Usable addresses =

16 - 2 = 14

addresses.

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

Key Concept

IPv4 Subnet Usable Host Calculation

Question 7Question →

A network engineer is assigned the IPv4 network block $10.140.0.0/18$ to provision subnets for standard branch offices. Each branch office requires a subnet capable of accommodating at least $110$ usable host IP addresses. What is the maximum number of equal-sized subnets supporting this host requirement that can be created from the $10.140.0.0/18$ block?

Show answer & explanation

Answer: 128

Answer

The maximum number of valid branch subnets that can be created is 128.

To support at least 110 usable hosts per subnet, 7 host bits are required (

2^7 - 2 = 126

usable hosts). Subtracting 7 host bits from 32 total IPv4 bits results in a

/25

prefix length. Borrowing bits from the allocated

/18

summary block gives

25 - 18 = 7

subnet bits. Therefore,

2^7 = 128

valid subnets can be created.

Step-by-Step Solution

1

Determine the required host bits for at least 110 usable host IP addresses.

7 host bits are required because

2^7 - 2 = 126 \ge 110

, whereas 6 host bits (

2^6 - 2 = 62

) are insufficient.

Every IPv4 subnet reserves two IP addresses for the network ID and the broadcast ID, making the usable host formula

2^h - 2

.

2

Determine the required CIDR prefix length for each subnet.

The prefix length is

/25

(

32 - 7 = 25

).

An IPv4 address consists of 32 bits; subtracting the 7 host bits leaves 25 network bits.

3

Calculate the number of borrowed subnet bits relative to the parent block.

7 subnet bits are borrowed (

25 - 18 = 7

).

Extending the parent network prefix from

/18

to

/25

borrows 7 bits for subnetting.

4

Calculate the total number of

/25

subnets within the

/18

parent block.

The total number of subnets is

2^7 = 128

.

Borrowing

n

subnet bits yields

2^n

distinct subnets.

Key Concept

Calculating IPv4 subnet capacity and prefix derivation based on host requirements

Estimated Time:2m 0s

Question 8Question →

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

Show answer & explanation

Answer: 6

Answer

The subnet provides 6 usable host IPv4 addresses.

With a prefix length of

/29

, there are

32 - 29 = 3

host bits available. The formula to calculate usable host addresses is

2^n - 2

, where

n

is the host bit count. Calculating

2^3 - 2 = 8 - 2 = 6

yields 6 usable host addresses.

Step-by-Step Solution

1

Determine host bits from CIDR prefix

3 host bits

An IPv4 address consists of 32 bits, so

32 - 29 = 3

bits remain for host identification.

2

Calculate total IP address space

8 total addresses

Total address space equals

2^n

where

n

is the host bit count (

2^3 = 8

).

3

Subtract reserved addresses

6 usable host addresses

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

Key Concept

IPv4 Usable Host Calculation

Estimated Time:45s

Question 9Question →

A network administrator is allocated the IPv4 address block $172.20.64.0/21$ for datacenter VLAN expansion. The architectural requirement mandates that the block be divided into the maximum possible number of equal-sized subnets such that each subnet can accommodate at least $50$ usable host addresses. If exactly $50$ hosts are assigned in every created subnet, how many usable host IP addresses across the entire $/21$ block remain unassigned?

Show answer & explanation

Answer: 384

Answer

384 usable host IP addresses remain unassigned across all created subnets.

To satisfy the requirement of at least

50

usable host addresses per subnet while maximizing the number of subnets,

6

host bits are required (

2^6 - 2 = 62

usable host addresses per subnet), resulting in a

/26

prefix length. Dividing the

/21

block into

/26

subnets produces

2^{26-21} = 32

subnets. With

50

host addresses assigned per subnet, each subnet has

62 - 50 = 12

unassigned usable host addresses. Across all

32

subnets, the total number of unassigned usable host IP addresses is

32 \times 12 = 384

.

Step-by-Step Solution

1

Determine the minimum host bits required to support at least 50 usable host addresses per subnet.

6 host bits (

h = 6

), which yields

2^6 - 2 = 62

usable host addresses per subnet and a prefix length of

/26

.

Using 5 host bits (

2^5 - 2 = 30

) is insufficient for 50 hosts. 6 host bits meet the minimum requirement while maximizing subnet creation.

2

Calculate the total number of

/26

subnets created from the

/21

summary block.

2^{26 - 21} = 2^5 = 32

subnets.

Borrowing 5 subnet bits from a

/21

block creates 32 subnets of size

/26

.

3

Determine the unassigned usable host addresses per subnet when 50 hosts are assigned.

62 - 50 = 12

unassigned usable host addresses per subnet.

Network and broadcast addresses are already excluded when calculating the 62 usable host addresses.

4

Multiply the unassigned usable host addresses per subnet by the total number of created subnets.

32 \text{ subnets} \times 12 \text{ addresses} = 384

total unassigned usable host IP addresses.

Aggregating the unused host capacity across all equal-sized subnets within the allocated block.

Key Concept

IPv4 Subnet Sizing and Usable Host Capacity Calculation

Question 10Question →

An organization is assigned the IPv4 address block $172.28.0.0/19$ to provision subnetting for a multi-building campus network. The system architecture mandates dividing this entire block into equal-sized subnets where each subnet must support at least $120$ usable host IP addresses. What is the maximum number of such equal-sized subnets that can be created from this $/19$ prefix?

Show answer & explanation

Answer: 64

Answer

64 subnets can be created.

To host at least 120 usable IP addresses, 7 host bits are required (

2^7 - 2 = 126

usable hosts). A block with 7 host bits corresponds to a

/25

prefix. Subnetting a

/19

block into

/25

subnets borrows

25 - 19 = 6

bits, resulting in

2^6 = 64

subnets.

Step-by-Step Solution

1

Determine the required host bits

7 host bits (

h = 7

)

The formula for usable IPv4 hosts per subnet is

2^h - 2

. For 120 hosts,

2^h - 2 \ge 120 \implies 2^h \ge 122

. The smallest power of two satisfying this is

2^7 = 128

(

128 - 2 = 126

usable host addresses).

2

Determine the required subnet prefix length

/25 prefix length

Subtracting 7 host bits from total 32 bits (

32 - 7 = 25

) gives a subnet mask of

/25

(

255.255.255.128

).

3

Calculate the number of subnets derived from the parent prefix

64 subnets

The parent prefix is

/19

and the subnet prefix is

/25

. The number of borrowed subnet bits is

25 - 19 = 6

bits. The total number of subnets created is

2^6 = 64

.

Key Concept

VLSM and Subnet Sizing Calculation

Question 11Question →

A network administrator is assigned the IPv4 address block $10.50.0.0/22$ to provision subnets for an enterprise IP telephony deployment. Each subnet requires support for at least $50$ usable host IP addresses. If the administrator subnets the entire $10.50.0.0/22$ block using a uniform prefix length that meets this host requirement with minimal wasted host space, how many total subnets can be created?

Show answer & explanation

Answer: 16

Answer

16 subnets can be created from the block.

To support at least 50 usable host IP addresses per subnet, 6 host bits are needed (

2^6 - 2 = 62

usable IPs), which corresponds to a /26 prefix length (

32 - 6 = 26

). Borrowing bits from the parent /22 prefix to create /26 subnets uses

26 - 22 = 4

subnet bits, resulting in

2^4 = 16

total subnets.

Step-by-Step Solution

1

Calculate the number of host bits required for 50 usable IP addresses

6 host bits are required

The formula for usable host IPs is

2^h - 2

. With

h = 5

,

2^5 - 2 = 30

hosts (insufficient). With

h = 6

,

2^6 - 2 = 62

usable hosts, which satisfies the requirement of at least 50 hosts.

2

Determine the new subnet prefix length

The prefix length is /26

Subtracting the 6 host bits from the 32-bit IPv4 address space yields

32 - 6 = 26

.

3

Calculate the number of subnets derived from the parent /22 block

16 subnets

Subtracting the original prefix (/22) from the new prefix (/26) gives

26 - 22 = 4

subnet bits. Borrowing 4 bits produces

2^4 = 16

subnets.

Key Concept

Subnet sizing and calculation of total subnets created using FLSM from a CIDR block.

Question 12Question →

A network architect is assigned the IPv4 address block $10.116.0.0/18$ to provision subnets for remote office sites. Each site requires a dedicated subnet that can support at least $120$ usable host IP addresses. What is the maximum number of subnets of this size that can be created from the allocated block?

Show answer & explanation

Answer: 128

Answer

The maximum number of subnets that can be created is 128.

To support at least 120 usable host IP addresses per subnet, 7 host bits are required because

2^7 - 2 = 126

usable hosts (

2^6 - 2 = 62

is insufficient). This results in a

/25

prefix (

32 - 7 = 25

). Borrowing bits from

/18

to

/25

gives

25 - 18 = 7

subnet bits. The total number of subnets created is

2^7 = 128

.

Step-by-Step Solution

1

Calculate the required host bits (

h

) for at least

120

usable host IP addresses.

7 host bits are required (

2^7 - 2 = 126 \ge 120

).

6 host bits provide only

2^6 - 2 = 62

usable hosts, which is insufficient. 7 host bits provide 126 usable hosts.

2

Determine the prefix length needed for 7 host bits.

The subnet prefix length is

/25

(

32 - 7 = 25

).

An IPv4 address consists of 32 bits, so subtracting 7 host bits leaves 25 network/subnet bits.

3

Determine the number of subnet bits borrowed from the allocated block.

7 bits are borrowed (

25 - 18 = 7

).

The original allocation is

/18

and the new subnets are

/25

.

4

Compute the total number of valid subnets.

The total number of subnets is

2^7 = 128

.

Borrowing 7 bits allows

2^7

distinct subnet combinations.

Key Concept

Subnet Sizing and Subnet Count Calculation

Estimated Time:1m 30s

Question 13Question →

A network engineer is allocated the IPv4 address block $172.16.64.0/21$ to provision point-to-point WAN links between branch offices. Each link is configured with a $/30$ subnet mask to conserve address space. What is the maximum number of $/30$ subnets that can be created from this $/21$ block?

Show answer & explanation

Answer: 512

Answer

The maximum number of

/30

subnets that can be created is 512.

To calculate the total number of subnets created, subtract the starting prefix length from the target prefix length:

30 - 21 = 9

borrowed subnet bits. Applying the formula

2^n

yields

2^9 = 512

subnets.

Step-by-Step Solution

1

Calculate the number of borrowed subnet bits.

9 subnet bits borrowed (

30 - 21 = 9

).

Subnetting a

/21

network block into

/30

subnets requires extending the prefix by 9 bits.

2

Calculate the total number of available subnets.

512 subnets (

2^9 = 512

).

Each borrowed bit doubles the number of created subnets, yielding

2^9 = 512

total subnets.

Key Concept

Calculating the total number of subnets created by borrowing host bits from an IPv4 CIDR prefix.

Question 14Question →

An enterprise network team is allocated the IPv4 address block $10.240.16.0/20$ to provision subnets for remote branch offices. Each branch subnet must support at least 60 usable host addresses. What is the maximum number of equal-sized subnets that can be created from this address block?

Show answer & explanation

Answer: 64

Answer

The maximum number of equal-sized subnets that can be created is 64.

To support 60 usable host addresses, a subnet requires 6 host bits because

2^6 - 2 = 62

usable addresses (

2^5 - 2 = 30

is insufficient). A subnet with 6 host bits corresponds to a prefix length of /26 (

32 - 6 = 26

). Subtracting the original block prefix (/20) from the subnet prefix (/26) yields 6 borrowed subnet bits. Calculating

2^6

gives 64 maximum equal-sized subnets.

Step-by-Step Solution

1

Calculate required host bits for 60 usable hosts

6 host bits (

2^6 - 2 = 62 \ge 60

)

Network and broadcast addresses are reserved and cannot be assigned to hosts, requiring subtraction of 2 from total addresses.

2

Determine target prefix length

Prefix length of /26

Subtracting 6 host bits from the 32 bits of an IPv4 address results in a 26-bit network prefix.

3

Calculate borrowed subnet bits

6 subnet bits

The difference between the new prefix (/26) and the allocated prefix (/20) represents the borrowed bits (

26 - 20 = 6

).

4

Calculate total usable subnets

64 subnets

The formula

2^s

, where

s

is the number of borrowed bits, gives

2^6 = 64

available subnets.

Key Concept

IPv4 Subnetting and Host Requirement Sizing

Question 15Question →

A network engineer is allocated the IPv4 address block $10.200.0.0/19$ to provision dedicated subnets for IP surveillance cameras across multiple industrial facilities. Each facility subnet must be configured using a $255.255.255.192$ subnet mask. How many distinct subnets can be created from this block?

Show answer & explanation

Answer: 128

Answer

128 subnets can be created.

To determine the number of subnets created, convert the subnet mask

255.255.255.192

to

/26

CIDR notation. Borrowing bits from the original

/19

prefix yields

26 - 19 = 7

subnet bits. Calculating

2^7

gives exactly 128 subnets.

Step-by-Step Solution

1

Determine the prefix length corresponding to the given subnet mask.

The dotted-decimal subnet mask

255.255.255.192

corresponds to a

/26

prefix (

24

network bits +

2

subnet bits in the fourth octet).

Converting the dotted-decimal mask to CIDR notation establishes the target prefix length.

2

Calculate the number of borrowed subnet bits.

26 \text{ bits (new prefix)} - 19 \text{ bits (original prefix)} = 7 \text{ borrowed subnet bits}

.

Subnetting extends the network prefix length from

/19

to

/26

.

3

Compute the total number of subnets.

2^7 = 128

total subnets.

The formula to calculate the number of subnets generated by borrowing

s

bits is

2^s

.

Key Concept

Subnet Count Calculation

Question 16Question →

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

Show answer & explanation

Answer: 32

Answer

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

Moving from a

/20

prefix to a

/25

prefix borrows 5 host bits (

25 - 20 = 5

). Calculating

2^5

gives 32 distinct subnets.

Step-by-Step Solution

1

Calculate the number of borrowed subnet bits

5 bits borrowed (

25 - 20 = 5

)

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

2

Calculate the total subnet count

32 subnets (

2^5 = 32

)

Each borrowed bit doubles the number of created subnets (

2^n

, where

n

is the number of borrowed bits).

Key Concept

Subnet counting using CIDR prefix borrowing

Question 17Question →

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

Show answer & explanation

Answer: 32

Answer

32 subnets can be created from the

10.100.0.0/21

block.

Subtracting the original prefix length (

/21

) from the new subnet prefix length (

/26

) yields

5

subnet bits (

26 - 21 = 5

). Calculating

2^5

gives

32

subnets available within the

10.100.0.0/21

parent block.

Step-by-Step Solution

1

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

Borrowed bits =

26 - 21 = 5

bits.

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

2

Calculate the total number of subnets using

2^n

, where

n

is the number of borrowed bits.

2^5 = 32

subnets.

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

Key Concept

Calculating Subnet Count from CIDR Prefix Differences

Estimated Time:1m 30s

Question 18Question →

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

Show answer & explanation

Answer: 510

Answer

510 usable host IPv4 addresses

To support 500 host devices, the number of host bits

h

must satisfy

2^h - 2 \ge 500

. A

/24

subnet (

h = 8

) yields only

2^8 - 2 = 254

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

/23

subnet (

h = 9

), which provides

2^9 - 2 = 510

usable host addresses.

Step-by-Step Solution

1

Determine the host bit requirement formula

Use

2^h - 2 \ge 500

, where

h

represents host bits.

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

2

Calculate the smallest number of host bits that accommodates 500 hosts

2^8 - 2 = 254

usable addresses (too small for 500 hosts);

2^9 - 2 = 510

usable addresses (meets the requirement).

Selecting 9 host bits leaves

32 - 9 = 23

network bits, resulting in a

/23

prefix length.

3

Calculate the usable host capacity

2^9 - 2 = 512 - 2 = 510

usable host addresses.

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

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