A network engineer is configuring an internal routing domain using the assigned IP address… — CompTIA Network+

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?

Answer: 64 subnets

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

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.

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.

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

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic