A network architect is implementing Variable Length Subnet Masking (VLSM) for an enterpris… — Cisco CCNA

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?

Answer: 4094 usable host addresses

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

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

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

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.

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.

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.

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

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