An enterprise network architect is assigned the IPv4 block 10.180.64.0/19. Using Variable … — CompTIA Network+

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?

Cevap: 4094 hosts

Cevap

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.

Adım Adım Çözüm

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.

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).

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.

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.

Anahtar Kavram

Variable Length Subnet Masking (VLSM) and Bit Boundary Alignment

Tahmini Süre:3m 0s

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru