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Zorluk: Çok zorIPv4 Addressing and Subnetting

A network architect is assigned the IPv4 block 172.16.32.0/20172.16.32.0/20 to provision subnets for four new departments in a branch office. The requirements for usable host IP addresses are as follows: Department A requires 500500 hosts, Department B requires 250250 hosts, Department C requires 100100 hosts, and Department D requires 5050 hosts. Using Variable Length Subnet Masking (VLSM) and allocating subnets contiguously starting from the lowest available network address in the block in order of largest to smallest host requirement, what is the broadcast address of Department D's subnet, and how many total IP addresses remain unallocated within the original /20/20 block?

  1. Broadcast address: 172.16.35.191172.16.35.191; Unallocated IP addresses: 31363136Cevap
  2. B
    Broadcast address: 172.16.35.190172.16.35.190; Unallocated IP addresses: 31363136
  3. C
    Broadcast address: 172.16.35.191172.16.35.191; Unallocated IP addresses: 31283128
  4. D
    Broadcast address: 172.16.35.255172.16.35.255; Unallocated IP addresses: 29442944

Cevap

Broadcast address: 172.16.35.191172.16.35.191; Unallocated IP addresses: 31363136
Evaluating host requirements with VLSM yields block sizes of 512512 (/23/23), 256256 (/24/24), 128128 (/25/25), and 6464 (/26/26). Sequential assignment starting from 172.16.32.0172.16.32.0 places Department D at 172.16.35.128/26172.16.35.128/26, which spans 172.16.35.128172.16.35.128 through 172.16.35.191172.16.35.191, making 172.16.35.191172.16.35.191 the broadcast address. Subtracting total allocated addresses (960960) from the total block size of 40964096 addresses leaves 31363136 unallocated IP addresses.

Adım Adım Çözüm

1
Determine total capacity of the parent block
A /20/20 network has 32−20=1232 - 20 = 12 host bits, yielding 212=40962^{12} = 4096 total IP addresses.
Calculating total initial pool size is required to find remaining unallocated space.
2
Calculate required prefix lengths and block sizes using VLSM in descending order
Department A (500500 hosts): Needs /23/23 (29−2=5102^9 - 2 = 510 usable, block size 512512). Department B (250250 hosts): Needs /24/24 (28−2=2542^8 - 2 = 254 usable, block size 256256). Department C (100100 hosts): Needs /25/25 (27−2=1262^7 - 2 = 126 usable, block size 128128). Department D (5050 hosts): Needs /26/26 (26−2=622^6 - 2 = 62 usable, block size 6464).
Subnet allocation requires accommodating host counts plus network and broadcast overhead (+2+2).
3
Assign contiguous subnet ranges from lowest available address
Dept A: 172.16.32.0/23172.16.32.0/23 (172.16.32.0−172.16.33.255172.16.32.0 - 172.16.33.255). Dept B: 172.16.34.0/24172.16.34.0/24 (172.16.34.0−172.16.34.255172.16.34.0 - 172.16.34.255). Dept C: 172.16.35.0/25172.16.35.0/25 (172.16.35.0−172.16.35.127172.16.35.0 - 172.16.35.127). Dept D: 172.16.35.128/26172.16.35.128/26 (172.16.35.128−172.16.35.191172.16.35.128 - 172.16.35.191).
Subnets must start on valid binary boundaries matching their block sizes.
4
Identify Department D broadcast address and calculate unallocated addresses
Broadcast address for Dept D is 172.16.35.191172.16.35.191. Total allocated addresses =512+256+128+64=960= 512 + 256 + 128 + 64 = 960. Remaining unallocated addresses =4096−960=3136= 4096 - 960 = 3136.
Broadcast is the last address of the subnet block, and unallocated capacity is total capacity minus used capacity.

Anahtar Kavram

Variable Length Subnet Masking (VLSM) and IPv4 Subnet Allocation
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IPv4 Addressing and Subnetting

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