Subnetting an ISP Allocation for 100 Organizations (120.60.4.0/22)

We are given an ISP address block keyword block of 120.60.4.0/22. The ISP must split it into sub-blocks sized so that each of 100 organizations receives just eight addresses.

We will:

1. Determine the size (number of addresses) of the original /22 block.
2. Choose the correct CIDR prefix length for “8 addresses”.
3. Allocate 100 equal sub-blocks sequentially.
4. Compute how many addresses remain unused.

We’ll use standard CIDR facts: an IPv4 block with prefix length $p$ contains $2^{32-p}$ addresses, and for subnet boundaries the increment in the last octet follows the subnet size.

1) Compute how many addresses are in 120.60.4.0/22

For IPv4, a CIDR block keyword is written as $/p$. The number of addresses in the block is:

$2^{32-p} = 2^{32-22} = 2^{10} = 1024$

So, 120.60.4.0/22 contains 1024 addresses.

Key idea: since $/22$ means the subnet mask covers the first 22 bits, the block spans in steps of $2^{10}=1024$ addresses, which corresponds to 256 in the last octet times 4 in the third octet (because $1024 = 4 \times 256$ addresses). Concretely, $120.60.4.0/22$ covers:

• from 120.60.4.0
• through 120.60.7.255

Thus, the usable range is: 120.60.4.0 – 120.60.7.255.

2) Find the CIDR prefix length that gives exactly 8 addresses

We need a sub-block with 8 addresses. For IPv4:

$2^{32-p} = 8 = 2^3 \Rightarrow 32-p = 3 \Rightarrow p = 29$

So, the correct sub-block size is /29, because:

• /29 contains $2^{32-29}=2^3=8$ addresses.

keyword Subnet mask for /29 is 255.255.255.248 (standard).
Therefore each organization should be assigned a /29 network: 8 total addresses per organization.

3) Sub-block design (slash notation for each sub-block)

Each organization gets one /29. The first few /29 networks from the base block are:

Continuing the pattern:

• Within a fixed third octet (here 4), the last octet starts at 0 and increases by 8: $0, 8, 16, \dots, 248$.
• There are $256/8 = 32$ distinct /29 networks per third-octet value.
• Since the original /22 spans third octets 4 through 7 (4 values), there are $4 \times 32 = 128$ total /29 subnets available in the whole /22.

So to allocate 100 subnets:

• Allocate 32 subnets in third octet 4 (Org 1–32)
• Allocate 32 subnets in third octet 5 (Org 33–64)
• Allocate 32 subnets in third octet 6 (Org 65–96)
• Allocate the remaining 4 subnets in third octet 7 (Org 97–100)

Exact slash notation blocks for Org 1–100

To keep this readable while still exact, use the following enumeration rules:

For third octet = 4 (Org 1–32):

• Org $k$ (where $k=1..32$) gets:
  • 120.60.4.$(8\\times (k-1))$/29

So Org 1–32 are:

• 120.60.4.0/29, 120.60.4.8/29, 120.60.4.16/29, ..., 120.60.4.248/29

For third octet = 5 (Org 33–64):

• Org $k$ (where $k=33..64$) gets:
  • 120.60.5.$(8\\times (k-33))$/29

So Org 33–64 are:

• 120.60.5.0/29, 120.60.5.8/29, 120.60.5.16/29, ..., 120.60.5.248/29

For third octet = 6 (Org 65–96):

• Org $k$ (where $k=65..96$) gets:
  • 120.60.6.$(8\\times (k-65))$/29

So Org 65–96 are:

• 120.60.6.0/29, 120.60.6.8/29, 120.60.6.16/29, ..., 120.60.6.248/29

For third octet = 7 (Org 97–100): Only the first 4 /29 blocks in that third octet:

• Org 97: 120.60.7.0/29
• Org 98: 120.60.7.8/29
• Org 99: 120.60.7.16/29
• Org 100: 120.60.7.24/29

4) How many addresses are still available after these allocations?

We computed:

• Original addresses in /22: $1024$
• Addresses allocated: $100 \\times 8 = 800$

So remaining addresses:

$1024 - 800 = 224$

Therefore, 224 addresses are still available inside 120.60.4.0/22 after allocating 100 organizations with /29 blocks.

Additionally, since each /29 uses 8 addresses, the number of unused /29 subnets is: $224/8 = 28$ So there are 28 unused /29 sub-blocks remaining.