CIDR Sub-block Design — 120.60.4.0/22

Given: ISP block = 120.60.4.0/22; 100 organisations; 8 addresses each.

Step 1 — Size of the ISP block

A /22 prefix leaves 32 − 22 = 10 host bits:

$$2^{10} = 1024 \text{ addresses}$$

Mask = 255.255.252.0, so the block spans 120.60.4.0 → 120.60.7.255 (third octet 4…7).

Step 2 — Size of each sub-block

Each organisation gets 8 = 2³ addresses → 3 host bits, hence the prefix length is

$$32 - 3 = 29 \;\Rightarrow\; \textbf{/29 per organisation}$$

A /29 block has a block size of 8 addresses, exactly what is required.

Step 3 — Design of the sub-blocks

Total requirement = 100 × 8 = 800 ≤ 1024 ✓, so all organisations fit. The /29 sub-blocks are carved consecutively from the beginning of the ISP block:

Organisation	Sub-block (CIDR)	Address range
Org 1	120.60.4.0/29	120.60.4.0 – 120.60.4.7
Org 2	120.60.4.8/29	120.60.4.8 – 120.60.4.15
Org 3	120.60.4.16/29	120.60.4.16 – 120.60.4.23
…	… (+8 each time)	…
Org 100	120.60.7.24/29	120.60.7.24 – 120.60.7.31

Position of Org 100: offset = (100 − 1) × 8 = 792 addresses from the start. 792 = 3 × 256 + 24, so the third octet advances by 3 (4 + 3 = 7) and the fourth octet is 24 → 120.60.7.24/29.

General form of the n-th sub-block (n = 1…100):
120.60.(4 + ⌊8(n−1)/256⌋).(8(n−1) mod 256)/29

Step 4 — Addresses still available

$$1024 - 800 = \textbf{224 addresses still available}$$

These are 120.60.7.32 → 120.60.7.255 (255 − 32 + 1 = 224 ✓). They remain with the ISP and can be given out later — e.g. as three more /29s plus a /27, a /26 and a /25 (32 + 64 + 128 = 224) — or used for the ISP's own infrastructure (routers, DNS, web servers).

Final answer

Item	Result
ISP block size	/22 = 1024 addresses (120.60.4.0 – 120.60.7.255)
Organisation sub-block	/29 (8 addresses)
Org 1 / Org 100	120.60.4.0/29 / 120.60.7.24/29
Allocated	800 addresses
Remaining	224 addresses (120.60.7.32 – 120.60.7.255)