(a) ISP Block 80.70.56.0/21 — Number and Range

Given: network address $= 80.70.56.0$, prefix $= /21$.

Step 1 — Number of addresses

$$2^{32-21} = 2^{11} = 2048 \text{ addresses}$$

Step 2 — Subnet mask: 21 bits = two full octets + 5 bits:

$$255.255.248.0$$

Step 3 — Range: the third octet advances in steps of $256 - 248 = 8$, starting at 56, so it ends at $56 + 8 - 1 = 63$.

• First address: 80.70.56.0
• Last (broadcast) address: 80.70.63.255
• Check: $8 \times 256 = 2048$ ✓

Answer (a): 2048 addresses, from 80.70.56.0 to 80.70.63.255.

(b) Allocation to Organisations and Unallocated Range

Sizing rule: give each organisation the next power of two block that can hold the requested number of usable addresses (usable $= 2^h - 2$ for a normal subnet).

Sequential allocation from 80.70.56.0 upward:

1. $2 \times 512 = 1024$ addresses
2. $2 \times 256 = 512$ addresses
3. $3 \times 64 = 192$ addresses

Total used $= 1024 + 512 + 192 = 1728$ addresses.

Unallocated addresses:

$$2048 - 1728 = 320 \text{ addresses} \Rightarrow \textbf{80.70.62.192 – 80.70.63.255}$$

(as 80.70.62.192/26 → 64 addresses, plus 80.70.63.0/24 → 256 addresses; $64 + 256 = 320$ ✓).

Answer (b): Org1 → 80.70.56.0–80.70.57.255, Org2 → 80.70.58.0–80.70.59.255, Org3 → 80.70.60.0–80.70.60.255, Org4 → 80.70.61.0–80.70.61.255, Org5 → 80.70.62.0–80.70.62.63, Org6 → 80.70.62.64–80.70.62.127, Org7 → 80.70.62.128–80.70.62.191; unallocated range 80.70.62.192 – 80.70.63.255 (320 addresses).