(a) Number and Range of Addresses in the ISP Block 80.70.56.0/21

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

Step 1 — Number of addresses

Formula: $\text{Number of addresses} = 2^{32 - \text{prefix length}}$

$$2^{32-21} = 2^{11} = 2048$$

The block contains 2048 addresses.

Step 2 — Subnet mask

/21 means the first 21 bits are 1s: two full octets plus 5 bits of the third octet.

$$11111111.11111111.11111000.00000000 = 255.255.248.0$$

Step 3 — Range of addresses

The third octet advances in steps of $256 - 248 = 8$. The block starts at the third octet value 56 and therefore ends at $56 + 8 - 1 = 63$.

• First (network) address = 80.70.56.0
• Last (broadcast) address = 80.70.63.255

Check: 8 values of the third octet (56 to 63) $\times 256 = 2048$ addresses $\checkmark$

Answer: the ISP block has 2048 addresses, ranging from 80.70.56.0 to 80.70.63.255 (first address 80.70.56.0, broadcast address 80.70.63.255).