# Q6(b)(ii) · CIDR block

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/papers/paper-2022-ai-final/q6bii

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/papers/paper-2022-ai-final--q6bii/

Exact historical variant. Current-syllabus questions solved; excluded topics and source limitations stated.

### Q6(b)(ii). One address in an allocated block is 210.90.12.7/20. Find its first address, last address and number of addresses. [5 marks]

### Answer

### Given

210.90.12.7/20

### To find

First and last addresses and total block size.

### Formula and why



$$
N=2^{32-p}
$$



p is the prefix length. The remaining host bits vary through every combination. The network address is the first address; setting all host bits to one gives the last address.

### Step 1. Find host bits

32 − 20 = 12. N = 4096.

### Step 2. Locate the aligned block

Mask = 255.255.240.0. The first varying octet is octet 3; its block width is 16. The supplied value 12 lies in the interval 0–15. Keep earlier octets fixed. Set any later octets to 0 for the first address and 255 for the last address.

### Step 3. Write the result

First = **210.90.0.0**; last = **210.90.15.255**; total = **4096** addresses. Usual usable host range: 210.90.0.1–210.90.15.254.