# Q7(b) · IPv4 /26 block

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/papers/paper-2025-feb-reexam/q7b

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/papers/paper-2025-feb-reexam--q7b/

Actual paper; 100 marks attempted. Current-syllabus alignment is stated per question.

### Q7(b). A block of addresses is granted to a small organization. One of the address is 210.16.36.56/26. Find the first address in the block, last address in the block and total number of addresses possible in the block. [5 marks]

### Answer

### Given

210.16.36.56/26

### 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 − 26 = 6. N = 64.

### Step 2. Locate the aligned block

Mask = 255.255.255.192. The first varying octet is octet 4; its block width is 64. The supplied value 56 lies in the interval 0–63. 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 = **210.16.36.0**; last = **210.16.36.63**; total = **64** addresses. Usual usable host range: 210.16.36.1–210.16.36.62.