# Q6(b) · IPv4 /26 block

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/papers/paper-2023-final/q6b

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/papers/paper-2023-final--q6b/

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

### Q6(b). A block of address is granted to a small organization and one of the address is 211.15.35.40/26. Determine the first address and last address in the block. Determine the total number of address in the block. [5 marks]

### Answer

### Given

211.15.35.40/26

### 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 40 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 = **211.15.35.0**; last = **211.15.35.63**; total = **64** addresses. Usual usable host range: 211.15.35.1–211.15.35.62.