Computer Networks · TEE · papers

Q7(b) · IPv4 /26 block

2025-02-19 · 2023-24 / 2024-25

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}LaTeX: 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.