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Q6(b)(ii) · CIDR block

2022-11-16 · AI-only Sem V

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.