An organization requires a range of IP address to assign one to each of its…

2020

An organization requires a range of IP address to assign one to each of its 1500 computers. The organization has approached an Internet Service Provider (ISP) for this task. The ISP uses CIDR and serves the requests from the available IP address space 202.61.0.0/17. The ISP wants to assign an address space to the organization which will minimize the number of routing entries in the ISP’s router using route aggregation. Which of the following address spaces are potential candidates from which the ISP can allot any one of the organization ?

I. 202.61.84.0 / 21
II. 202.61.104.0 / 21
III. 202.61.64.0 / 21
IV. 202.61.144.0 / 21 

  1. A.

    I and II only

  2. B.

    II and III onlY

  3. C.

    III and IV only

  4. D.

    I and IV only

Attempted by 79 students.

Show answer & explanation

Correct answer: B

Key points: determine smallest prefix that fits 1500 hosts, check valid /21 boundaries, and ensure the prefix lies inside 202.61.0.0/17.

  • Address count: A /21 prefix provides 2^(32-21) = 2^11 = 2048 addresses, which is sufficient for 1500 hosts. A /22 provides only 1024 addresses and is insufficient.

  • Valid /21 boundaries: The mask 255.255.248.0 means the third octet must be a multiple of 8 (0, 8, 16, 24, ..., 120, 128, ...). A candidate like 202.61.84.0/21 is not valid because 84 is not a multiple of 8.

  • ISP allocation range: 202.61.0.0/17 covers addresses from 202.61.0.0 through 202.61.127.255 (third-octet values 0–127). Any assigned /21 must start within this range.

Evaluate each candidate:

  • 202.61.84.0/21 — Not valid: 84 is not a multiple of 8, so this is not a correct /21 network boundary.

  • 202.61.104.0/21 — Valid: 104 is a multiple of 8 and lies within the ISP's /17 range, and the prefix provides 2048 addresses.

  • 202.61.64.0/21 — Valid: 64 is a multiple of 8 and lies within the ISP's /17 range, and the prefix provides 2048 addresses.

  • 202.61.144.0/21 — Not valid for this ISP allocation: 144 is outside the third-octet range 0–127 covered by 202.61.0.0/17.

Conclusion: the only suitable networks from the list are 202.61.104.0/21 and 202.61.64.0/21.

Explore the full course: Gate Guidance By Sanchit Sir