--- title: "Class C pool for 32, 65, 10, 21, 9" subject: "Computer Networks and Security" chapter: 4 key: "semester_7/computer_networks_and_security:subnetting-class-c-five-depts" source: https://www.bctengineer.com/q/semester_7/computer_networks_and_security/subnetting-class-c-five-depts site: BCT Engineering --- # Class C pool for 32, 65, 10, 21, 9 Computer Networks and Security · Chapter 4: Network Layer - Asked in: 2074 Chaitra, 8 marks - Video: [Variable length Subnet masks (VLSM) example](https://www.youtube.com/watch?v=OgjeNzuRBp0) — CS-IT Tech Studio - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/computer_networks_and_security/subnetting-class-c-five-depts - Whole chapter: https://www.bctengineer.com/md/semester_7/computer_networks_and_security/chapter-4 ## Question How can you dedicate 32, 65, 10, 21 and 9 public IP addresses to the departments A, B, C, D and E respectively from the pool of class C IP addresses with minimum loss? Explain. ## Short answer Use VLSM, largest first from 200.1.1.0/24: B gets 200.1.1.0/25, A 200.1.1.128/26, D 200.1.1.192/27, C 200.1.1.224/28, E 200.1.1.240/28. Total host-address wastage is 109. ## Answer Assumption: the Class C pool is 200.1.1.0/24 (256 addresses) and each number is the count of hosts to be addressed. **Method (VLSM):** each department needs *hosts + 2* addresses (network address and broadcast address cannot be given to hosts). Choose the smallest block 2ⁿ ≥ hosts + 2, where n is the number of host bits and prefix = 32 − n. Allocate the **largest block first**, one after another without overlap, so every block starts on a boundary that is a multiple of its size and the waste is minimum. #### Step 1: Block size for each department | Dept | Hosts | Hosts + 2 | Block 2ⁿ | n | Prefix | Subnet mask | |---|---|---|---|---|---|---| | A | 32 | 34 | 64 | 6 | /26 | 255.255.255.192 | | B | 65 | 67 | 128 | 7 | /25 | 255.255.255.128 | | C | 10 | 12 | 16 | 4 | /28 | 255.255.255.240 | | D | 21 | 23 | 32 | 5 | /27 | 255.255.255.224 | | E | 9 | 11 | 16 | 4 | /28 | 255.255.255.240 | #### Step 2: Allocation (largest first) from 200.1.1.0/24 | Dept | Network address | Usable host range | Broadcast address | Mask | Unused hosts | |---|---|---|---|---|---| | B | 200.1.1.0/25 | 200.1.1.1 – 200.1.1.126 | 200.1.1.127 | 255.255.255.128 | 61 | | A | 200.1.1.128/26 | 200.1.1.129 – 200.1.1.190 | 200.1.1.191 | 255.255.255.192 | 30 | | D | 200.1.1.192/27 | 200.1.1.193 – 200.1.1.222 | 200.1.1.223 | 255.255.255.224 | 9 | | C | 200.1.1.224/28 | 200.1.1.225 – 200.1.1.238 | 200.1.1.239 | 255.255.255.240 | 4 | | E | 200.1.1.240/28 | 200.1.1.241 – 200.1.1.254 | 200.1.1.255 | 255.255.255.240 | 5 | #### Step 3: Wastage | Dept | Usable addresses | Needed | Wasted | |---|---|---|---| | B | 126 | 65 | 61 | | A | 62 | 32 | 30 | | D | 30 | 21 | 9 | | C | 14 | 10 | 4 | | E | 14 | 9 | 5 | | **Total** | 246 | 137 | 109 | Addresses allocated: 128 + 64 + 32 + 16 + 16 = 256 of 256, so the Class C pool is used exactly, with no leftover block. **Answer: A: 200.1.1.128/26; B: 200.1.1.0/25; C: 200.1.1.224/28; D: 200.1.1.192/27; E: 200.1.1.240/28.** Total wastage = 109 host addresses. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which prefix does B (65 hosts) get, and why? A. /26 since 64 is close B. /27 since it is a small department C. /25 since 67 addresses need 128 D. /24 since it is a Class C Answer: C. 65 + 2 = 67 > 64, so 128 = 2^7 is needed, giving /25. (Key point: Block for B) 2. What mask does D (21 hosts) use? A. 255.255.255.192 B. 255.255.255.240 C. 255.255.255.248 D. 255.255.255.224 Answer: D. 23 addresses need 32 = 2^5, so /27 = 255.255.255.224. (Key point: Block for D) 3. What is the total of all block sizes, and what does that mean? A. 512, so another pool is needed B. 128 only, half the pool is free C. 200, leaving 56 free D. 256, so the Class C pool is used exactly Answer: D. 128 + 64 + 32 + 16 + 16 = 256, using the whole /24. (Key point: Pool usage) 4. What is the total wastage? A. 61 B. 109 C. 246 D. 137 Answer: B. 61 + 30 + 9 + 4 + 5 = 109. (Key point: Total wastage) ## Close variants set in other exams - [Class C pool for 10, 12, 8, 14](https://www.bctengineer.com/question.md?key=semester_7%2Fcomputer_networks_and_security%3Asubnetting-class-c-four-depts) — Computer Networks and Security, chapter 4, asked 2× ## Related in Computer Networks and Security - [Subnetting for Banijya bank departments](https://www.bctengineer.com/question.md?key=semester_7%2Fcomputer_networks_and_security%3Asubnetting-banijya-bank) — Computer Networks and Security, chapter 4 ## Related in other subjects - [AVL tree for 52, 65, 72, 15, 11, 32, 20](https://www.bctengineer.com/question.md?key=semester_4%2Fdata_structure_and_algorithm%3Aavl-build-52-65-72) — Data Structure and Algorithm, chapter 6