--- title: "Class C pool for 10, 12, 8, 14" subject: "Computer Networks and Security" chapter: 4 key: "semester_7/computer_networks_and_security:subnetting-class-c-four-depts" source: https://www.bctengineer.com/q/semester_7/computer_networks_and_security/subnetting-class-c-four-depts site: BCT Engineering --- # Class C pool for 10, 12, 8, 14 Computer Networks and Security · Chapter 4: Network Layer - Asked in: 2070 Chaitra, 8 marks - 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-four-depts - Whole chapter: https://www.bctengineer.com/md/semester_7/computer_networks_and_security/chapter-4 ## Question How can you dedicate 10, 12, 8 and 14 public IP addresses to departments A, B, C and D respectively from the pool of class C with minimum losses of IP? Explain. ## Short answer VLSM: all departments need /28 blocks (16 addresses). D: 200.1.1.0/28, B: 200.1.1.16/28, A: 200.1.1.32/28, C: 200.1.1.48/28. Total wastage 12 host addresses. ## Answer Assumption: the public Class C pool is 200.1.1.0/24 and each number is the count of hosts. **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 | 10 | 12 | 16 | 4 | /28 | 255.255.255.240 | | B | 12 | 14 | 16 | 4 | /28 | 255.255.255.240 | | C | 8 | 10 | 16 | 4 | /28 | 255.255.255.240 | | D | 14 | 16 | 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 | |---|---|---|---|---|---| | D | 200.1.1.0/28 | 200.1.1.1 – 200.1.1.14 | 200.1.1.15 | 255.255.255.240 | 0 | | B | 200.1.1.16/28 | 200.1.1.17 – 200.1.1.30 | 200.1.1.31 | 255.255.255.240 | 2 | | A | 200.1.1.32/28 | 200.1.1.33 – 200.1.1.46 | 200.1.1.47 | 255.255.255.240 | 4 | | C | 200.1.1.48/28 | 200.1.1.49 – 200.1.1.62 | 200.1.1.63 | 255.255.255.240 | 6 | #### Step 3: Wastage | Dept | Usable addresses | Needed | Wasted | |---|---|---|---| | D | 14 | 14 | 0 | | B | 14 | 12 | 2 | | A | 14 | 10 | 4 | | C | 14 | 8 | 6 | | **Total** | 56 | 44 | 12 | Addresses allocated: 64 of 256. The remaining 192 addresses (200.1.1.64/26, 200.1.1.128/25) stay free for future departments and are not counted as wastage. **Answer: A: 200.1.1.32/28; B: 200.1.1.16/28; C: 200.1.1.48/28; D: 200.1.1.0/28.** Total wastage = 12 host addresses. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does D (14 hosts) get a /28? A. 14 needs a /27 B. /28 has 16 usable hosts C. 14 + 2 = 16 fits exactly in 2^4 D. 14 hosts need /29 Answer: C. A /28 has 16 addresses with 14 usable, zero waste. (Key point: Block for 14 hosts) 2. Why do departments with 8, 10 and 12 hosts also get /28? A. Each needs 16 hosts B. Smallest power of 2 >= hosts + 2 is 16 for each C. They share D's block D. /28 is the only mask allowed Answer: B. 10, 12 and 14 addresses all round up to 16; 8 + 2 = 10 also rounds to 16. (Key point: Same block for all) 3. What is A's (10 hosts) network address? A. 200.1.1.48/28 B. 200.1.1.16/28 C. 200.1.1.32/28 D. 200.1.1.0/28 Answer: C. D .0, B .16, A .32, C .48. (Key point: Network address of A) 4. What is the total wastage? A. 56 B. 64 C. 12 D. 44 Answer: C. 0 + 2 + 4 + 6 = 12. (Key point: Total wastage) ## Close variants set in other exams - [Class C pool for 32, 65, 10, 21, 9](https://www.bctengineer.com/question.md?key=semester_7%2Fcomputer_networks_and_security%3Asubnetting-class-c-five-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 - [K-map Σ(1,2,3,8,9,10,11,14)+d(0,4,12), NAND](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Akmap-nand-1238-dc-0412) — Digital Logic, chapter 3, asked 5× - [Σ(0,1,3,6,10,12,14) using 8×1 MUX](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Amux-8x1-function-0136) — Digital Logic, chapter 4