--- title: "Subnet 10.10.10.0/24 for 5, 16, 23, 27" subject: "Computer Networks and Security" chapter: 4 key: "semester_7/computer_networks_and_security:subnetting-10-10-10-0" source: https://www.bctengineer.com/q/semester_7/computer_networks_and_security/subnetting-10-10-10-0 site: BCT Engineering --- # Subnet 10.10.10.0/24 for 5, 16, 23, 27 Computer Networks and Security · Chapter 4: Network Layer - Asked in: 2071 Chaitra, 10 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-10-10-10-0 - Whole chapter: https://www.bctengineer.com/md/semester_7/computer_networks_and_security/chapter-4 ## Question You are given the following address space 10.10.10.0/24. You have to assign addresses to 4 departments with the following hosts 5, 16, 23 and 27 respectively. Perform the subnetting in such a way that the IP address wastage in each department is minimum. Also find out the subnet mask, network address, broadcast address and unassigned range in each department. ## Short answer VLSM: largest first, block sizes /27 for D, C, B and /29 for A. D: 10.10.10.0/27, C: 10.10.10.32/27, B: 10.10.10.64/27, A: 10.10.10.96/29. Total wastage 25 host addresses. ## Answer Here unassigned range means the usable addresses left unused in each department's block. **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 | 5 | 7 | 8 | 3 | /29 | 255.255.255.248 | | B | 16 | 18 | 32 | 5 | /27 | 255.255.255.224 | | C | 23 | 25 | 32 | 5 | /27 | 255.255.255.224 | | D | 27 | 29 | 32 | 5 | /27 | 255.255.255.224 | #### Step 2: Allocation (largest first) from 10.10.10.0/24 | Dept | Network address | Usable host range | Broadcast address | Mask | Assigned / spare range | |---|---|---|---|---|---| | D | 10.10.10.0/27 | 10.10.10.1 – 10.10.10.30 | 10.10.10.31 | 255.255.255.224 | 10.10.10.1 – 10.10.10.27 used; spare 10.10.10.28 – 10.10.10.30 | | C | 10.10.10.32/27 | 10.10.10.33 – 10.10.10.62 | 10.10.10.63 | 255.255.255.224 | 10.10.10.33 – 10.10.10.55 used; spare 10.10.10.56 – 10.10.10.62 | | B | 10.10.10.64/27 | 10.10.10.65 – 10.10.10.94 | 10.10.10.95 | 255.255.255.224 | 10.10.10.65 – 10.10.10.80 used; spare 10.10.10.81 – 10.10.10.94 | | A | 10.10.10.96/29 | 10.10.10.97 – 10.10.10.102 | 10.10.10.103 | 255.255.255.248 | 10.10.10.97 – 10.10.10.101 used; spare 10.10.10.102 | #### Step 3: Wastage | Dept | Usable addresses | Needed | Wasted | |---|---|---|---| | D | 30 | 27 | 3 | | C | 30 | 23 | 7 | | B | 30 | 16 | 14 | | A | 6 | 5 | 1 | | **Total** | 96 | 71 | 25 | Addresses allocated: 104 of 256. The remaining 152 addresses (10.10.10.104/29, 10.10.10.112/28, 10.10.10.128/25) stay free for future departments and are not counted as wastage. **Answer: A: 10.10.10.96/29; B: 10.10.10.64/27; C: 10.10.10.32/27; D: 10.10.10.0/27.** Total wastage = 25 host addresses. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which prefix does A (5 hosts) get? A. /27 B. /30 C. /28 D. /29 Answer: D. 7 addresses need 8 = 2^3, so /29 (6 usable). (Key point: Block for 5 hosts) 2. Why does C (23 hosts) get a /27? A. 23 hosts need 23 addresses only B. 25 addresses need 32 C. Because B uses /27 D. A /28 holds 23 Answer: B. Smallest 2^n >= 25 is 32, so /27. (Key point: Block for 23 hosts) 3. What is A's unused usable address? A. 10.10.10.104 B. 10.10.10.102 C. 10.10.10.96 D. 10.10.10.103 Answer: B. A 10.10.10.96/29 has usable .97-.102; hosts use .97-.101, leaving .102 spare (.103 is broadcast). (Key point: Unassigned range for A) 4. Which department is placed first at 10.10.10.0? A. D (27 hosts), the largest B. C (23 hosts) C. B (16 hosts) D. A (5 hosts), the smallest Answer: A. Largest first, so D at 10.10.10.0/27. (Key point: Allocation order) 5. What is the total wastage? A. 71 B. 96 C. 104 D. 25 Answer: D. 3 + 7 + 14 + 1 = 25. (Key point: Total wastage) ## Close variants set in other exams - [Subnet 202.70.91.0/24 for 25, 16, 29, 11](https://www.bctengineer.com/question.md?key=semester_7%2Fcomputer_networks_and_security%3Asubnetting-202-70-91) — Computer Networks and Security, chapter 4, asked 2× ## Related in Computer Networks and Security - [Subnet 202.77.19.0/24 for 30, 24, 25, 20](https://www.bctengineer.com/question.md?key=semester_7%2Fcomputer_networks_and_security%3Asubnetting-202-77-19) — Computer Networks and Security, chapter 4 - [Subnet 120.89.96.0/19 for 100-800 hosts](https://www.bctengineer.com/question.md?key=semester_7%2Fcomputer_networks_and_security%3Asubnetting-120-89-96-four-depts) — Computer Networks and Security, chapter 4, asked 2× - [Subnet 202.70.64.0/19 for 100-800 hosts](https://www.bctengineer.com/question.md?key=semester_7%2Fcomputer_networks_and_security%3Asubnetting-202-70-64-19) — Computer Networks and Security, chapter 4, asked 2× - [Subnet 120.89.96.0/19 for 500-2000 hosts](https://www.bctengineer.com/question.md?key=semester_7%2Fcomputer_networks_and_security%3Asubnetting-120-89-96-500-2000) — Computer Networks and Security, chapter 4 - [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