--- title: "Subnet 202.70.91.0/24 for 25, 16, 29, 11" subject: "Computer Networks and Security" chapter: 4 key: "semester_7/computer_networks_and_security:subnetting-202-70-91" source: https://www.bctengineer.com/q/semester_7/computer_networks_and_security/subnetting-202-70-91 site: BCT Engineering --- # Subnet 202.70.91.0/24 for 25, 16, 29, 11 Computer Networks and Security · Chapter 4: Network Layer - Asked in: 2073 Chaitra, 8 marks - Video: [Subnetting Explained: Networking Basics](https://www.youtube.com/watch?v=hbdT_Q9DM8w) — WhiteboardDoodles - 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-202-70-91 - Whole chapter: https://www.bctengineer.com/md/semester_7/computer_networks_and_security/chapter-4 ## Question Suppose we have 4 departments A, B, C and D having 25 hosts, 16 hosts, 29 hosts and 11 hosts respectively. You are given a network 202.70.91.0/24. Perform the subnetting in such a way that the IP address wastage in each department is minimum and find out the subnet mask, network address, broadcast address and usable host range in each department. ## Short answer VLSM: allocate largest block first, block sizes /27 for A,B,C (32 addresses) and /28 for D (16 addresses). C: 202.70.91.0/27, A: 202.70.91.32/27, B: 202.70.91.64/27, D: 202.70.91.96/28. Total wastage 23 host addresses. ## Answer **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 | 25 | 27 | 32 | 5 | /27 | 255.255.255.224 | | B | 16 | 18 | 32 | 5 | /27 | 255.255.255.224 | | C | 29 | 31 | 32 | 5 | /27 | 255.255.255.224 | | D | 11 | 13 | 16 | 4 | /28 | 255.255.255.240 | #### Step 2: Allocation (largest first) from 202.70.91.0/24 | Dept | Network address | Usable host range | Broadcast address | Mask | Assigned / spare range | |---|---|---|---|---|---| | C | 202.70.91.0/27 | 202.70.91.1 – 202.70.91.30 | 202.70.91.31 | 255.255.255.224 | 202.70.91.1 – 202.70.91.29 used; spare 202.70.91.30 | | A | 202.70.91.32/27 | 202.70.91.33 – 202.70.91.62 | 202.70.91.63 | 255.255.255.224 | 202.70.91.33 – 202.70.91.57 used; spare 202.70.91.58 – 202.70.91.62 | | B | 202.70.91.64/27 | 202.70.91.65 – 202.70.91.94 | 202.70.91.95 | 255.255.255.224 | 202.70.91.65 – 202.70.91.80 used; spare 202.70.91.81 – 202.70.91.94 | | D | 202.70.91.96/28 | 202.70.91.97 – 202.70.91.110 | 202.70.91.111 | 255.255.255.240 | 202.70.91.97 – 202.70.91.107 used; spare 202.70.91.108 – 202.70.91.110 | #### Step 3: Wastage | Dept | Usable addresses | Needed | Wasted | |---|---|---|---| | C | 30 | 29 | 1 | | A | 30 | 25 | 5 | | B | 30 | 16 | 14 | | D | 14 | 11 | 3 | | **Total** | 104 | 81 | 23 | Addresses allocated: 112 of 256. The remaining 144 addresses (202.70.91.112/28, 202.70.91.128/25) stay free for future departments and are not counted as wastage. **Answer: A: 202.70.91.32/27; B: 202.70.91.64/27; C: 202.70.91.0/27; D: 202.70.91.96/28.** Total wastage = 23 host addresses. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does B (16 hosts) get a /27 and not a /28? A. A /27 is the default for Class C B. A /28 is reserved C. A /28 has only 14 usable hosts, less than 16 D. 16 hosts need 16 addresses exactly Answer: C. /28 gives 16 addresses, 14 usable; 18 addresses are needed so /27 is required. (Key point: Block for 16 hosts) 2. Which prefix does D (11 hosts) get? A. /26 B. /28 C. /27 D. /29 Answer: B. 13 addresses need 16 = 2^4, so /28. (Key point: Block for 11 hosts) 3. What is the correct allocation of C (29 hosts)? A. 202.70.91.64/27 B. 202.70.91.32/27 C. 202.70.91.0/27 D. 202.70.91.96/28 Answer: C. C is allocated first (largest block, 29 hosts), from 202.70.91.0/27, with broadcast .31. (Key point: Final allocation) 4. What is the total wastage? A. 23 B. 144 C. 81 D. 104 Answer: A. 1 + 5 + 14 + 3 = 23. (Key point: Total wastage) ## Close variants set in other exams - [Subnet 10.10.10.0/24 for 5, 16, 23, 27](https://www.bctengineer.com/question.md?key=semester_7%2Fcomputer_networks_and_security%3Asubnetting-10-10-10-0) — Computer Networks and Security, chapter 4, asked 2× ## Related in Computer Networks and Security - [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 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× - [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 - [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