--- title: "Subnet 120.89.96.0/19 for 100-800 hosts" subject: "Computer Networks and Security" chapter: 4 key: "semester_7/computer_networks_and_security:subnetting-120-89-96-four-depts" source: https://www.bctengineer.com/q/semester_7/computer_networks_and_security/subnetting-120-89-96-four-depts site: BCT Engineering --- # Subnet 120.89.96.0/19 for 100-800 hosts Computer Networks and Security · Chapter 4: Network Layer - Asked in: 2074 Asoj, 8 marks - Video: [Learning Subnetting Part 3 - Dividing a Network into Subnets](https://www.youtube.com/watch?v=fTTfbxrYRE4) — danscourses - 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-120-89-96-four-depts - Whole chapter: https://www.bctengineer.com/md/semester_7/computer_networks_and_security/chapter-4 ## Question Suppose you are a private consultant hired by a company to set up the network for their enterprise and you are given a large number of consecutive IP addresses starting at 120.89.96.0/19. Suppose that four departments A, B, C and D request 100, 500, 800 and 400 addresses respectively. How can the subnetting be performed so that address wastage will be minimum? ## Short answer Use VLSM, allocating largest blocks first: C gets 120.89.96.0/22, B 120.89.100.0/23, D 120.89.102.0/23, A 120.89.104.0/25. Total host-address wastage is 368. ## Answer A /19 block contains 2^(32−19) = 2^13 = 8192 addresses, so the pool is large enough. **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 | 100 | 102 | 128 | 7 | /25 | 255.255.255.128 | | B | 500 | 502 | 512 | 9 | /23 | 255.255.254.0 | | C | 800 | 802 | 1024 | 10 | /22 | 255.255.252.0 | | D | 400 | 402 | 512 | 9 | /23 | 255.255.254.0 | #### Step 2: Allocation (largest first) from 120.89.96.0/19 | Dept | Network address | Usable host range | Broadcast address | Mask | Unused hosts | |---|---|---|---|---|---| | C | 120.89.96.0/22 | 120.89.96.1 – 120.89.99.254 | 120.89.99.255 | 255.255.252.0 | 222 | | B | 120.89.100.0/23 | 120.89.100.1 – 120.89.101.254 | 120.89.101.255 | 255.255.254.0 | 10 | | D | 120.89.102.0/23 | 120.89.102.1 – 120.89.103.254 | 120.89.103.255 | 255.255.254.0 | 110 | | A | 120.89.104.0/25 | 120.89.104.1 – 120.89.104.126 | 120.89.104.127 | 255.255.255.128 | 26 | #### Step 3: Wastage | Dept | Usable addresses | Needed | Wasted | |---|---|---|---| | C | 1022 | 800 | 222 | | B | 510 | 500 | 10 | | D | 510 | 400 | 110 | | A | 126 | 100 | 26 | | **Total** | 2168 | 1800 | 368 | Addresses allocated: 2176 of 8192. The remaining 6016 addresses (120.89.104.128/25, 120.89.105.0/24, 120.89.106.0/23, 120.89.108.0/22, 120.89.112.0/20) stay free for future departments and are not counted as wastage. **Answer: A: 120.89.104.0/25; B: 120.89.100.0/23; C: 120.89.96.0/22; D: 120.89.102.0/23.** Total wastage = 368 host addresses. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which rule gives the block size for each department? A. Largest 2^n <= hosts + 2 B. Smallest 2^n >= hosts - 2 C. Smallest 2^n >= hosts D. Smallest 2^n >= hosts + 2 Answer: D. The network and broadcast addresses cannot be assigned, so hosts + 2 addresses are needed. (Key point: Block size rule) 2. Which prefix does Dept C (800 hosts) get? A. /22 B. /23 C. /24 D. /21 Answer: A. 802 addresses need 1024 = 2^10, so prefix /22. (Key point: Block for Dept C) 3. What is Dept B's (500 hosts) network address? A. 120.89.104.0/25 B. 120.89.96.0/23 C. 120.89.100.0/23 D. 120.89.102.0/23 Answer: C. C takes 96.0-99.255, so B starts at 120.89.100.0/23; D follows at 102.0/23 and A at 104.0/25. (Key point: Final allocation) 4. What is the total wastage of unused host addresses? A. 368 B. 222 C. 2168 D. 110 Answer: A. Waste is 222 + 10 + 110 + 26 = 368. (Key point: Total wastage) 5. How many addresses are in a /19 block? A. 2048 B. 4096 C. 8192 D. 16384 Answer: C. 2^(32-19) = 2^13 = 8192. (Key point: Pool size) ## Close variants set in other exams - [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× ## Related in Computer Networks and Security - [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 - [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