--- title: "Subnet 120.89.96.0/19 for 500-2000 hosts" subject: "Computer Networks and Security" chapter: 4 key: "semester_7/computer_networks_and_security:subnetting-120-89-96-500-2000" source: https://www.bctengineer.com/q/semester_7/computer_networks_and_security/subnetting-120-89-96-500-2000 site: BCT Engineering --- # Subnet 120.89.96.0/19 for 500-2000 hosts Computer Networks and Security · Chapter 4: Network Layer - Asked in: 2070 Asar, 8 marks - Video: [16. How to Find the Number of Subnets Valid Hosts](https://www.youtube.com/watch?v=uyRtYUg6bnw) — System Engineer - 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-120-89-96-500-2000 - Whole chapter: https://www.bctengineer.com/md/semester_7/computer_networks_and_security/chapter-4 ## Question A large number of consecutive IP addresses are available starting at 120.89.96.0/19. Suppose that four organizations A, B, C and D request 500, 1000, 1500 and 2000 addresses respectively. How can the subnetting be performed so that address wastage is minimum? ## Short answer Use VLSM, largest first: D gets 120.89.96.0/21, C 120.89.104.0/21, B 120.89.112.0/22, A 120.89.116.0/23. Total host-address wastage is 624. ## Answer A /19 block contains 2^(32−19) = 2^13 = 8192 addresses, so the pool is large enough. **Method (VLSM):** each organization 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 organization | Org | Hosts | Hosts + 2 | Block 2ⁿ | n | Prefix | Subnet mask | |---|---|---|---|---|---|---| | A | 500 | 502 | 512 | 9 | /23 | 255.255.254.0 | | B | 1000 | 1002 | 1024 | 10 | /22 | 255.255.252.0 | | C | 1500 | 1502 | 2048 | 11 | /21 | 255.255.248.0 | | D | 2000 | 2002 | 2048 | 11 | /21 | 255.255.248.0 | #### Step 2: Allocation (largest first) from 120.89.96.0/19 | Org | Network address | Usable host range | Broadcast address | Mask | Unused hosts | |---|---|---|---|---|---| | D | 120.89.96.0/21 | 120.89.96.1 – 120.89.103.254 | 120.89.103.255 | 255.255.248.0 | 46 | | C | 120.89.104.0/21 | 120.89.104.1 – 120.89.111.254 | 120.89.111.255 | 255.255.248.0 | 546 | | B | 120.89.112.0/22 | 120.89.112.1 – 120.89.115.254 | 120.89.115.255 | 255.255.252.0 | 22 | | A | 120.89.116.0/23 | 120.89.116.1 – 120.89.117.254 | 120.89.117.255 | 255.255.254.0 | 10 | #### Step 3: Wastage | Org | Usable addresses | Needed | Wasted | |---|---|---|---| | D | 2046 | 2000 | 46 | | C | 2046 | 1500 | 546 | | B | 1022 | 1000 | 22 | | A | 510 | 500 | 10 | | **Total** | 5624 | 5000 | 624 | Addresses allocated: 5632 of 8192. The remaining 2560 addresses (120.89.118.0/23, 120.89.120.0/21) stay free for future organizations and are not counted as wastage. **Answer: A: 120.89.116.0/23; B: 120.89.112.0/22; C: 120.89.104.0/21; D: 120.89.96.0/21.** Total wastage = 624 host addresses. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which prefix is needed for 2000 hosts? A. /21 B. /20 C. /22 D. /23 Answer: A. 2002 addresses need 2048 = 2^11, so /21. (Key point: Block for 2000 hosts) 2. Why does the organisation with 1500 hosts get a /21 block? A. 1500 needs exactly 1500 addresses B. 1502 addresses need 2048, as 1024 is too small C. Because /22 holds 1500 hosts D. Because it is the second largest Answer: B. Smallest power of 2 >= 1502 is 2048, giving /21; waste is 546. (Key point: Block for 1500 hosts) 3. Where does the largest block (D) start, and what follows? A. D at 120.89.112.0/21, then C at 120.89.96.0/21 B. D at 120.89.96.0/21, then C at 120.89.98.0/21 C. D at 120.89.96.0/22, then C at 120.89.100.0/22 D. D at 120.89.96.0/21, then C at 120.89.104.0/21 Answer: D. Blocks are placed largest first: D 96.0-103.255, C 104.0-111.255, B 112.0/22, A 116.0/23. (Key point: Allocation start) 4. What is the total wastage? A. 624 B. 5000 C. 5624 D. 546 Answer: A. 46 + 546 + 22 + 10 = 624. (Key point: Total wastage) ## Related in Computer Networks and Security - [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×