--- title: "Subnet 202.70.64.0/19 for 100-800 hosts" subject: "Computer Networks and Security" chapter: 4 key: "semester_7/computer_networks_and_security:subnetting-202-70-64-19" source: https://www.bctengineer.com/q/semester_7/computer_networks_and_security/subnetting-202-70-64-19 site: BCT Engineering --- # Subnet 202.70.64.0/19 for 100-800 hosts Computer Networks and Security · Chapter 4: Network Layer - Asked in: 2073 Shrawan, 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-202-70-64-19 - Whole chapter: https://www.bctengineer.com/md/semester_7/computer_networks_and_security/chapter-4 ## Question You are a private contractor hired by a large company to set up the network for their enterprise and you are given a large number of consecutive IP addresses starting at 202.70.64.0/19. 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, largest first: C gets 202.70.64.0/22, B 202.70.68.0/23, D 202.70.70.0/23, A 202.70.72.0/25. 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 202.70.64.0/19 | Dept | Network address | Usable host range | Broadcast address | Mask | Unused hosts | |---|---|---|---|---|---| | C | 202.70.64.0/22 | 202.70.64.1 – 202.70.67.254 | 202.70.67.255 | 255.255.252.0 | 222 | | B | 202.70.68.0/23 | 202.70.68.1 – 202.70.69.254 | 202.70.69.255 | 255.255.254.0 | 10 | | D | 202.70.70.0/23 | 202.70.70.1 – 202.70.71.254 | 202.70.71.255 | 255.255.254.0 | 110 | | A | 202.70.72.0/25 | 202.70.72.1 – 202.70.72.126 | 202.70.72.127 | 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 (202.70.72.128/25, 202.70.73.0/24, 202.70.74.0/23, 202.70.76.0/22, 202.70.80.0/20) stay free for future departments and are not counted as wastage. **Answer: A: 202.70.72.0/25; B: 202.70.68.0/23; C: 202.70.64.0/22; D: 202.70.70.0/23.** Total wastage = 368 host addresses. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In what order should blocks be allocated to keep waste minimum and blocks aligned? A. Smallest block first B. Largest block first, one after another without overlap C. Alphabetical by department D. Random order with gaps Answer: B. Largest first keeps each block starting on a multiple of its size, avoiding gaps. (Key point: Allocation order) 2. Which prefix does Dept A (100 hosts) get? A. /27 B. /26 C. /25 D. /24 Answer: C. 102 addresses need 128 = 2^7, so prefix /25. (Key point: Block for Dept A) 3. What is Dept D's (400 hosts) network address? A. 202.70.64.0/22 B. 202.70.68.0/23 C. 202.70.70.0/23 D. 202.70.72.0/25 Answer: C. C takes 64.0/22, B 68.0/23, then D gets 202.70.70.0/23. (Key point: Final allocation) 4. What is the total wastage? A. 6016 B. 110 C. 2176 D. 368 Answer: D. 222 + 10 + 110 + 26 = 368 unused hosts. (Key point: Total wastage) ## Close variants set in other exams - [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× ## 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 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× - [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