--- title: "Subnet 202.77.19.0/24 for 30, 24, 25, 20" subject: "Computer Networks and Security" chapter: 4 key: "semester_7/computer_networks_and_security:subnetting-202-77-19" source: https://www.bctengineer.com/q/semester_7/computer_networks_and_security/subnetting-202-77-19 site: BCT Engineering --- # Subnet 202.77.19.0/24 for 30, 24, 25, 20 Computer Networks and Security · Chapter 4: Network Layer - Asked in: 2072 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-202-77-19 - Whole chapter: https://www.bctengineer.com/md/semester_7/computer_networks_and_security/chapter-4 ## Question Explain how you can allocate 30, 24, 25 and 20 IP addresses to the four different departments of ABC company with minimum wastage. Specify the range of IP addresses, broadcast address, network address and subnet mask for each department from the given address pool 202.77.19.0/24. ## Short answer VLSM: all departments require /27 blocks (32 addresses). A: 202.77.19.0/27, C: 202.77.19.32/27, B: 202.77.19.64/27, D: 202.77.19.96/27. Total wastage 21 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 | 30 | 32 | 32 | 5 | /27 | 255.255.255.224 | | B | 24 | 26 | 32 | 5 | /27 | 255.255.255.224 | | C | 25 | 27 | 32 | 5 | /27 | 255.255.255.224 | | D | 20 | 22 | 32 | 5 | /27 | 255.255.255.224 | #### Step 2: Allocation (largest first) from 202.77.19.0/24 | Dept | Network address | Usable host range | Broadcast address | Mask | Assigned / spare range | |---|---|---|---|---|---| | A | 202.77.19.0/27 | 202.77.19.1 – 202.77.19.30 | 202.77.19.31 | 255.255.255.224 | 202.77.19.1 – 202.77.19.30 used; no spare | | C | 202.77.19.32/27 | 202.77.19.33 – 202.77.19.62 | 202.77.19.63 | 255.255.255.224 | 202.77.19.33 – 202.77.19.57 used; spare 202.77.19.58 – 202.77.19.62 | | B | 202.77.19.64/27 | 202.77.19.65 – 202.77.19.94 | 202.77.19.95 | 255.255.255.224 | 202.77.19.65 – 202.77.19.88 used; spare 202.77.19.89 – 202.77.19.94 | | D | 202.77.19.96/27 | 202.77.19.97 – 202.77.19.126 | 202.77.19.127 | 255.255.255.224 | 202.77.19.97 – 202.77.19.116 used; spare 202.77.19.117 – 202.77.19.126 | #### Step 3: Wastage | Dept | Usable addresses | Needed | Wasted | |---|---|---|---| | A | 30 | 30 | 0 | | C | 30 | 25 | 5 | | B | 30 | 24 | 6 | | D | 30 | 20 | 10 | | **Total** | 120 | 99 | 21 | Addresses allocated: 128 of 256. The remaining 128 addresses (202.77.19.128/25) stay free for future departments and are not counted as wastage. **Answer: A: 202.77.19.0/27; B: 202.77.19.64/27; C: 202.77.19.32/27; D: 202.77.19.96/27.** Total wastage = 21 host addresses. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does A (30 hosts) get a /27? A. 32 addresses = 30 hosts + network + broadcast B. /27 has 32 usable hosts C. 30 is a power of 2 D. 30 hosts need only a /28 Answer: A. 30 + 2 = 32 = 2^5, so /27 fits with zero waste. (Key point: Block for 30 hosts) 2. Which mask do all four departments use? A. 255.255.255.224 B. 255.255.255.240 C. 255.255.255.128 D. 255.255.255.192 Answer: A. Each needs 22 to 32 addresses, so all get 32-address /27 blocks. (Key point: Same mask for all) 3. What is B's (24 hosts) network address? A. 202.77.19.96/27 B. 202.77.19.32/27 C. 202.77.19.64/27 D. 202.77.19.0/27 Answer: C. A at .0, C at .32, B at .64, D at .96. (Key point: Network address of B) 4. How much of the /24 stays free? A. 128 addresses (202.77.19.128/25) B. 64 addresses C. 96 addresses D. 0 addresses Answer: A. 4 x 32 = 128 allocated; 202.77.19.128/25 is left free. 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