- Asked 2/11
- 2074 Chaitra · 8 marks
Class C pool for 32, 65, 10, 21, 9
Short answer
Use VLSM, largest first from 200.1.1.0/24: B gets 200.1.1.0/25, A 200.1.1.128/26, D 200.1.1.192/27, C 200.1.1.224/28, E 200.1.1.240/28. Total host-address wastage is 109.
How can you dedicate 32, 65, 10, 21 and 9 public IP addresses to the departments A, B, C, D and E respectively from the pool of class C IP addresses with minimum loss? Explain.
Assumption: the Class C pool is 200.1.1.0/24 (256 addresses) and each number is the count of hosts to be addressed.
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 | 32 | 34 | 64 | 6 | /26 | 255.255.255.192 |
| B | 65 | 67 | 128 | 7 | /25 | 255.255.255.128 |
| C | 10 | 12 | 16 | 4 | /28 | 255.255.255.240 |
| D | 21 | 23 | 32 | 5 | /27 | 255.255.255.224 |
| E | 9 | 11 | 16 | 4 | /28 | 255.255.255.240 |
Step 2: Allocation (largest first) from 200.1.1.0/24
| Dept | Network address | Usable host range | Broadcast address | Mask | Unused hosts |
|---|---|---|---|---|---|
| B | 200.1.1.0/25 | 200.1.1.1 – 200.1.1.126 | 200.1.1.127 | 255.255.255.128 | 61 |
| A | 200.1.1.128/26 | 200.1.1.129 – 200.1.1.190 | 200.1.1.191 | 255.255.255.192 | 30 |
| D | 200.1.1.192/27 | 200.1.1.193 – 200.1.1.222 | 200.1.1.223 | 255.255.255.224 | 9 |
| C | 200.1.1.224/28 | 200.1.1.225 – 200.1.1.238 | 200.1.1.239 | 255.255.255.240 | 4 |
| E | 200.1.1.240/28 | 200.1.1.241 – 200.1.1.254 | 200.1.1.255 | 255.255.255.240 | 5 |
Step 3: Wastage
| Dept | Usable addresses | Needed | Wasted |
|---|---|---|---|
| B | 126 | 65 | 61 |
| A | 62 | 32 | 30 |
| D | 30 | 21 | 9 |
| C | 14 | 10 | 4 |
| E | 14 | 9 | 5 |
| Total | 246 | 137 | 109 |
Addresses allocated: 128 + 64 + 32 + 16 + 16 = 256 of 256, so the Class C pool is used exactly, with no leftover block.
Answer: A: 200.1.1.128/26; B: 200.1.1.0/25; C: 200.1.1.224/28; D: 200.1.1.192/27; E: 200.1.1.240/28. Total wastage = 109 host addresses.
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- Class C pool for 10, 12, 8, 14 · Computer Networks and Security, chapter 4 · asked 2×
- Subnetting for Banijya bank departments · Computer Networks and Security, chapter 4
- AVL tree for 52, 65, 72, 15, 11, 32, 20 · Data Structure and Algorithm, chapter 6