- Asked 2/11
- 2070 Chaitra · 8 marks
Class C pool for 10, 12, 8, 14
Short answer
VLSM: all departments need /28 blocks (16 addresses). D: 200.1.1.0/28, B: 200.1.1.16/28, A: 200.1.1.32/28, C: 200.1.1.48/28. Total wastage 12 host addresses.
How can you dedicate 10, 12, 8 and 14 public IP addresses to departments A, B, C and D respectively from the pool of class C with minimum losses of IP? Explain.
Assumption: the public Class C pool is 200.1.1.0/24 and each number is the count of hosts.
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 | 10 | 12 | 16 | 4 | /28 | 255.255.255.240 |
| B | 12 | 14 | 16 | 4 | /28 | 255.255.255.240 |
| C | 8 | 10 | 16 | 4 | /28 | 255.255.255.240 |
| D | 14 | 16 | 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 |
|---|---|---|---|---|---|
| D | 200.1.1.0/28 | 200.1.1.1 – 200.1.1.14 | 200.1.1.15 | 255.255.255.240 | 0 |
| B | 200.1.1.16/28 | 200.1.1.17 – 200.1.1.30 | 200.1.1.31 | 255.255.255.240 | 2 |
| A | 200.1.1.32/28 | 200.1.1.33 – 200.1.1.46 | 200.1.1.47 | 255.255.255.240 | 4 |
| C | 200.1.1.48/28 | 200.1.1.49 – 200.1.1.62 | 200.1.1.63 | 255.255.255.240 | 6 |
Step 3: Wastage
| Dept | Usable addresses | Needed | Wasted |
|---|---|---|---|
| D | 14 | 14 | 0 |
| B | 14 | 12 | 2 |
| A | 14 | 10 | 4 |
| C | 14 | 8 | 6 |
| Total | 56 | 44 | 12 |
Addresses allocated: 64 of 256. The remaining 192 addresses (200.1.1.64/26, 200.1.1.128/25) stay free for future departments and are not counted as wastage.
Answer: A: 200.1.1.32/28; B: 200.1.1.16/28; C: 200.1.1.48/28; D: 200.1.1.0/28. Total wastage = 12 host addresses.
✓ Answer checked by Claude Opus 5.5 (high effort)
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