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- 2071 Chaitra · 10 marks
Subnet 10.10.10.0/24 for 5, 16, 23, 27
Short answer
VLSM: largest first, block sizes /27 for D, C, B and /29 for A. D: 10.10.10.0/27, C: 10.10.10.32/27, B: 10.10.10.64/27, A: 10.10.10.96/29. Total wastage 25 host addresses.
You are given the following address space 10.10.10.0/24. You have to assign addresses to 4 departments with the following hosts 5, 16, 23 and 27 respectively. Perform the subnetting in such a way that the IP address wastage in each department is minimum. Also find out the subnet mask, network address, broadcast address and unassigned range in each department.
Here unassigned range means the usable addresses left unused in each department's block.
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 | 5 | 7 | 8 | 3 | /29 | 255.255.255.248 |
| B | 16 | 18 | 32 | 5 | /27 | 255.255.255.224 |
| C | 23 | 25 | 32 | 5 | /27 | 255.255.255.224 |
| D | 27 | 29 | 32 | 5 | /27 | 255.255.255.224 |
Step 2: Allocation (largest first) from 10.10.10.0/24
| Dept | Network address | Usable host range | Broadcast address | Mask | Assigned / spare range |
|---|---|---|---|---|---|
| D | 10.10.10.0/27 | 10.10.10.1 – 10.10.10.30 | 10.10.10.31 | 255.255.255.224 | 10.10.10.1 – 10.10.10.27 used; spare 10.10.10.28 – 10.10.10.30 |
| C | 10.10.10.32/27 | 10.10.10.33 – 10.10.10.62 | 10.10.10.63 | 255.255.255.224 | 10.10.10.33 – 10.10.10.55 used; spare 10.10.10.56 – 10.10.10.62 |
| B | 10.10.10.64/27 | 10.10.10.65 – 10.10.10.94 | 10.10.10.95 | 255.255.255.224 | 10.10.10.65 – 10.10.10.80 used; spare 10.10.10.81 – 10.10.10.94 |
| A | 10.10.10.96/29 | 10.10.10.97 – 10.10.10.102 | 10.10.10.103 | 255.255.255.248 | 10.10.10.97 – 10.10.10.101 used; spare 10.10.10.102 |
Step 3: Wastage
| Dept | Usable addresses | Needed | Wasted |
|---|---|---|---|
| D | 30 | 27 | 3 |
| C | 30 | 23 | 7 |
| B | 30 | 16 | 14 |
| A | 6 | 5 | 1 |
| Total | 96 | 71 | 25 |
Addresses allocated: 104 of 256. The remaining 152 addresses (10.10.10.104/29, 10.10.10.112/28, 10.10.10.128/25) stay free for future departments and are not counted as wastage.
Answer: A: 10.10.10.96/29; B: 10.10.10.64/27; C: 10.10.10.32/27; D: 10.10.10.0/27. Total wastage = 25 host addresses.
✓ Answer checked by Claude Opus 5.5 (high effort)
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