- Asked 2/11
- 2073 Chaitra · 8 marks
Subnet 202.70.91.0/24 for 25, 16, 29, 11
Short answer
VLSM: allocate largest block first, block sizes /27 for A,B,C (32 addresses) and /28 for D (16 addresses). C: 202.70.91.0/27, A: 202.70.91.32/27, B: 202.70.91.64/27, D: 202.70.91.96/28. Total wastage 23 host addresses.
Suppose we have 4 departments A, B, C and D having 25 hosts, 16 hosts, 29 hosts and 11 hosts respectively. You are given a network 202.70.91.0/24. Perform the subnetting in such a way that the IP address wastage in each department is minimum and find out the subnet mask, network address, broadcast address and usable host range in each department.
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 | 25 | 27 | 32 | 5 | /27 | 255.255.255.224 |
| B | 16 | 18 | 32 | 5 | /27 | 255.255.255.224 |
| C | 29 | 31 | 32 | 5 | /27 | 255.255.255.224 |
| D | 11 | 13 | 16 | 4 | /28 | 255.255.255.240 |
Step 2: Allocation (largest first) from 202.70.91.0/24
| Dept | Network address | Usable host range | Broadcast address | Mask | Assigned / spare range |
|---|---|---|---|---|---|
| C | 202.70.91.0/27 | 202.70.91.1 – 202.70.91.30 | 202.70.91.31 | 255.255.255.224 | 202.70.91.1 – 202.70.91.29 used; spare 202.70.91.30 |
| A | 202.70.91.32/27 | 202.70.91.33 – 202.70.91.62 | 202.70.91.63 | 255.255.255.224 | 202.70.91.33 – 202.70.91.57 used; spare 202.70.91.58 – 202.70.91.62 |
| B | 202.70.91.64/27 | 202.70.91.65 – 202.70.91.94 | 202.70.91.95 | 255.255.255.224 | 202.70.91.65 – 202.70.91.80 used; spare 202.70.91.81 – 202.70.91.94 |
| D | 202.70.91.96/28 | 202.70.91.97 – 202.70.91.110 | 202.70.91.111 | 255.255.255.240 | 202.70.91.97 – 202.70.91.107 used; spare 202.70.91.108 – 202.70.91.110 |
Step 3: Wastage
| Dept | Usable addresses | Needed | Wasted |
|---|---|---|---|
| C | 30 | 29 | 1 |
| A | 30 | 25 | 5 |
| B | 30 | 16 | 14 |
| D | 14 | 11 | 3 |
| Total | 104 | 81 | 23 |
Addresses allocated: 112 of 256. The remaining 144 addresses (202.70.91.112/28, 202.70.91.128/25) stay free for future departments and are not counted as wastage.
Answer: A: 202.70.91.32/27; B: 202.70.91.64/27; C: 202.70.91.0/27; D: 202.70.91.96/28. Total wastage = 23 host addresses.
✓ Answer checked by Claude Opus 5.5 (high effort)
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