- 2072 Chaitra · 8 marks
Subnet 202.77.19.0/24 for 30, 24, 25, 20
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.
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.
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.
✓ Answer checked by Claude Opus 5.5 (high effort)
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