- 2070 Asar · 8 marks
Subnet 120.89.96.0/19 for 500-2000 hosts
Short answer
Use VLSM, largest first: D gets 120.89.96.0/21, C 120.89.104.0/21, B 120.89.112.0/22, A 120.89.116.0/23. Total host-address wastage is 624.
A large number of consecutive IP addresses are available starting at 120.89.96.0/19. Suppose that four organizations A, B, C and D request 500, 1000, 1500 and 2000 addresses respectively. How can the subnetting be performed so that address wastage is minimum?
A /19 block contains 2^(32−19) = 2^13 = 8192 addresses, so the pool is large enough.
Method (VLSM): each organization 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 organization
| Org | Hosts | Hosts + 2 | Block 2ⁿ | n | Prefix | Subnet mask |
|---|---|---|---|---|---|---|
| A | 500 | 502 | 512 | 9 | /23 | 255.255.254.0 |
| B | 1000 | 1002 | 1024 | 10 | /22 | 255.255.252.0 |
| C | 1500 | 1502 | 2048 | 11 | /21 | 255.255.248.0 |
| D | 2000 | 2002 | 2048 | 11 | /21 | 255.255.248.0 |
Step 2: Allocation (largest first) from 120.89.96.0/19
| Org | Network address | Usable host range | Broadcast address | Mask | Unused hosts |
|---|---|---|---|---|---|
| D | 120.89.96.0/21 | 120.89.96.1 – 120.89.103.254 | 120.89.103.255 | 255.255.248.0 | 46 |
| C | 120.89.104.0/21 | 120.89.104.1 – 120.89.111.254 | 120.89.111.255 | 255.255.248.0 | 546 |
| B | 120.89.112.0/22 | 120.89.112.1 – 120.89.115.254 | 120.89.115.255 | 255.255.252.0 | 22 |
| A | 120.89.116.0/23 | 120.89.116.1 – 120.89.117.254 | 120.89.117.255 | 255.255.254.0 | 10 |
Step 3: Wastage
| Org | Usable addresses | Needed | Wasted |
|---|---|---|---|
| D | 2046 | 2000 | 46 |
| C | 2046 | 1500 | 546 |
| B | 1022 | 1000 | 22 |
| A | 510 | 500 | 10 |
| Total | 5624 | 5000 | 624 |
Addresses allocated: 5632 of 8192. The remaining 2560 addresses (120.89.118.0/23, 120.89.120.0/21) stay free for future organizations and are not counted as wastage.
Answer: A: 120.89.116.0/23; B: 120.89.112.0/22; C: 120.89.104.0/21; D: 120.89.96.0/21. Total wastage = 624 host addresses.
✓ Answer checked by Claude Opus 5.5 (high effort)
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