- Asked 2/11
- 2073 Shrawan · 8 marks
Subnet 202.70.64.0/19 for 100-800 hosts
Short answer
Use VLSM, largest first: C gets 202.70.64.0/22, B 202.70.68.0/23, D 202.70.70.0/23, A 202.70.72.0/25. Total host-address wastage is 368.
You are a private contractor hired by a large company to set up the network for their enterprise and you are given a large number of consecutive IP addresses starting at 202.70.64.0/19. Suppose that four departments A, B, C and D request 100, 500, 800 and 400 addresses respectively. How can the subnetting be performed so that address wastage will be minimum?
A /19 block contains 2^(32−19) = 2^13 = 8192 addresses, so the pool is large enough.
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 | 100 | 102 | 128 | 7 | /25 | 255.255.255.128 |
| B | 500 | 502 | 512 | 9 | /23 | 255.255.254.0 |
| C | 800 | 802 | 1024 | 10 | /22 | 255.255.252.0 |
| D | 400 | 402 | 512 | 9 | /23 | 255.255.254.0 |
Step 2: Allocation (largest first) from 202.70.64.0/19
| Dept | Network address | Usable host range | Broadcast address | Mask | Unused hosts |
|---|---|---|---|---|---|
| C | 202.70.64.0/22 | 202.70.64.1 – 202.70.67.254 | 202.70.67.255 | 255.255.252.0 | 222 |
| B | 202.70.68.0/23 | 202.70.68.1 – 202.70.69.254 | 202.70.69.255 | 255.255.254.0 | 10 |
| D | 202.70.70.0/23 | 202.70.70.1 – 202.70.71.254 | 202.70.71.255 | 255.255.254.0 | 110 |
| A | 202.70.72.0/25 | 202.70.72.1 – 202.70.72.126 | 202.70.72.127 | 255.255.255.128 | 26 |
Step 3: Wastage
| Dept | Usable addresses | Needed | Wasted |
|---|---|---|---|
| C | 1022 | 800 | 222 |
| B | 510 | 500 | 10 |
| D | 510 | 400 | 110 |
| A | 126 | 100 | 26 |
| Total | 2168 | 1800 | 368 |
Addresses allocated: 2176 of 8192. The remaining 6016 addresses (202.70.72.128/25, 202.70.73.0/24, 202.70.74.0/23, 202.70.76.0/22, 202.70.80.0/20) stay free for future departments and are not counted as wastage.
Answer: A: 202.70.72.0/25; B: 202.70.68.0/23; C: 202.70.64.0/22; D: 202.70.70.0/23. Total wastage = 368 host addresses.
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- Subnet 120.89.96.0/19 for 100-800 hosts · Computer Networks and Security, chapter 4 · asked 2×
- Subnet 202.77.19.0/24 for 30, 24, 25, 20 · Computer Networks and Security, chapter 4
- Subnet 202.70.91.0/24 for 25, 16, 29, 11 · Computer Networks and Security, chapter 4 · asked 2×
- Subnet 120.89.96.0/19 for 500-2000 hosts · Computer Networks and Security, chapter 4