# Q6(b) · ISP VLSM allocation

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/papers/paper-2023-jan-ai-reexam/q6b

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/papers/paper-2023-jan-ai-reexam--q6b/

Exact historical variant. Current-syllabus questions solved; excluded topics and source limitations stated.

### Q6(b). An ISP has 190.100.0.0/16 (65536 addresses). Group 1: 64 customers, 256 addresses each. Group 2: 128 customers, 128 addresses each. Group 3: 128 customers, 64 addresses each. Design aligned customer sub-blocks and find how many addresses remain. [10 marks]

### Answer

### Given

190.100.0.0/16; Group 1: 64 customers × 256 addresses; Group 2: 128 customers × 128 addresses; Group 3: 128 customers × 64 addresses

**Source correction/assumption:** All requests are total addresses, not usable hosts; therefore no extra two addresses are added.

### To find

Each customer prefix, every group range, first/last customer blocks, and unallocated addresses.

### Formula and why



$$
p=32-\log_2(\text{addresses per customer}),\quad A_{group}=n\times A_{customer}
$$



These are requested total addresses, not usable-host counts. Allocate customer blocks on their prefix boundary. A group with a non-power-of-two customer count need not be forced into one larger, wasteful CIDR aggregate.

### Step 1. Compute sizes and allocate consecutive aligned customer blocks

| Group | Customers | Addresses each | Customer prefix | Used | Group range |
| --- | --- | --- | --- | --- | --- |
| 1 | 64 | 256 | /24 | 16384 | 190.100.0.0–190.100.63.255 |
| 2 | 128 | 128 | /25 | 16384 | 190.100.64.0–190.100.127.255 |
| 3 | 128 | 64 | /26 | 8192 | 190.100.128.0–190.100.159.255 |

### Step 2. State first and last customer ranges

| Group | First customer | Last customer |
| --- | --- | --- |
| 1 | 190.100.0.0–190.100.0.255 | 190.100.63.0–190.100.63.255 |
| 2 | 190.100.64.0–190.100.64.127 | 190.100.127.128–190.100.127.255 |
| 3 | 190.100.128.0–190.100.128.63 | 190.100.159.192–190.100.159.255 |

### Step 3. Count the unused addresses

Allocated = 16384 + 16384 + 8192 = 40960. Unused = 65536 − 40960 = **24576**.

Unused range: 190.100.160.0–190.100.255.255.