# Q3(b) · VLSM customer allocation

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/papers/paper-2025-feb-reexam/q3b

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/papers/paper-2025-feb-reexam--q3b/

Actual paper; 100 marks attempted. Current-syllabus alignment is stated per question.

### Q3(b). An ISP is allocated a block of addresses starting with 180.50.0.0/16 (65,536 addresses). The ISP needs to distribute these addresses among three customer groups as follows: i) The first group consists of 80 customers; each requires 256 addresses. ii) The second group consists of 150 customers; each requires 256 addresses. iii) The third group consists of 32 customers; each requires 128 addresses. Design the sub blocks and determine the first address and last address of first customer and last customer for each group. Calculate how many addresses are still available after these allocations. [10 marks]

### Answer

### Given

180.50.0.0/16; Group 1: 80 customers × 256 addresses; Group 2: 150 customers × 256 addresses; Group 3: 32 customers × 128 addresses

### 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 | 80 | 256 | /24 | 20480 | 180.50.0.0–180.50.79.255 |
| 2 | 150 | 256 | /24 | 38400 | 180.50.80.0–180.50.229.255 |
| 3 | 32 | 128 | /25 | 4096 | 180.50.230.0–180.50.245.255 |

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

| Group | First customer | Last customer |
| --- | --- | --- |
| 1 | 180.50.0.0–180.50.0.255 | 180.50.79.0–180.50.79.255 |
| 2 | 180.50.80.0–180.50.80.255 | 180.50.229.0–180.50.229.255 |
| 3 | 180.50.230.0–180.50.230.127 | 180.50.245.128–180.50.245.255 |

### Step 3. Count the unused addresses

Allocated = 20480 + 38400 + 4096 = 62976. Unused = 65536 − 62976 = **2560**.

Unused range: 180.50.246.0–180.50.255.255.