# Q1D · Address block /26

Subject: Computer Networks

Exam: M2

Reader: https://deep-learning-study-room.vercel.app/cn/#m2/papers/paper-2025-final/q1d

Canonical: https://deep-learning-study-room.vercel.app/read/cn/m2/papers/paper-2025-final--q1d/

Actual TEE paper · current-M2 matching parts; original marks

### Actual PYQ. A block of address is granted to a small organization and one of the address is 200.140.64.48/26. Determine the first address and last address in the block. Determine the total number of address in the block. [5 marks]

#### Answer

## Given

IP/prefix = `200.140.64.48/26`.

## To find

The complete containing address block and its usable host range.

## Formula and why



$$
h=32-p,\qquad N=2^h
$$





$$
\text{step}=256-\text{changing mask octet}
$$



Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.

## Step-by-step solution

#### Step 1. Count host bits and addresses

h=32−26=6; N=26=64.

#### Step 2. Construct the mask

Mask=`255.255.255.192`; binary=`11111111.11111111.11111111.11000000`.

#### Step 3. Find the changing-octet interval

The changing octet is number 4. Its mask value is 192, so step=256−192=64. The IP value 48 lies in 0…63.

#### Step 4. Form the first/network address

Keep earlier octets; place 0 in octet 4; set any later octets to zero. Network=**200.140.64.0**.

#### Step 5. Show the AND operation

| Octet | IP bits | Mask bits | AND result | Decimal |
| --- | --- | --- | --- | --- |
| 1 | 11001000 | 11111111 | 11001000 | 200 |
| 2 | 10001100 | 11111111 | 10001100 | 140 |
| 3 | 01000000 | 11111111 | 01000000 | 64 |
| 4 | 00110000 | 11000000 | 00000000 | 0 |

#### Step 6. Form the last/broadcast address

Use 63 in the changing octet and 255 in any later octets. Broadcast=**200.140.64.63**.

#### Step 7. State usable hosts

First usable=200.140.64.1; last usable=200.140.64.62; count=64−2=62.

**Final answer:** Network 200.140.64.0; broadcast 200.140.64.63; 64 total, 62 ordinary usable; range 200.140.64.1–200.140.64.62.

#### What this result means

The first address in a block means its network address. It is different from the first usable host address.