Categories of Networks

Categories of Networks (Based on Geographic Scope)

Networks are broadly categorized based on their geographical coverage, size, and ownership.

1. Personal Area Network (PAN)

• Covers the smallest area, typically within a person's reach (a few meters).
• Used for connecting personal devices like smartphones, laptops, Bluetooth headsets.
• Example: Bluetooth connection between a phone and earbuds.

2. Local Area Network (LAN)

• Covers a limited geographic area such as a room, building, or campus.
• High data transfer speed, privately owned.
• Example: Network within a school, office, or lab.
• Technologies: Ethernet, Wi-Fi.

3. Metropolitan Area Network (MAN)

• Covers a city or town (range: 5 to 50 km).
• Larger than LAN but smaller than WAN.
• Often used by ISPs or government bodies to connect multiple LANs.
• Example: Cable TV network across a city.

4. Wide Area Network (WAN)

• Covers a large geographic area such as a country or the entire world.
• Uses public/private communication links.
• Slower than LAN, higher latency.
• Example: The Internet, banking networks.

5. Wireless Network (WLAN/WWAN)

• Networks that use wireless transmission (radio waves, infrared) instead of cables.
• Can be local (Wi-Fi) or wide area (cellular 4G/5G).
• Example: Wi-Fi hotspot, mobile data network.

Summary Table

These categories help network designers choose the appropriate technology, topology, and protocols based on the scale and purpose of the network.

MAC Address and CRC Calculation

MAC Address (2 marks)

A MAC (Media Access Control) address is a unique physical hardware address permanently assigned to a Network Interface Card (NIC) by its manufacturer. It operates at the Data Link Layer (Layer 2) of the OSI model and is used to uniquely identify a device on a local network.

Key features:

• It is a 48-bit (6-byte) address, normally written in hexadecimal.
• Example: 00:1A:2B:3C:4D:5E
• The first 3 bytes (24 bits) form the OUI (Organizationally Unique Identifier) identifying the manufacturer.
• The last 3 bytes identify the specific device.
• Also called the physical / hardware address; usually burned into the NIC's ROM and globally unique.

CRC Calculation (3 marks)

Given data

• Message $M$ = 1011011
• Generator $G(x) = x^3 + x^2 + 1 \Rightarrow$ binary 1101
• Degree $r = 3$

Step 1: Append $r = 3$ zeros

$$ M' = 1011011\underbrace{000}_{3\text{ zeros}} = 1011011000 $$

Step 2: Modulo-2 (XOR) division by 1101

Dividend bits: 1 0 1 1 0 1 1 0 0 0

      1101010   <-- quotient (not required)
    ┌───────────
1101│1011011000
     1101
     ────
     0110 1        bring down bit → 1101
     1101
     ────
     0000 11       bring down bits → 0011
          (0011 < 1101) → q=0, bring down → 0111
          (0111 < 1101) → q=0, bring down → 1110
          1101
          ────
          0011 0    bring down → 0110
          (0110 < 1101) → q=0  (no more bits)
     Remainder = 010

Working the XOR steps cleanly:

Let me track the final bits exactly. After processing all 10 dividend bits:

• Start: 1011 → XOR 1101 = 0110
• +bit(0): 1100 → XOR 1101 = 0001
• +bit(1): 0011 → < divisor, q=0
• +bit(1): 0111 → < divisor, q=0
• +bit(0): 1110 → XOR 1101 = 0011
• +bit(0): 0110 → < divisor, q=0
• +bit(0): 1100 → XOR 1101 = 0001

Final 3-bit remainder (CRC) = 001

Verification

Divide the 3 remaining bits carefully; the last register content after all 10 bits gives remainder = 001.

Check: transmitted frame $T = 1011011,001$. Dividing $1011011001$ by 1101 must give remainder 000:

1011011001
1101
────
0110 → 1100 → 0001 → 0011 → 0111 → 1110→0011 → 0110 → 1101 → 0000

$1101 \oplus 1101 = 0000$ ✓ remainder is 0, so the CRC is correct.

Step 3: Transmitted frame

$$ T = \text{Message} + \text{CRC} = 1011011 , \big| , 001 = \boxed{1011011001} $$

CRC (remainder) = 001 Transmitted encoded frame = 1011011001

Pure ALOHA Throughput and Vulnerable Time

STEP 1 - EXTRACT: Given Data

• Frame size = 200 bits
• Channel bandwidth = not numerically given (question says "shared channel of bandwidth" with the number missing)
• Frame production rate = not numerically given (question says "produces frames per second" with the number missing)

Missing data note: The problem statement is incomplete. Two numeric values that are essential for a concrete throughput answer are absent:

1. The channel bandwidth (e.g., "200 kbps").
2. The frame generation rate (e.g., "1000 frames per second").

Without these, a specific numeric throughput cannot be computed. I will therefore:

• Answer the fully solvable part (vulnerable time) generally, and
• Present the throughput method, showing what would be computed once the missing numbers are supplied.

No values such as "200 bps" or "1000 frames/s" are assumed here, since the question does not supply them.

STEP 2 - SOLVE

Frame Transmission Time

$$T_t = \frac{\text{Frame size}}{\text{Bandwidth}} = \frac{200 \text{ bits}}{B \text{ bps}}$$

This cannot be evaluated because $B$ is not given.

Vulnerable Time of Pure ALOHA (solvable in general)

In pure ALOHA a frame collides if another transmission starts within one frame time before or after the start of the frame. Hence:

$$\boxed{T_{vulnerable} = 2 \times T_t}$$

If, for example, $B = 200$ bps then $T_t = 1$ s and $T_{vulnerable} = 2$ s. This depends entirely on the (missing) bandwidth.

Throughput (method only, cannot evaluate numerically)

For pure ALOHA:

$$S = G , e^{-2G}$$

where the offered load is

$$G = (\text{frame rate}) \times T_t = \lambda \times \frac{200}{B}$$

Once $\lambda$ (frames/second) and $B$ (bps) are provided:

1. Compute $T_t = 200/B$.
2. Compute $G = \lambda \cdot T_t$.
3. Compute $S = G e^{-2G}$.

Standard reference result: The maximum possible throughput of pure ALOHA occurs at $G = 0.5$:

$$S_{max} = 0.5 , e^{-1} = 0.5 \times 0.368 = 0.184 = 18.4%$$

This maximum is a property of the protocol and does not require the missing data.

Summary

Note on the missing data: The bandwidth and frame rate are needed before a single numeric throughput can be given. If the intended values were $B = 200$ bps and $\lambda = 1000$ frames/s, then $T_t = 1$ s, $G = 1000$, and $S = 1000 , e^{-2000} \approx 0$ (essentially zero, since the channel is enormously overloaded). Mixing $G=1000$ with the separate $G=0.5$ maximum would be internally inconsistent.