Consider a point-to-point link 4 km in length. At what bandwidth would propagation delay (at a speed of 2 × 108m/s) equal transmit delay for 100-byte packets? What about 512-byte packets?

To solve this problem, we'll determine the bandwidth at which the propagation delay equals the transmission delay for two different packet sizes: 100 bytes and 512 bytes.

### Definitions and Formulas

1. **Propagation Delay (\(T_p\))**:
   \[
   T_p = \frac{\text{Length of the link}}{\text{Propagation speed}}
   \]

2. **Transmission Delay (\(T_t\))**:
   \[
   T_t = \frac{\text{Packet size}}{\text{Bandwidth}}
   \]

### Given Values:

- Length of the link = 4 km = 4000 meters
- Propagation speed = \(2 \times 10^8\) m/s
- Packet size = 100 bytes or 512 bytes (we will calculate for both)
- Bandwidth = \(B\) (we need to find this value)

### Step 1: Calculate the Propagation Delay (\(T_p\))

\[
T_p = \frac{4000 \text{ meters}}{2 \times 10^8 \text{ meters/second}} = 20 \times 10^{-6} \text{ seconds} = 20 \text{ microseconds}
\]

### Step 2: Equating Propagation Delay and Transmission Delay

We equate \(T_p\) and \(T_t\) to find the bandwidth \(B\).

\[
T_p = T_t
\]
\[
20 \times 10^{-6} \text{ seconds} = \frac{\text{Packet size (in bits)}}{B}
\]

### For a 100-byte packet:

1 byte = 8 bits, so 100 bytes = \(100 \times 8 = 800\) bits.

\[
20 \times 10^{-6} = \frac{800}{B}
\]

Solving for \(B\):

\[
B = \frac{800}{20 \times 10^{-6}} = \frac{800}{0.00002} = 40 \times 10^6 \text{ bits per second} = 40 \text{ Mbps}
\]

### For a 512-byte packet:

512 bytes = \(512 \times 8 = 4096\) bits.

\[
20 \times 10^{-6} = \frac{4096}{B}
\]

Solving for \(B\):

\[
B = \frac{4096}{20 \times 10^{-6}} = \frac{4096}{0.00002} = 204.8 \times 10^6 \text{ bits per second} = 204.8 \text{ Mbps}
\]

### Final Answer:

- For 100-byte packets, the bandwidth at which propagation delay equals transmission delay is **40 Mbps**.
- For 512-byte packets, the bandwidth at which propagation delay equals transmission delay is **204.8 Mbps**.

Would you like to go into more detail or have any other questions? Here are some related questions to consider:

1. How would the result change if the length of the link was doubled?
2. What would be the propagation delay if the speed were instead \(3 \times 10^8\) m/s?
3. How do different propagation speeds affect the required bandwidth for the given delays?
4. What happens to transmission delay if the bandwidth is increased to 1 Gbps?
5. How do different packet sizes impact network performance in terms of delays?

**Tip:** In high-speed networks, the transmission delay often becomes negligible compared to propagation delay, especially with large bandwidths.

Explore how to calculate bandwidth where propagation delay equals transmission delay for different packet sizes (100 bytes and 512 bytes) over a 4 km link at a speed of 2 × 10^8 m/s. Learn the formulas and steps involved in solving these network performance problems.

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