(i) Consider a transmitter employing sliding window protocol for flow control with a window size set to 5. If the link used is a satellite link with a data rate of 20kbps and link propagation delay of 500ms. Assume each data frame is of size 200 bytes, calculate the link utilisation and effective data rate.
(i) Consider a transmitter employing sliding window protocol for flow control with a window size set to 5. If the link used is a satellite link with a data rate of 20kbps and link propagation delay of 500ms. Assume each data frame is of size 200 bytes, calculate the link utilisation and effective data rate.
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### Exercise: Calculating Link Utilization and Effective Data Rate
**Problem Statement:**
1. Consider a transmitter employing a sliding window protocol for flow control with a window size set to 5.
2. The link used is a satellite link with a data rate of 20 kbps and a link propagation delay of 500 ms.
3. Each data frame is of size 200 bytes.
**Task:**
Calculate the link utilization and effective data rate.
**Solution:**
1. **Identify Given Data:**
- Window size (W): 5 frames
- Data rate (R): 20 kbps (kilo bits per second)
- Propagation delay (Tp): 500 milliseconds (ms)
- Frame size (F): 200 bytes
2. **Convert Units:**
- Data rate (R): \( 20 \text{ kbps} = 20,000 \text{ bps} \)
- Frame size (F): \( 200 \text{ bytes} = 200 \times 8 = 1600 \text{ bits} \)
- Propagation delay (Tp): \( 500 \text{ ms} = 0.5 \text{ s} \)
3. **Transmission Time per Frame (Tt):**
\[
Tt = \frac{\text{Frame Size}}{\text{Data Rate}} = \frac{1600 \text{ bits}}{20,000 \text{ bps}} = 0.08 \text{ seconds}
\]
4. **Round Trip Time (RTT):**
\[
RTT = 2 \times \text{Propagation Delay} = 2 \times 0.5 \text{ seconds} = 1 \text{ second}
\]
5. **Link Utilization (U):**
\[
U = \frac{\text{Window Size} \times \text{Frame Transmission Time}}{\text{Round Trip Time}} = \frac{5 \times 0.08 \text{ seconds}}{1 \text{ second}} = 0.4
\]
6. **Effective Data Rate (Reff):**
\[
Reff = \text{Link Utilization} \times \text{Data Rate} = 0.4 \times 20,000 \text{ bps} =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F039f7f42-bea1-420c-8e74-506e9aacc740%2F949682ff-470e-4156-bb7c-665f8441d6b2%2Fcloa3a8_processed.png&w=3840&q=75)
Transcribed Image Text:---
### Exercise: Calculating Link Utilization and Effective Data Rate
**Problem Statement:**
1. Consider a transmitter employing a sliding window protocol for flow control with a window size set to 5.
2. The link used is a satellite link with a data rate of 20 kbps and a link propagation delay of 500 ms.
3. Each data frame is of size 200 bytes.
**Task:**
Calculate the link utilization and effective data rate.
**Solution:**
1. **Identify Given Data:**
- Window size (W): 5 frames
- Data rate (R): 20 kbps (kilo bits per second)
- Propagation delay (Tp): 500 milliseconds (ms)
- Frame size (F): 200 bytes
2. **Convert Units:**
- Data rate (R): \( 20 \text{ kbps} = 20,000 \text{ bps} \)
- Frame size (F): \( 200 \text{ bytes} = 200 \times 8 = 1600 \text{ bits} \)
- Propagation delay (Tp): \( 500 \text{ ms} = 0.5 \text{ s} \)
3. **Transmission Time per Frame (Tt):**
\[
Tt = \frac{\text{Frame Size}}{\text{Data Rate}} = \frac{1600 \text{ bits}}{20,000 \text{ bps}} = 0.08 \text{ seconds}
\]
4. **Round Trip Time (RTT):**
\[
RTT = 2 \times \text{Propagation Delay} = 2 \times 0.5 \text{ seconds} = 1 \text{ second}
\]
5. **Link Utilization (U):**
\[
U = \frac{\text{Window Size} \times \text{Frame Transmission Time}}{\text{Round Trip Time}} = \frac{5 \times 0.08 \text{ seconds}}{1 \text{ second}} = 0.4
\]
6. **Effective Data Rate (Reff):**
\[
Reff = \text{Link Utilization} \times \text{Data Rate} = 0.4 \times 20,000 \text{ bps} =
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