Question

A 1-km-long, 10-Mbps CSMA.CD 1.AN (not \(802.3\) ) has a propagation speed of \(200 \mathrm{~m} / \mathrm{jecc}\). Repeaters are not allowed in this system. Data frames are 256 bits long. including 32 bits of header, checksum, and other overhead. The first bit stot after al saccessful transmission is reserved for the receiver to capture the channel in order lo send a 32-bit acknowledyement frame. What is the effective data rate, excluding overhead, assuming that there are no collisaons?

Short answer

The effective data rate is approximately 2.24 Mbps.

Step-by-step solution

Calculate Frame Transmission Time

First, determine the time taken to transmit a full data frame. The frame is 256 bits long and the speed is 10 Mbps. The transmission time \( t_t \) is given by \( t_t = \frac{256}{10 \times 10^6} \) seconds.

Calculate Propagation Delay

Next, calculate the propagation delay \( t_p \) which is the time taken for a signal to travel from the sender to the receiver over the 1 km distance. The propagation speed is 200 meters per microsecond, therefore, \( t_p = \frac{1000}{200\times 10^6} \) seconds.

Calculate Acknowledgement Transmission Time

An acknowledgement frame consists of 32 bits. Calculate the time to transmit this frame using the same speed of 10 Mbps. The time is \( t_{ack} = \frac{32}{10 \times 10^6} \) seconds.

Calculate Total Time for Data and Acknowledgement Exchange

Now, calculate the total time taken for one cycle of sending a data frame and receiving an acknowledgment. This total time \( t_{total} \) includes the transmission time for the data frame, the propagation time for the data frame and acknowledgment, and the transmission time for the acknowledgment frame. So, \( t_{total} = t_t + t_p + t_{ack} + t_p \).

Calculate Effective Data Rate

Finally, determine the effective data rate by considering only the useful data of 224 bits per 256-bit frame, as the rest are overhead (32 bits). The effective data rate \( R_{eff} \) is calculated by dividing the useful data (224 bits) by the total time calculated in Step 4: \( R_{eff} = \frac{224}{t_{total}} \).

Key concepts

CSMA/CD Protocol

The Carrier Sense Multiple Access with Collision Detection (CSMA/CD) protocol is a network mediation technique primarily used in Ethernet networks to manage data exchange. In this system, all devices connected to the same network segment can detect when carriers are sending data. Before a device tries to transmit data, it listens to check if the channel is clear (carrier sense). If the channel is busy, the device waits before retrying. This avoids data collisions, ensuring efficient data transmission. However, if two devices simultaneously detect a free channel and attempt to send data, a collision occurs. This is where collision detection comes into play. Devices stop transmitting when they detect a collision, and they wait for a random time period before reattempting data transmission. This method is vital for ensuring low-latency communication and efficient bandwidth usage in Ethernet networks. CSMA/CD is efficient, but works best in low-traffic situations as collisions become more frequent in heavier traffic.

Propagation Delay

Propagation delay is a crucial factor in network communication. It is the time it takes for a signal to travel from the sender to the receiver through the network medium. In our exercise, signals travel at a speed of 200 meters per microsecond, meaning that over a 1-kilometer distance, the propagation delay is calculated as follows: distance divided by speed. \[ t_p = \frac{1000 \text{ meters}}{200 \times 10^6 \text{ meters per second}} \]This gives a result in seconds, representing the time it takes for the signal to reach its destination. It is important to understand that propagation delay is influenced by the physical medium's properties and distance. Shorter distances and faster propagation speeds result in lower delays, improving network performance. Propagation delay can vary depending on medium, with optical fiber generally providing lower delay compared to copper cabling.

Frame Transmission Time

The frame transmission time is the period required to send a network data frame. This is determined by the size of the frame and the network's transmission speed. In our scenario, we are dealing with a 256-bit data frame being sent at a rate of 10 Mbps.The calculation for transmission time is as follows:\[ t_t = \frac{256 \text{ bits}}{10 \times 10^6 \text{ bits per second}} \]The result shows how long it takes to transmit the entire frame from the sender. Within this frame, only a part of the bits carry the useful data, while the rest include required overhead for managing and checking data integrity. Efficient utilization of frame transmission time is crucial for achieving optimal data rates as it directly impacts how fast data can be sent across the network.

Acknowledgement Transmission

Acknowledgement transmission is a key component in reliable data communication processes. It involves sending back a signal from the receiver to confirm data receipt. In the exercise, the receiver sends a 32-bit acknowledgement frame after capturing the channel post-data transmission. This occurs at the same 10 Mbps speed used for sending the data frames. The computation for the time to send this 32-bit acknowledgement is:\[ t_{ack} = \frac{32 \text{ bits}}{10 \times 10^6 \text{ bits per second}} \]This time period is essential for maintaining the flow of accurate data exchange between devices. In larger systems, efficient acknowledgement processes help in minimizing errors and network congestion, ultimately contributing to effective data rates. Acknowledgement frames ensure that the sender is aware of receipts, allowing it to continue without resending, which enhances throughput and system reliability.