Question

Consider a protocol that uses piggybacking, a sending window size of 4, and 400 -bit frames. This protocol is used to transfer data over a \(200 \mathrm{kbps}\) channel with a \(4 \mathrm{msec}\) one-way propagation delay. Unfortunately, the receiver has no data to send back. It needs to send its acknowledgements in separate frames. What is the maximum amount of time the receiver can wait before sending, such that the bandwidth efficiency does not drop below \(50 \%\) ?

Short answer

The receiver can wait a maximum of 8 ms before sending acknowledgments to maintain 50% efficiency.

Step-by-step solution

Calculate Round-Trip Time (RTT)

The total round-trip time (RTT) is calculated by considering the one-way propagation delay, which is given as 4 ms one-way. Thus, the RTT is double that because the data must go to the receiver and back as an acknowledgment. Hence, the RTT is \[ 2 \times 4 \text{ ms} = 8 \text{ ms} \].

Calculate Bandwidth-Delay Product (BDP)

The bandwidth-delay product (BDP) tells us the amount of data that can be in transit in the network. Using the formula \( \text{BDP} = \text{Bandwidth} \times \text{RTT} \), where the bandwidth is 200 kbps and the RTT is 8 ms (converted to seconds for calculation), \[ \text{BDP} = 200,000 \text{ bps} \times 0.008 \text{ s} = 1600 \text{ bits} \].

Calculate Maximum Number of Frames in Transit

Each frame is 400 bits. Therefore, the maximum number of frames that can be in transit (without exceeding the BDP) is \[ \frac{1600 \text{ bits}}{400 \text{ bits/frame}} = 4 \text{ frames} \]. This confirms that the sending window size of 4 is appropriate.

Define Efficiency

Efficiency (\(E\)) is calculated as the ratio of the actual utilization of the channel to its potential, where potential utilization is getting ACKs back with no idle time. Since acknowledgment frames are used without piggybacking, the efficiency is given by\[ E = \frac{\text{Transmission time}}{\text{Transmission time} + \text{Idle time}} \geq 0.5 \].

Calculate Transmission Time

The transmission time for one frame is the time to send a frame divided by the bandwidth, which is \[ \frac{400 \text{ bits}}{200,000 \text{ bps}} = 2 \text{ ms} \]. So, to send 4 frames, the total transmission time is \(4 \times 2 \text{ ms} = 8 \text{ ms} \).

Solve for Maximum Idle Time

To retain 50% efficiency, we set up the inequality: \[ \frac{8}{8 + \text{Idle time}} \geq 0.5 \]. Solving yields,\[ 8 \geq 0.5 \times (8 + \text{Idle time}) \]\[ 8 \geq 4 + 0.5 \times \text{Idle time} \]\[ 4 \geq 0.5 \times \text{Idle time} \]\[ \text{Idle time} \leq 8 \text{ ms} \].

Key concepts

Window Size

The window size in networking refers to the number of frames or packets that can be sent before waiting for an acknowledgment. This concept is crucial, particularly in protocols that use piggybacking, where acknowledgments are sent along with outgoing data frames to improve efficiency. In this exercise, the window size is set to 4 frames. With a larger window size, more data can be transmitted before requiring an acknowledgment, allowing for better utilization of the network's bandwidth. However, choosing an appropriate window size is essential as it impacts the flow of data. Too small of a window leads to inefficiencies by underutilizing the available bandwidth, while too large may overwhelm the receiver or increase latency. Essentially, the window size is a balancing act that directly affects the efficiency of data transfer.

Bandwidth Efficiency

Bandwidth efficiency describes how effectively a network's bandwidth is being utilized. It's calculated as the ratio of the amount of data successfully transmitted to the potential throughput of the network if it were being fully utilized.

In our exercise, achieving at least 50% bandwidth efficiency means that half of the network's potential is used for transmitting actual data, while the other half might be idle or used for sending acknowledgments. A higher efficiency indicates a well-optimized network where most of the bandwidth is devoted to data transmission rather than waiting or acknowledging.

• Factors influencing bandwidth efficiency include window size, propagation delay, and the method of acknowledging frames, such as piggybacking versus separate acknowledgment frames.
• Having efficient bandwidth usage is essential for minimizing transmission delays and maximizing throughput, which is critical for both small and large scale networks.

Round-Trip Time (RTT)

Round-trip time (RTT) is the duration it takes for a signal to go from the sender to the receiver and back again. This measurement is critical in network communications as it impacts how quickly acknowledgments can be received and subsequently, how effectively the data flow can continue without waiting unnecessarily. RTT in this scenario is calculated as twice the given one-way propagation delay, which means it totals 8 milliseconds (since it's 4 ms each way).

• RTT influences the timing of acknowledgments and thus affects the windowing mechanism, as quicker acknowledgment allows for faster opening of the window for the next set of frames.
• Network protocols often need to adapt to RTT changes to maintain efficiency, especially in dynamic environments where delays might fluctuate considerably.

Bandwidth-Delay Product (BDP)

The Bandwidth-Delay Product (BDP) is a key concept that illustrates the capacity of the network — essentially how much data can be "in-flight" at any moment given the current bandwidth and the delay in the network. It's calculated by multiplying the bandwidth of the channel with the round-trip time (RTT).

For instance, in this case, with a bandwidth of 200 kbps and an RTT of 8 milliseconds, the BDP is 1600 bits. This means the network can hold 1600 bits of unacknowledged data at one time. Understanding BDP helps in defining window sizes and ensuring that the network is neither underutilized nor overwhelmed.

Achieving an ideal BDP ensures optimal utilization of available resources, balancing the load between sending too much data (leading to congestion) or too little (leading to inefficiency).

In scenarios where latency is high or bandwidth is large, aligning the window size to the BDP allows smoother and more seamless data transmission with minimal hold-ups.