Question

A channel has a bit rate of \(4 \mathrm{kbps}\) and a propagation delay of \(20 \mathrm{msec}\). For what range of frame sizes does stop-and-wait give an efficiency of at least \(50 \%\) ?

Short answer

The frame size should be at least 160 bits for at least 50% efficiency.

Step-by-step solution

Understanding the Parameters

Start by identifying the parameters given in the task: the bit rate is given as 4 kbps, which can be represented as \( R = 4000 \text{ bits per second} \), and the propagation delay is \( t_p = 20 \text{ milliseconds} = 0.020 \text{ seconds} \).

Define Stop-and-Wait Efficiency

The efficiency \( E \) of the stop-and-wait ARQ protocol is defined as the ratio of the time used for transmitting the frame to the total time for transmitting and waiting for the acknowledgment: \[ E = \frac{T_{transmission}}{T_{transmission} + 2t_p} \] Where \( T_{transmission} \) is the time taken to transmit the frame, and \( 2t_p \) accounts for the round-trip propagation delay.

Expressing Transmission Time

The transmission time \( T_{transmission} \) can be expressed in terms of frame size \( L \) (in bits) as: \[ T_{transmission} = \frac{L}{R} \] Substitute \( R = 4000 \text{ bps} \) to obtain \[ T_{transmission} = \frac{L}{4000} \]

Setting the Efficiency Condition

For the stop-and-wait protocol to have an efficiency of at least 50%, set the efficiency equation to: \[ \frac{L/4000}{L/4000 + 0.040} \geq 0.5 \] Where the denominator includes \( 2t_p = 0.040 \text{ s} \).

Solving the Efficiency Inequality

Multiply both sides by the denominator to simplify the inequality: \[ L \geq 0.5 \cdot (L + 160) \] Solve this by distributing and isolating \( L \): \[ L \geq 0.5L + 80 \] Subtract \( 0.5L \) from both sides: \[ 0.5L \geq 80 \] Finally, multiply both sides by 2 to find \( L \): \[ L \geq 160 \text{ bits} \]

Finalizes the Frame Size Range

Therefore, for the stop-and-wait protocol to be at least 50% efficient, the frame size \( L \) must be greater than or equal to 160 bits. There is technically no upper limit as larger frame sizes generally decrease the impact of propagation delay.

Key concepts

Bit Rate

The bit rate of a communication channel is a fundamental parameter that describes how quickly data is transmitted over that channel. It is expressed in bits per second (bps) and determines how many bits are sent each second.
In the context of our exercise, the bit rate is given as 4 kilobits per second (kbps), which is equivalent to 4000 bits per second. The higher the bit rate, the faster the data is transmitted. This means that a higher bit rate will result in a shorter transmission time for a given amount of data.
Understanding the relationship between bit rate and other factors, like frame size and network efficiency, is essential because it influences how effectively a network operates. In stop-and-wait ARQ, efficiency is directly impacted by the bit rate, as it affects how the data is sent and acknowledged within the network.

Propagation Delay

Propagation delay refers to the time it takes for a signal to travel from the sender to the receiver in a network. It is influenced by the distance the signal must travel and the speed at which it moves through the medium (such as a cable or air).
In our example, the propagation delay is given as 20 milliseconds (ms), which is 0.020 seconds. This delay is a crucial part of the network's total latency, affecting how long it takes for a confirmation (or acknowledgment) of transmission to return to the sender.
Propagation delay is important in stop-and-wait ARQ because it contributes to the total time the sender must wait to receive an acknowledgment before sending the next frame. If the propagation delay is significant compared to the transmission time, it can severely reduce network efficiency.

Transmission Time

Transmission time is the amount of time required to send all the bits of a data frame from the sender to the receiver, without considering the time it takes for the acknowledgment to return.
This is calculated using the formula:\[ T_{\text{transmission}} = \frac{L}{R} \]where \( L \) is the frame size in bits and \( R \) is the bit rate in bits per second.
In our situation, if the frame size, denoted as \( L \), needs to be determined, this formula becomes crucial. It shows that as the frame size increases, the transmission time also increases, assuming the bit rate remains constant. This relationship is pivotal when calculating the network's efficiency under the stop-and-wait ARQ protocol.

Frame Size

Frame size is the number of bits in a single data packet or frame that is transmitted over a network. This size affects both how long it takes to send the frame and how long the sender must wait for an acknowledgment.
In the context of stop-and-wait, a larger frame size generally means more efficient use of the channel because a larger frame's transmission time can help mask the impact of the propagation delay. Therefore, there's less idle time waiting for an acknowledgment.
In our problem setup, we've determined that to achieve at least 50% efficiency, the frame size \( L \) must be greater than or equal to 160 bits. This emphasizes the importance of choosing appropriate frame sizes in network design to optimize performance.

Network Efficiency

Network efficiency, especially in the context of the Stop-and-Wait ARQ protocol, is a measure of how effectively the network can transmit data without being hindered by delays or idle times.
The efficiency \( E \) in stop-and-wait is given by:\[ E = \frac{T_{\text{transmission}}}{T_{\text{transmission}} + 2t_p} \]where \( T_{\text{transmission}} \) is the transmission time and \( 2t_p \) accounts for the round-trip propagation delay.
Achieving higher network efficiency means maximizing the amount of time data is transmitted compared to the time spent on waiting for acknowledgments. In our example, to maintain at least 50% efficiency, the frame size must meet or exceed 160 bits, subsequently balancing transmission time and propagation delay.