Question

The distance from earth to a distant planet is approximately \(9 \times 10^{10} \mathrm{~m}\). What is the channel utilization if a stop-and-wait protocol is used for frame transmission on a 64 Mbps point-to-point link? Assume that the frame size is \(32 \mathrm{~KB}\) and the speed of light is \(3 \times 10^{8} \mathrm{~m} / \mathrm{s}\).

Short answer

The channel utilization is approximately 0.000683\%.

Step-by-step solution

Determine the Round-Trip Time (RTT)

Calculate the one-way trip time by dividing the distance by the speed of light: \[ \text{One-way time} = \frac{9 \times 10^{10} \text{ m}}{3 \times 10^8 \text{ m/s}} = 300 \text{ seconds} \]Since the signal must travel to the planet and back, the Round-Trip Time (RTT) is twice the one-way time: \[ \text{RTT} = 2 \times 300 \text{ seconds} = 600 \text{ seconds} \]

Calculate Transmission Time of the Frame

First, convert the frame size from kilobytes to bits as the link speed is in bits per second:\[ 32 \text{ KB} = 32 \times 1024 \times 8 \text{ bits} = 262144 \text{ bits} \]Then, calculate the transmission time using the link speed of 64 Mbps:\[ \text{Transmission Time} = \frac{262144 \text{ bits}}{64 \times 10^6 \text{ bits/second}} \approx 0.0041 \text{ seconds} \]

Calculate Total Time per Frame in Stop-and-Wait Protocol

The total time required for one frame to be sent and for the acknowledgment to be received is the sum of the transmission time and the RTT:\[ \text{Total Time per Frame} = \text{Transmission Time} + \text{RTT} = 0.0041 \text{ seconds} + 600 \text{ seconds} \approx 600.0041 \text{ seconds} \]

Compute Channel Utilization

Channel utilization is the ratio of the transmission time to the total time per cycle of the stop-and-wait protocol:\[ \text{Utilization} = \frac{\text{Transmission Time}}{\text{Total Time per Frame}} = \frac{0.0041}{600.0041} \approx 0.00000683 \]Convert this to percentage:\[ \text{Utilization (percentage)} = 0.00000683 \times 100 \approx 0.000683\% \]

Key concepts

Channel Utilization

Channel utilization is a critical concept when discussing data transmission, particularly in contexts like the stop-and-wait protocol. It measures how effectively a communication channel is used when data is transmitted. In simple terms, it tells us how much time the channel spends doing "useful work" versus being idle.

In the context of a stop-and-wait protocol, channel utilization can often be very low. This is because, after sending a frame, the sender must wait until it receives an acknowledgment before sending the next frame. This waiting period can be quite long compared to the time it takes to actually send each frame, especially when signals must travel long distances, such as from Earth to a distant planet.

**Key Points:**

• The transmission time is short, while the waiting time (Round-Trip Time) can be lengthy, leading to inefficiencies.
• Utilization is calculated as the ratio of the time spent transmitting data to the total cycle time, which includes waiting time.
• In this exercise, the channel utilization was calculated to be a mere 0.000683%, indicating the vast majority of time the channel is idle.

Round-Trip Time (RTT)

Round-Trip Time (RTT) is an essential concept in network communications. It refers to the time it takes for a signal to travel from the sender to the receiver and back again. RTT is crucial because it directly impacts the efficiency of protocols like stop-and-wait.

For this specific problem involving a distant planet, RTT becomes the dominating factor in the total time equation. To calculate it, we first determine the one-way trip time. This is done by dividing the distance by the speed of light (given in the problem). Since the next step is to multiply by two (to account for the return trip), we get the RTT.

**Why RTT Matters:**

• Stop-and-wait protocols can suffer from long RTTs because the sender must wait for acknowledgment before proceeding.
• High RTT values result in large amounts of time where the channel is not actively used for data transmission.
• In our example, the RTT was calculated to be 600 seconds, which vastly overshadows the actual transmission time.

Transmission Time

Transmission time is the period it takes to send a single data frame over the communication channel. This is a significant factor because it is the "active" time during which the channel is being utilized for transmission. When assessing channel utilization, transmission time plays a crucial role in showing how effective the data link is.

To find transmission time, the frame size must be converted to bits. Since communication speeds are often given in bits per second, the equation becomes simple: frame size in bits divided by the link speed (also in bits per second). In our case, this calculation resulted in approximately 0.0041 seconds.

**Core Insights:**

• Transmission time is generally quite short in comparison to the RTT, especially for distant communications.
• The effectiveness of data transmission can be enhanced by having a lower RTT in place or employing strategies like pipelining.
• In this problem, the transmission time displayed how minimal the actual data-sending period is relative to the idle periods induced by RTT.