Question

Suppose a 100-Mbps point-to-point link is being set up between Earth and a new lunar colony. The distance from the moon to Earth is approximately \(385,000 \mathrm{~km}\), and data travels over the link at the speed of light-3 \(\times 10^{8} \mathrm{~m} / \mathrm{s}\). (a) Calculate the minimum RTT for the link. (b) Using the RTT as the delay, calculate the delay \(\times\) bandwidth product for the link. (c) What is the significance of the delay \(\times\) bandwidth product computed in (b)? (d) A camera on the lunar base takes pictures of Earth and saves them in digital format to disk. Suppose Mission Control on Earth wishes to download the most current image, which is \(25 \mathrm{MB}\). What is the minimum amount of time that will elapse between when the request for the data goes out and the transfer is finished?

Short answer

The minimum RTT is 2.566 seconds. The delay × bandwidth product is 256.6 megabits. The image transfer will take a minimum of 4.566 seconds.

Step-by-step solution

- Calculate the Minimum RTT

First, calculate the time it takes for a signal to travel one way between Earth and the moon. Use the formula: \[ \text{Time} = \frac{\text{Distance}}{\text{Speed of light}} \]Substitute the given values: \[ \text{Time} = \frac{385,000 \text{ km}}{3 \times 10^5 \text{ km/s}} = 1.283 \text{ seconds} \]The RTT (Round-Trip Time) is twice the one-way time: \[ \text{RTT} = 1.283 \text{ seconds} \times 2 = 2.566 \text{ seconds} \]

- Calculate the Delay × Bandwidth Product

The delay × bandwidth product is calculated using the formula: \[ \text{Delay × Bandwidth} = \text{RTT} \times \text{Bandwidth} \]Given the bandwidth is 100 Mbps, convert it to bits per second: \[ 100 \text{ Mbps} = 100 \times 10^6 \text{ bps} \]Substitute the values: \[ \text{Delay × Bandwidth} = 2.566 \text{ seconds} \times 100 \times 10^6 \text{ bps} = 256,600,000 \text{ bits} \]

- Understand the Significance of Delay × Bandwidth Product

The delay × bandwidth product represents the amount of data that can be 'in flight' on the link at any given time. It indicates the maximum amount of unacknowledged data that can be on the network.

- Calculate the Minimum Time for Image Transfer

First, calculate the time required to transmit the image. Convert the size of the image to bits: \[ 25 \text{ MB} = 25 \times 10^6 \text{ bytes} = 2 \times 25 \times 10^6 \times 10^3 \text{ bits} = 200 \times 10^6 \text{ bits} \]Time to transmit the image: \[ \text{Transmit Time} = \frac{\text{Image Size}}{\text{Bandwidth}} = \frac{200 \times 10^6 \text{ bits}}{100 \times 10^6 \text{ bps}} = 2 \text{ seconds} \]Add the one-way delay to send the request and the RTT for the response: \[ \text{Total Time} = \text{RTT} + \text{Transmit Time} = 2.566 \text{ seconds} + 2 \text{ seconds} = 4.566 \text{ seconds} \]

Key concepts

RTT calculation

RTT (Round-Trip Time) calculation is essential in understanding delays in data transmission over long distances. RTT is the time it takes for a signal to travel from the source to the destination and back again. To calculate RTT, you need to know the distance between the two points and the speed at which the signal travels. For instance, considering the Earth-Moon link, the one-way distance is approximately 385,000 km. Signals travel at the speed of light, roughly 300,000 km/s. The one-way travel time is calculated by dividing the distance by the speed: \[\text{Time} = \frac{385,000 \text{ km}}{300,000 \text{ km/s}} = 1.283 \text{ seconds}\] Since RTT is a round trip, we multiply this time by 2: \[\text{RTT} = 1.283 \times 2 = 2.566 \text{ seconds}\] Therefore, the minimum RTT for the link is 2.566 seconds.

delay × bandwidth product

The delay × bandwidth product is a critical concept in networking. This product helps to determine the amount of data that can be in transit (in-flight) in the network at any given moment. It is calculated using the formula: \[\text{Delay × Bandwidth} = \text{RTT} \times \text{Bandwidth}\] Given a 100-Mbps (Megabits per second) link and RTT of 2.566 seconds, we can convert the bandwidth to bits per second: \[100 \text{ Mbps} = 100 \times 10^6 \text{ bps}\] Using the values, we get: \[\text{Delay × Bandwidth} = 2.566 \times 100 \times 10^6 \text{ bps} = 256,600,000 \text{ bits}\] This product represents the maximum amount of unacknowledged data that can be present in the network. It is crucial for understanding the efficiency and throughput of data transmission channels.

data transmission delay

Data transmission delay is the amount of time taken for data to travel from the source to the destination. This delay consists of several components, including propagation delay, transmission delay, queuing delay, and processing delay. In our scenario between Earth and the moon, the primary concern is the propagation and transmission delay.
Propagation delay is the time it takes for a signal to travel through a medium, which can be calculated using: \[\text{Propagation Delay} = \frac{\text{Distance}}{\text{Speed of Light}}\] Transmission delay, on the other hand, depends on the size of the data and the bandwidth of the link: \[\text{Transmission Delay} = \frac{\text{Data Size}}{\text{Bandwidth}}\] Combining both, and considering a 100-Mbps link, the delay for transmitting a 25-MB image is: \[25 \text{ MB} = 25 \times 10^6 \text{ bytes} = 200 \times 10^6 \text{ bits}\] \[\text{Transmission Delay} = \frac{200 \times 10^6 \text{ bits}}{100 \times 10^6 \text{bps}} = 2 \text{ seconds}\] Thus, adding the RTT, the total data transmission delay becomes more significant in understanding the communication time required.

point-to-point link

A point-to-point link is a direct communication pathway between two endpoints. Unlike networks with multiple intermediary devices, a point-to-point link connects two nodes directly, providing a dedicated communication channel. This setup can be highly efficient for long-distance data transmission, as seen in the Earth-Moon link example.
Some of the main benefits of point-to-point links include:

• Reduced interference and congestion since the link is dedicated to two endpoints.
• Higher reliability because the communication path is straightforward.
• Ease of troubleshooting due to the simple architectural layout.

Understanding the properties and benefits of point-to-point links is crucial for designing efficient, long-distance communication networks.

speed of light in networking

The speed of light in networking is a fundamental metric defining the maximum speed at which data can travel through a medium. It significantly impacts the propagation delay and ultimately the RTT. For most practical purposes, the speed of light in a vacuum is approximately 300,000 km/s. However, in networking cables like fiber optics, this speed can be slightly less due to the refractive index of the medium.
In the Earth-Moon example, signals travel at roughly the speed of light, impacting the total communication time:

• One-way propagation delay: Calculated from distance and speed of light.
• Round-Trip Time (RTT): Twice the one-way propagation delay.

Understanding this helps network engineers optimize and anticipate delays in long-distance communication setups, ensuring that data transmission occurs as efficiently as possible.