Chapter 1: Problem 16
Suppose a 128-Kbps point-to-point link is set up between Earth and a rover on Mars. The distance from Earth to Mars (when they are closest together) is approximately \(55 \mathrm{Gm}\), 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) Calculate the delay \(\times\) bandwidth product for the link. (c) A camera on the rover takes pictures of its surroundings and sends these to Earth. How quickly after a picture is taken can it reach Mission Control on Earth? Assume that each image is \(5 \mathrm{Mb}\) in size.
Short Answer
Expert verified
Minimum RTT is 366.67 seconds. The delay-bandwidth product is 23.466 Mb. It takes 222.39 seconds for a picture to reach Earth.
Step by step solution
01
- Calculate the minimum RTT (Round Trip Time)
Round Trip Time (RTT) is the time it takes for a signal to travel to the destination and back. Since we know the speed of light is approximately \(3 \times 10^{8} \mathrm{~m} / \mathrm{s}\), and the distance from Earth to Mars when they are closest is \(55 \mathrm{Gm}\) (\(55 \times 10^{9} \mathrm{~m}\)), we can calculate the one-way trip time as follows:\[ \text{One-way trip time} \ = \frac{55 \times 10^{9} \mathrm{~m}}{3 \times 10^{8} \mathrm{~m/s}} \ = 183.33 \text{ seconds} \ \text{Round Trip Time (RTT)} = 2 \times 183.33 \text{ seconds} \ = 366.67 \text{ seconds} \] Thus, the minimum RTT for the link is 366.67 seconds.
02
- Calculate the delay-bandwidth product
The delay-bandwidth product is calculated by multiplying the link's bandwidth by the propogation delay (one-way trip time). Given the bandwidth is \(128 \mathrm{Kbps}\) (\(128 \times 10^{3} \mathrm{bps}\)), and the one-way delay calculated before as 183.33 seconds, we get: \[ \text{Delay-Bandwidth Product} \ = (128 \times 10^{3} \mathrm{bps}) \times 183.33 \mathrm{s} \ = 23.466 \times 10^{6} \mathrm{bits} \ = 23.466 \mathrm{Mb} \]
03
- Calculate the time taken for a picture to reach Earth
To calculate how quickly a picture reaches Mission Control on Earth, we add the transmission time to the one-way propagation delay. Given that the picture size is \(5 \mathrm{Mb}\) and bandwidth is \(128 \mathrm{Kbps} \ (128 \times 10^{3} \mathrm{bps})\), the transmission time is: \[ \text{Transmission time} \ = \frac{5 \times 10^{6} \mathrm{bits}}{128 \times 10^{3} \mathrm{bps}} \ = \frac{5 \times 10^{6}}{128 \times 10^{3}} \ = 39.06 \text{ seconds} \] The total time to send a picture to Earth: \[ \text{Total time} \ = 183.33 \text{ seconds} \ (\text{propagation delay}) + 39.06 \text{ seconds} \ (\text{transmission time}) \ = 222.39 \text{ seconds} \] Therefore, it takes 222.39 seconds after a picture is taken for it to reach Mission Control on Earth.
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Round Trip Time (RTT)
When communicating over a network, the Round Trip Time (RTT) represents how long it takes for a signal to travel from the sender to the receiver and back again. In our example with the Mars rover, the RTT is critical for understanding the delay in communication due to the vast distance between Earth and Mars. Since the speed of light in a vacuum is approximately 300,000,000 meters per second, and the closest distance between Earth and Mars is 55,000,000,000 meters, we found that the one-way trip takes 183.33 seconds. Multiplying by two gives the minimum RTT of 366.67 seconds.
delay-bandwidth product
The delay-bandwidth product is a key concept in networking that helps determine the amount of data that can be 'in flight' within the network. It is calculated by multiplying the link's bandwidth by the propagation delay. For the Martian link, with a bandwidth of 128 Kbps (128,000 bps) and a one-way delay of 183.33 seconds, the delay-bandwidth product is 23.466 megabits. This value tells us how much data could be in the network during the delay period, providing insight into the efficiency and data handling capacity of the link.
propagation delay
Propagation delay refers to the time it takes for a signal to travel from the sender to the receiver in a network. This delay is influenced by the distance the signal must travel and the speed at which it propagates. In the scenario of transmitting data between Earth and Mars, the propagation delay is significant because the distance is immense (55 million kilometers), and the speed of light, while extremely fast, results in a one-way trip taking 183.33 seconds.
transmission time
Transmission time is the amount of time it takes to send all the bits of a message from the sender to the receiver. It is determined by dividing the size of the data by the bandwidth of the link. In our exercise, sending a 5 megabit image from the Mars rover to Earth over a 128 Kbps link results in a transmission time of 39.06 seconds. This is separate from the propagation delay and must be added to it for the total time a message takes to reach its destination.
data transmission
Data transmission encompasses the entire process of sending data from one point to another. This includes both the transmission time and the propagation delay. Using the Martian example, once the rover captures a picture, it must send this data through the entire communication link to Mission Control on Earth. The total time taken includes the 39.06 seconds for the actual data transfer (transmission time) plus the 183.33 seconds it takes for the data to travel the distance (propagation delay), thus making the total time 222.39 seconds. Understanding these components is crucial for designing efficient and reliable communication systems.
