Question

A \(100-\mathrm{km}\)-long cable runs at the \(\mathrm{T} 1\) data rate. The propagation speed in the cable is \(2 / 3\) the speed of light in vacuum. How many bits fit in the cable?

Short answer

The cable can hold 772 bits at the T1 data rate.

Step-by-step solution

Identify Given Values

We start by identifying the parameters given in the problem. The length of the cable is 100 km, and the propagation speed is \( \frac{2}{3} \) of the speed of light in vacuum. The T1 data rate is approximately \( 1.544 \text{ Mbps} \).

Calculate Propagation Speed

The speed of light in vacuum is approximately \( 3 \times 10^8 \text{ m/s} \). The propagation speed through the cable is \( \frac{2}{3} \times 3 \times 10^8 = 2 \times 10^8 \text{ m/s} \).

Convert Cable Length to Meters

The length of the cable is given as 100 km. We convert this to meters: \( 100 \times 1000 = 100,000 \text{ meters} \).

Calculate Propagation Time Through the Cable

We find the time it takes for a signal to propagate from one end of the cable to the other using the formula: \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100,000}{2 \times 10^8} \text{ seconds} = 5 \times 10^{-4} \text{ seconds} \).

Calculate Number of Bits in Transit

To find out how many bits fit into the cable, we multiply the data rate by the propagation time: \( 1.544 \times 10^6 \text{ bits/s} \times 5 \times 10^{-4} \text{ s} = 772 \text{ bits} \).

Key concepts

Propagation Speed

When discussing networking, one of the most significant factors is propagation speed. This is the speed at which signals can travel through a medium, such as a cable or fiber optic line. In the given problem, the propagation speed is specified to be \( \frac{2}{3} \) of the speed of light in a vacuum.
The speed of light is a universal constant, approximately \( 3 \times 10^8 \text{ m/s} \). Therefore, the propulsion speed in the cable is calculated as: \[(\frac{2}{3}) \times (3 \times 10^8) = 2 \times 10^8 \text{ m/s} \] This means signals will travel through the cable at a speed of 200,000,000 meters per second.

Understanding propagation speed is crucial in networking because it affects how quickly data can be transmitted across distances. In real-world applications, materials used for cables may affect speed, but in theoretical models like this one, it's often a fraction of the speed of light.

T1 Data Rate

The T1 data rate is a standard in telecommunications that refers to the rate at which data is transmitted over a network. It is specifically defined at about \( 1.544 \text{ Mbps} \) (Megabits per second).
This rate means that 1.544 million bits can be transmitted each second. T1 lines have been essential in providing consistent and reliable data transmission over long distances. They are often replaced by faster technologies in many areas now, but understanding T1 is foundational in comprehending digital data rates.

Knowing the data rate is vital because it helps you determine how much data can be sent, thus planning the required bandwidth and assessing the network's capability.

Calculating Bits in Transit

To determine the number of bits that can fit in a cable, we need to consider the amount of data that is "in flight" at any given moment. This depends on two things: the propagation speed and the data rate.
First, calculate how long it takes for a bit to travel from one end of the cable to the other, known as the propagation time. Using the formula: \[\text{Time} = \frac{\text{Distance}}{\text{Speed}} \]For our 100 km cable (100,000 meters) with a propagation speed of \( 2 \times 10^8 \text{ m/s} \), the time is: \[5 \times 10^{-4} \text{ seconds} \] Next, multiply the propagation time by the data rate to find the number of bits in transit.
And using the T1 data rate:\[1.544 \times 10^6 \text{ bits/s} \times 5 \times 10^{-4} \text{ s} = 772 \text{ bits} \]So, 772 bits are "on the line" at any moment, traveling through that 100 km cable. Knowing this helps optimize and map the current network's efficiency. It's a fundamental aspect of understanding latency and data throughput.