Question

Consider a point-to-point link \(50 \mathrm{~km}\) in length. At what bandwidth would propagation delay (at a speed of \(2 \times 10^{8} \mathrm{~m} / \mathrm{s}\) ) equal transmit delay for 100 -byte packets? What about 512 -byte packets?

Short answer

For 100-byte packets: 3.2 Mbps. For 512-byte packets: 16.384 Mbps.

Step-by-step solution

- Find the Propagation Delay

The propagation delay can be found using the formula \(\text{Propagation Delay} = \frac{\text{Distance}}{\text{Propagation Speed}}\). The given distance is 50 km or 50,000 meters, and the speed is \(2 \times 10^8 \text{ m/s}\). Thus, \[ \text{Propagation Delay} = \frac{50,000 \text{ m}}{2 \times 10^8 \text{ m/s}} = 0.00025 \text{ s} = 250 \text{ µs} \]

- Define Transmission Delay

Transmission delay is given by \(\text{Transmission Delay} = \frac{\text{Packet Size}}{\text{Bandwidth}}\). We will use this formula to equate it with the propagation delay and solve for the bandwidth.

- Calculate Bandwidth for 100-byte Packets

For 100-byte packets, we have the size as 100 bytes or 800 bits. Set the transmission delay equal to the propagation delay: \( \frac{800 \text{ bits}}{\text{Bandwidth (bps)}} = 0.00025 \text{ s} \). Solving for bandwidth, we get \[ \text{Bandwidth} = \frac{800 \text{ bits}}{0.00025 \text{ s}} = 3.2 \times 10^6 \text{ bps} = 3.2 \text{ Mbps} \]

- Calculate Bandwidth for 512-byte Packets

For 512-byte packets, we have the size as 512 bytes or 4096 bits. Set the transmission delay equal to the propagation delay: \( \frac{4096 \text{ bits}}{\text{Bandwidth (bps)}} = 0.00025 \text{ s} \). Solving for bandwidth, we get \[ \text{Bandwidth} = \frac{4096 \text{ bits}}{0.00025 \text{ s}} = 1.6384 \times 10^7 \text{ bps} = 16.384 \text{ Mbps} \]

Key concepts

Propagation delay

Propagation delay is the time it takes for a signal to travel from the sender to the receiver over a given medium. This delay is crucial for understanding network performance. It's affected by the distance between the two points and the speed at which the signal travels.

This speed is often close to the speed of light, but it can vary based on the medium (like fiber optics or copper wires). In this exercise, the propagation speed is given as \(2 \times 10^8 \text{ m/s}\).

To calculate propagation delay, use the formula: \[\text{Propagation Delay} = \frac{\text{Distance}}{\text{Propagation Speed}} \]
For example, for a 50 km link: \[\text{Propagation Delay} = \frac{50,000 \text{ m}}{2 \times 10^8 \text{ m/s}} = 0.00025 \text{ s} \]
This is 250 microseconds.

Transmission delay

Transmission delay refers to the time required to push all the packet's bits onto the wire. It is determined by the packet size and the bandwidth of the link. Understanding transmission delay helps identify how long it takes to send data.

The formula to calculate transmission delay is: \[\text{Transmission Delay} = \frac{\text{Packet Size}}{\text{Bandwidth}} \frac{\text{Packet Size (bits)}}{\text{Bandwidth (bps)}} \]
In this exercise, we see that for different packet sizes, the bandwidth required changes. For 100-byte packets (which is 800 bits): \[\text{Transmission Delay} = \frac{800 \text{ bits}}{0.00025 \text{ s}} = 3.2 \times 10^6 \text{ bps} = 3.2 \text{ Mbps} \]
For 512-byte packets (which is 4096 bits): \[\text{Transmission Delay} = \frac{4096 \text{ bits}}{0.00025 \text{ s}} = 16.384 \text{ Mbps} \frac{\text{Packet Size (bits)}}{\text{Bandwidth (bps)}}\]This shows how larger packets need higher bandwidth to keep the same delay.

Bandwidth calculation

Bandwidth is the maximum rate of data transfer across a given path. It is a key factor in network performance as it dictates how much data can be sent over the network in a given time.

In scenarios like the current exercise, bandwidth calculation helps us ensure that the network can handle the desired data rates with minimal delays. We equate propagation and transmission delays to solve for bandwidth.

For 100-byte packets: \[\text{Bandwidth} = \frac{800 \text{ bits}}{0.00025 \text{ s}} = 3.2 \times 10^6 \text{ bps} = 3.2 \text{ Mbps} \]
And for 512-byte packets: \[\text{Bandwidth} = \frac{4096 \text{ bits}}{0.00025 \text{ s}} = 16.384 \text{ Mbps} \]
This shows that to maintain a balance with the propagation delay, larger packet sizes require higher bandwidth.