Question

Consider a closed-loop network (e.g., token ring) with bandwidth \(100 \mathrm{Mbps}\) and propagation speed of \(2 \times 10^{8} \mathrm{~m} / \mathrm{s}\). What would the circumference of the loop be to exactly contain one 250 -byte packet, assuming nodes do not introduce delay? What would the circumference be if there was a node every \(100 \mathrm{~m}\), and each node introduced 10 bits of delay?

Short answer

The loop circumference without node delay is 4000 meters. The loop circumference with node delays is 4800 meters.

Step-by-step solution

- Convert Packet Size to Bits

The size of the packet is given in bytes. Convert it to bits since the bandwidth is given in bits per second. 1 byte = 8 bits Therefore, 250 bytes = 250 * 8 = 2000 bits.

- Calculate Transmission Time

Determine the time it takes to transmit the packet using the bandwidth. Transmission time = \( \frac{ \text{Packet size (bits)} }{ \text{Bandwidth (bits/second)} } \) \[ \text{Transmission time} = \frac{2000}{100 \times 10^6} = 20 \mu s \]

- Calculate Distance Without Node Delay

Using the propagation speed, calculate the distance the signal travels in the transmission time. Distance = Propagation speed * Transmission time \[ \text{Distance} = 2 \times 10^8 \times 20 \times 10^{-6} = 4000 \text{ meters} \]Therefore, the circumference of the loop without node delay is 4000 meters.

- Calculate Additional Delay Due to Nodes

Calculate the delay introduced by the nodes. The network has a node every 100 meters, so the number of nodes around the loop is: \[ \frac{ \text{Circumference} }{ \text{Distance between nodes} } = \frac{4000}{100} = 40 \]Each node introduces 10 bits = \( \frac{10}{100 \times 10^6} = 0.1 \text{ microseconds} \). Total node delay = 40 * 0.1 microseconds = 4 microseconds.

- Calculate Effective Propagation Time with Node Delays

Include the node delays in the total effective propagation time. Total transmission time with delays = Transmission time + Total node delay \[ \text{Total transmission time} = 20 \text{ microseconds} + 4 \text{ microseconds} = 24 \text{ microseconds} \]

- Calculate New Circumference with Node Delays

Using the new total transmission time after accounting for node delays, calculate the new distance. Distance (new circumference) = Propagation speed * Total transmission time \[ \text{New distance} = 2 \times 10^8 \times 24 \times 10^{-6} = 4800 \text{ meters} \] Therefore, the circumference of the loop with node delays is 4800 meters.

Key concepts

token ring network

A token ring network is a type of communication protocol where each node is connected in a closed-loop configuration. Data travels in a circular manner around the ring from node to node. Every node checks if the data packet is addressed to it and takes action accordingly.
The primary benefit of this network is that each device gets a fair chance to transmit data. However, if one node fails, it can disrupt the entire network. Understanding token ring networks is crucial as they play a significant role in network topologies, ensuring orderly data transmission and reducing collisions.

bandwidth

Bandwidth refers to the maximum rate at which data can be transmitted over a network channel. In simpler terms, it is the data transfer capacity of a network.
In our problem, the network has a bandwidth of 100 Mbps (Megabits per second). This means the network can transmit 100 million bits of data per second. High bandwidth allows more data to be transferred in less time, making the network faster.
To calculate the time required to transmit a packet, you need to know the size of the packet in bits and then divide it by the bandwidth. For example, with a packet size of 2000 bits and a bandwidth of 100 Mbps, the transmission time is: \[ \text{Transmission time} = \frac{2000}{100 \times 10^6} = 20 \times 10^{-6} \text{ seconds} = 20 \text{ microseconds} \]

propagation speed

Propagation speed is the rate at which a signal travels through a medium. In the context of networking, it is how fast the data travels from one point to another along the network path.
In our exercise, the propagation speed is given as 2 x 10^8 meters per second. This is close to the speed of light in a vacuum, but in real-life networking, it can vary depending on the medium (e.g., copper wire, fiber optic).
To calculate the distance a signal can travel in a given time, multiply the propagation speed by the transmission time. For instance, with a propagation speed of 2 x 10^8 m/s and transmission time of 20 microseconds, the distance is: \[ \text{Distance} = 2 \times 10^8 \times 20 \times 10^{-6} = 4000 \text{ meters} \]

packet transmission time

Packet transmission time is the time it takes for a packet to be completely sent from the source to the destination over the network. It depends on both the packet size and the network bandwidth.
To find the transmission time, you can use the formula:
\[ \text{Transmission time} = \frac{\text{Packet size (bits)}}{\text{Bandwidth (bits/second)}} \]
For our 250-byte packet (which is 2000 bits) and a network with 100 Mbps bandwidth, the transmission time is calculated as follows: \[ \text{Transmission time} = \frac{2000}{100 \times 10^6} = 20 \text{ microseconds} \]
Transmission time is crucial to understand because, in real-time applications, lower transmission times are desirable to ensure efficient data flow.

node delay

Node delay refers to the additional time it takes for a data packet to be processed as it passes through each node in the network. This delay can be due to various factors, such as processing time at the node, buffering, and queuing.
In the provided exercise, each node introduces a delay of 10 bits. To convert this into time, given the bandwidth, we use the formula:\[ \text{Node Delay Time} = \frac{\text{Node Delay (bits)}}{\text{Bandwidth (bits/second)}} \]
So, 10 bits at 100 Mbps is: \[ \frac{10}{100 \times 10^6} = 0.1 \text{ microseconds per node} \]
With 40 nodes around the loop, the total node delay is: \[ \text{Total Node Delay} = 40 \times 0.1 = 4 \text{ microseconds} \]
This node delay must be added to the total transmission time to accurately calculate the new circumference of the network with node delays.