Question

For a 100-Mbps token ring network with a token rotation time of \(200 \mu\) s that allows each station to transmit one 1-KB packet each time it possesses the token, calculate the maximum effective throughput rate that any one host can achieve. Do this assuming (a) immediate release and (b) delayed release.

Short answer

Immediate release: 40.96 Mbps, Delayed release: 29.06 Mbps

Step-by-step solution

Title - Determine the packet size in bits

Each packet is 1-KB. Convert this to bits: \( 1 \text{ KB} = 1024 \text{ bytes} = 1024 \times 8 \text{ bits} = 8192 \text{ bits} \)

Title - Calculate transmission time for one packet

The transmission rate is 100 Mbps. Determine the time to send one packet: \( \text{Transmission time} = \frac{8192 \text{ bits}}{100 \text{ Mbps}} = \frac{8192}{100 \times 10^6} \text{ s} = 81.92 \mu\text{s} \)

Title - Calculate maximum effective throughput (immediate release)

For immediate release, the host can send another packet as soon as it finishes the current one. The effective throughput is the packet size divided by the token rotation time (since the transmission time is less than the rotation time): \( \text{Effective throughput} = \frac{8192 \text{ bits}}{200 \mu\text{s}} = 40.96 \text{ Mbps} \)

Title - Calculate maximum effective throughput (delayed release)

For delayed release, the host can send only one packet per token rotation: \( \text{Effective throughput} = \frac{8192 \text{ bits}}{200 \mu\text{s} + 81.92 \mu\text{s}} \approx \frac{8192 \text{ bits}}{281.92 \mu\text{s}} \approx 29.06 \text{ Mbps} \)

Key concepts

Token Ring Network

A token ring network is a type of local area network (LAN) architecture where all computers are connected in a ring or star topology, and data passes sequentially between nodes.
The key element here is the 'token,' a small data packet that circulates around the network.
Only the computer that holds the token is allowed to send data, ensuring that only one device transmits at a time, which helps to avoid collisions.
In a token ring network, a token travels around the ring, and as each device gets the token, it either sends data if it has any to send or passes the token along to the next device.

• **Advantages:**
  • Predictable performance due to token-passing mechanism.
  • Reduces collisions since only one device transmits at a time.
• **Disadvantages:**
  • Can be relatively slow, especially as the number of connected devices increases.
  • If any device or connection in the ring fails, it can disrupt the entire network.

Transmission Time

Transmission time is the amount of time required to send a packet of data from one point to another in a network.
It is calculated based on the size of the data packet and the data transmission rate of the network.

• The formula for transmission time is given as:
  \text{Transmission time} = \frac{\text{Packet size (bits)}}{\text{Transmission rate (bits per second)}} \[...\]
• In the context of the exercise with a 100-Mbps token ring network, for a single 1-KB packet (which is 8192 bits), the transmission time comes out to be 81.92 microseconds \[ ...\].

This transmission time indicates how swiftly data can be sent across the network. Shorter transmission times are better as they imply faster data transfer.

Effective Throughput

Effective throughput is the actual rate at which useful data is transmitted over a network.
It considers various factors like the transmission time, token rotation time, and network overhead.

• **Immediate Release:** This refers to the scenario where the host can send another packet as soon as it finishes sending the current one.
  • Here, the maximum effective throughput is calculated using the formula:
    \[ \text{Effective throughput} = \frac{\text{Packet size (bits)}}{\text{Token rotation time}} \]=\(40.96 \text{ Mbps}\)
• **Delayed Release:** In this scenario, the host can only send one packet per token rotation.
  • The maximum effective throughput is recalculated to consider additional waiting times:
    \[ \text{Effective throughput} = \frac{8192 \text{ bits}}{200 \mu\text{s} + 81.92 \mu\text{s}} \approx 29.06 \text{ Mbps} \]
    This leads to a lower effective throughput because of the additional waiting time per transmission cycle.

Understanding effective throughput helps in determining the actual performance and efficiency of network data transmissions.