Question

Consider a token ring with a ring latency of \(200 \mu\). Assuming that the delayed token release strategy is used, what is the effective throughput rate that can be achieved if the ring has a bandwidth of \(4 \mathrm{Mbps}\) ? What is the effective throughput rate that can be achieved if the ring has a bandwidth of \(100 \mathrm{Mbps}\) ? Answer for both a single active host and for "many" hosts; for the latter, assume there are sufficiently many hosts transmitting that the time spent advancing the token can be ignored. Assume a packet size of \(1 \mathrm{~KB}\).

Short answer

For bandwidth 4 Mbps: \ - Single host: 39.6 kbps \ - Many hosts: 4 Mbps \[For bandwidth 100 Mbps: \ - Single host: 39.99 kbps \ - Many hosts: 100 Mbps\]

Step-by-step solution

Understand the delayed token release strategy

The delayed token release strategy means the token isn't released until the sender is finished transmitting the packet. This effectively adds the transmission time of the packet to the latency.

Calculate the transmission time for a packet

Given a packet size of 1 KB (which is 8,000 bits), use the formula \(\text{Transmission time} = \frac{\text{Packet size (bits)}}{\text{Bandwidth (bits/sec)}}\).

Compute for 4 Mbps (single host)

\[\text{Transmission time} = \frac{8,000 \text{ bits}}{4 \times 10^6 \text{ bits/sec}} = 2 \mu s\]With a ring latency of 200 \(\mu\)s, the effective bandwidth for a single host is computed as the size of the packet divided by the total time (latency plus the transmission time). \ \[\text{Effective throughput} = \frac{8 \text{ kb}}{200 \mu \text{s} + 2 \mu \text{s}} = \frac{8 \text{ kb}}{202 \mu \text{s}} \approx 39.6 \text{ kbps}\]

Compute for 4 Mbps (many hosts)

With many hosts, the ring latency becomes negligible. Hence, the effective throughput is just the transmission time for the packet. \[\text{Effective throughput} = \frac{8 \text{ kb}}{2 \mu \text{s}} = 4 \text{ Mbps}\]

Compute for 100 Mbps (single host)

\[\text{Transmission time} = \frac{8,000 \text{ bits}}{100 \times 10^6 \text{ bits/sec}} = 0.08 \mu s\]With a ring latency of 200 \(\mu\)s, the effective bandwidth for a single host is computed as \ \[\text{Effective throughput} = \frac{8 \text{ kb}}{200 \mu \text{s} + 0.08 \mu \text{s}} = \frac{8 \text{ kb}}{200.08 \mu \text{s}} \approx 39.99 \text{ kbps}\]

Compute for 100 Mbps (many hosts)

With many hosts, the ring latency becomes negligible. Hence, the effective throughput is just the transmission time for the packet. \[\text{Effective throughput} = \frac{8 \text{ kb}}{0.08 \mu \text{s}} = 100 \text{ Mbps}\]

Key concepts

Delayed Token Release Strategy

The delayed token release strategy is crucial in a token ring network. Here, the token—which signals network permission—is not released until the sender has finished transmitting its entire packet. This mechanism ensures that no other host can send packets during this time, reducing packet collisions. It essentially extends the network's latency since the total time includes the packet's transmission time. By adding this delay, the network can manage transmissions more efficiently but at the cost of increased latency.

Ring Latency

Ring latency in a token ring network refers to the time it takes for the token to circulate the entire ring once. This includes the transmission time of packets and any intrinsic delays due to the network's physical properties. It can significantly impact the effective throughput rate, especially in environments where the delayed token release strategy is used. The length of the ring, the number of devices, and the speed at which the token travels all contribute to the total latency.

Network Bandwidth

Network bandwidth is the maximum rate at which data can be transmitted over the network path. This is measured in bits per second (bps). For instance, in the given problem, we have bandwidths of 4 Mbps and 100 Mbps. The bandwidth limits how much data can be sent within a specific period and directly affects the transmission time of packets. Higher bandwidth allows for faster data transmission, reducing the time a packet takes to circulate the network, thereby increasing the network's effective throughput.

Effective Throughput Rate

The effective throughput rate is the actual data transfer rate achieved, accounting for network delays like latency and token release strategies. It's calculated by considering the total time a packet takes, including both transmission and any additional delays. For a single host with 4 Mbps bandwidth, the effective throughput is significantly lower than the nominal bandwidth due to latency. However, when multiple hosts are transmitting, the impact of ring latency diminishes, allowing throughput to approach the nominal bandwidth.

Single Active Host

When there's only one active host in the network, the token effectively waits for this host to finish its transmission before being released. This means that the ring latency combined with the packet's transmission time creates a bottleneck. For example, with a 4 Mbps bandwidth, the additional delay significantly drops the effective throughput rate to around 39.6 kbps due to the token's delayed release and ring's inherent latency. Even with a 100 Mbps bandwidth, the effective throughput is only about 39.99 kbps because the packet transmission time becomes significantly shorter, but the latency remains.

Multiple Hosts

In a scenario with multiple hosts actively transmitting data, the effect of ring latency can be minimized. Because the network continuously transmits data, the delayed token release strategy's impact is less pronounced. With many hosts, the dominant factor becomes the transmission time of the packets. For instance, with 4 Mbps and 100 Mbps bandwidths, the effective throughput rates can achieve near nominal speeds (4 Mbps and 100 Mbps respectively) because the time taken to advance the token becomes negligible. This scenario maximizes the network's capacity, making it efficient for heavy traffic loads.