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Chapter 2: Problem 59

Suppose a \(100-\mathrm{Mbps}\) delayed-release token ring has 10 stations, a ring latency of \(30 \mu \mathrm{s}\), and an agreed-upon TTRT of \(350 \mu \mathrm{s}\). (a) How many synchronous frame bytes could each station send, assuming all are allocated the same amount? (b) Assume stations \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) are in increasing order on the ring. Due to uniform synchronous traffic, the TRT without asynchronous data is \(300 \mu \mathrm{s}\). B sends a \(200-\mu \mathrm{s}(2.5-\mathrm{Kb})\) asynchronous frame. What TRT will \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) then see on their next measurement? Who may transmit such a frame next?

Short Answer

Expert verified
(a) Each station can send 400 synchronous frame bytes. (b) TRT for A, B, and C will be 500 μs. Station C may transmit next.

Step by step solution

01

Calculate Total Time Available for Synchronous Frames

The Total Token Rotation Time (TTRT) is given as 350 \(\mu \mathrm{s}\). Subtract the ring latency to find the total time available for sending synchronous frames:\[ \text{Total Time for Synchronous Frames} = \text{TTRT} - \text{Ring Latency} = 350 \mu \mathrm{s} - 30 \mu \mathrm{s} = 320 \mu \mathrm{s} \]
02

Allocate Time Equally to All Stations

Divide the available time for synchronous frames equally among the 10 stations:\[ \text{Time per Station} = \frac{320 \mu \mathrm{s}}{10} = 32 \mu \mathrm{s} \]
03

Convert Time to Frame Bytes

Convert the time each station gets for synchronous frames into bytes. Use the data rate of 100-Mbps:\[ \text{Bytes per Station} = \frac{100 \times 10^6 \text{ bits/sec} \times 32 \times 10^{-6} \text{ sec}}{8 \text{ bits/byte}} = 400 \text{ bytes} \]
04

Determine Base Token Rotation Time

Given the base Token Rotation Time is 300 \( \mu \mathrm{s} \) without asynchronous frames.
05

Calculate New TRT for Station A

Station B sends a 200-\( \mu \mathrm{s} \) asynchronous frame. Given the TRT without asynchronous frames is 300 \( \mu \mathrm{s} \), the new TRT for Station A will be:\[ \text{TRT}_{A} = 300 \mu \mathrm{s} + 200 \mu \mathrm{s} = 500 \mu \mathrm{s} \]
06

Calculate New TRT for Station B

Station B adds the 200-\( \mu \mathrm{s} \) asynchronous frame to the previous TRT. Therefore, the TRT for Station B: \[ \text{TRT}_{B} = 500 \mu \mathrm{s} \]
07

Calculate New TRT for Station C

Since Station C is next in the order, it will also see the TRT as 500 \( \mu \mathrm{s} \). \[ \text{TRT}_{C} = 500 \mu \mathrm{s} \]
08

Determine Next Eligible Station for Asynchronous Frame

Given the ring continuity and the equal allocation concept, after Station B has sent an asynchronous frame, the next station to send may logically be Station C. However, this depends on the fairness algorithm employed by the token ring.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Synchronous Frames
In a token ring network, synchronous frames are data packets sent during a predefined time window, where each station gets an equal chance to transmit. This method ensures that all stations can send critical data without being affected by network congestion. Synchronous frames are especially important for applications requiring consistent and predictable communication. The total available time for synchronous frames can be calculated by subtracting the ring latency from the Total Token Rotation Time (TTRT). For example, in a network with a TTRT of 350 µs and a ring latency of 30 µs, the available time for synchronous frames is 320 µs. If divided equally among 10 stations, each station gets 32 µs. Given a data rate of 100 Mbps, each station can send 400 bytes during their allocated time.
Token Rotation Time (TRT)
Token Rotation Time (TRT) is the time taken for the token to make a complete circle around the ring to all stations. This metric is crucial for understanding how efficiently the network operates. A base TRT can be established, which is the TRT without any asynchronous data frames. For instance, if the base TRT is 300 µs, it sets the standard by which the impact of additional frames is measured. When asynchronous frames are introduced, they add to the TRT, potentially affecting the fairness and efficiency of the network. Therefore, network planners must consider TRT adjustments to maintain optimal performance.
Ring Latency
Ring latency refers to the time it takes for a signal to travel once around the entire ring network. It includes the propagation delays between stations and any processing delay within the nodes. In our example, the ring latency is 30 µs. This latency is a crucial factor when calculating the time available for sending synchronous frames. High ring latency can reduce network throughput and efficiency, making it important to minimize latency through good network design, ensuring that each station gets a fair opportunity to transmit data.
Asynchronous Frames
Asynchronous frames are data packets sent when the token becomes available, outside the predefined synchronous window. These frames can introduce variability in the Token Rotation Time (TRT). For example, if Station B sends a 200-µs asynchronous frame, it extends the total TRT to 500 µs for the next round of measurements. While asynchronously sent data can optimize overall network utilization, it may also lead to longer waiting times for other stations, especially if the network is already near its capacity. Therefore, balancing synchronous and asynchronous frames is crucial for maintaining network performance and fairness.
Network Throughput
Network throughput refers to the amount of data successfully transmitted through the network in a given time period. In a token ring network, throughput is influenced by multiple factors including ring latency, TRT, and the ratio of synchronous to asynchronous frames. Throughput can be optimized by ensuring that the TRT is minimized and that ring latency is kept as low as possible. Additionally, adequate allocation of time for both synchronous and asynchronous frames ensures balanced utilization of the network's bandwidth, leading to higher throughput and efficient network performance. Asynchronous frames can enhance throughput but must be managed carefully to avoid congestion and ensure fairness.

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Most popular questions from this chapter

Assume that a SONET receiver resynchronizes its clock whenever a 1 bit appears; otherwise, the receiver samples the signal in the middle of what it believes is the bit's time slot. (a) What relative accuracy of the sender's and receiver's clocks is required in order to receive correctly 480 bytes (one ATM AAL. 5 cell's worth) in a row? (b) Consider a forwarding station A on a SONET STS-1 line, receiving frames from the downstream end \(B\) and retransmitting them upstream. What relative accuracy of A's and B's clocks is required to keep A from accumulating more than one extra frame per minute?

Suppose the round-trip propagation delay for Ethernet is \(46.4 \mu \mathrm{s}\). This yields a minimum packet size of 512 bits ( 464 bits corresponding to propagation delay \(+\) 48 bits of jam signal). (a) What happens to the minimum packet size if the delay time is held constant, and the signalling rate rises to \(100 \mathrm{Mbps}\) ? (b) What are the drawbacks to so large a minimum packet size? (c) If compatibility were not an issue, how might the specifications be written so as to permit a smaller minimum packet size?

Coaxial cable Ethernet was limited to a maximum of \(500 \mathrm{~m}\) between repeaters, which regenerate the signal to \(100 \%\) of its original amplitude. Along one \(500-\mathrm{m}\) segment, the signal could decay to no less than \(14 \%\) of its original value \((8.5 \mathrm{~dB})\). Along \(1500 \mathrm{~m}\), then, the decay might be \((0.14)^{3}=0.3 \%\). Such a signal, even along \(2500 \mathrm{~m}\), is still strong enough to be read; why then are repeaters required every \(500 \mathrm{~m}\) ? $$ \begin{array}{l|l} \hline \text { Item } & \text { Delay } \\ \hline \text { Coaxial cable } & \text { propagation speed } .77 c \\ \text { Link/drop cable } & \text { propagation speed .65c } \\ \text { Repeaters } & \text { approximately 0.6 } \mu \text { s each } \\ \text { Transceivers } & \text { approximately 0.2 } \mu \text { s each } \\ \hline \end{array} $$

In stop-and-wait transmission, suppose that both sender and receiver retransmit their last frame immediately on receipt of a duplicate ACK or data frame; such a strategy is superficially reasonable because receipt of such a duplicate is most likely to mean the other side has experienced a timeout. (a) Draw a timeline showing what will happen if the first data frame is somehow duplicated, but no frame is lost. How long will the duplications continue? This situation is known as the Sorcerer's Apprentice bug. (b) Suppose that, like data, ACKs are retransmitted if there is no response within the timeout period. Suppose also that both sides use the same timeout interval. Identify a reasonably likely scenario for triggering the Sorcerer's Apprentice bug.

An IEEE \(802.5\) token ring has five stations and a total wire length of \(230 \mathrm{~m}\). How many bits of delay must the monitor insert into the ring? Do this for both \(4 \mathrm{Mbps}\) and \(16 \mathrm{Mbps}\); use a propagation rate of \(2.3 \times 10^{8} \mathrm{~m} / \mathrm{s}\).
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