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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 \(2.3 \times 10^{8} \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
For 4 Mbps, insert 4 bits. For 16 Mbps, insert 16 bits.

Step by step solution

01

Determine the propagation delay

Calculate the propagation delay using the formula \[ \text{Propagation Delay} = \frac{\text{Total Wire Length}}{\text{Propagation Speed}} \] The total wire length is \( 230 \text{ m} \) and the propagation speed is \( 2.3 \times 10^8 \text{ m/s} \). Therefore, the propagation delay is: \[ \text{Propagation Delay} = \frac{230 \text{ m}}{2.3 \times 10^8 \text{ m/s}} = 10^{-6} \text{ s} \]
02

Convert the propagation delay to a more convenient unit

The propagation delay is \(10^{-6} \text{ s}\), which is equivalent to \(1 \text{ microsecond (} \text{μs} \text{)} \). So, the propagation delay is \(1 \text{ μs} \).
03

Calculate the bit delay for 4 Mbps

At \(4 \text{ Mbps}\), the amount of delay in terms of bits can be calculated using the formula: \[ \text{Bit Delay} = \text{Bit Rate} \times \text{Propagation Delay} \] Plugging in the values, we get: \[ \text{Bit Delay} = 4 \text{ Mbps} \times 1 \text{ μs} = 4 \text{ bits} \] Therefore, for \(4 \text{ Mbps}\), the monitor must insert \(4\) bits of delay.
04

Calculate the bit delay for 16 Mbps

At \(16 \text{ Mbps}\), the amount of delay in terms of bits can be calculated using the same formula: \[ \text{Bit Delay} = \text{Bit Rate} \times \text{Propagation Delay} \] Plugging in the values, we get: \[ \text{Bit Delay} = 16 \text{ Mbps} \times 1 \text{ μs} = 16 \text{ bits} \] Therefore, for \(16 \text{ Mbps}\), the monitor must insert \(16\) bits of delay.

Key Concepts

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

Propagation Delay
Propagation delay is the time it takes for a signal to travel from one point to another. This is a crucial concept in digital communications and network performance analysis. We calculate it using the formula: \( \text{Propagation Delay} = \frac{\text{Total Wire Length}}{\text{Propagation Speed}} \).

This helps us understand how quickly data can move through a network. For example, in the exercise, we found the total wire length to be 230 meters and the propagation speed to be \( 2.3 \times 10^8 \) meters per second. Using these values, the propagation delay was calculated to be 1 microsecond \( (10^{-6} \text{ s}) \).

Knowing the propagation delay is essential in designing networks that meet specific performance criteria. Lower propagation delays are desirable as they indicate faster signal transmission times.
Bit Rate
Bit rate refers to the number of bits that are transmitted per unit of time in a digital communication system. It is usually measured in Megabits per second (Mbps).

In the exercise, we calculated bit delays for two bit rates: 4 Mbps and 16 Mbps. The bit rate is directly related to how quickly data can be sent through the network. Using the formula \( \text{Bit Delay} = \text{Bit Rate} \times \text{Propagation Delay} \), we determined that at 4 Mbps, the bit delay is 4 bits, and at 16 Mbps, the bit delay is 16 bits.

This tells us that higher bit rates result in more bits being in transit at any given time, which can impact the design and efficiency of the network.
IEEE 802.5
IEEE 802.5 is a standard for token ring networks. Token ring is a type of local area network where devices are connected in a ring topology. Each device passes a token, a special type of frame, to its neighbor when it is done transmitting data.

This standard includes mechanisms for managing access to the network, improving performance, and reducing collisions. In our exercise, we determined how many bits of delay the monitor must insert into the ring, which helps in synchronizing network communication and maintaining performance.

The IEEE 802.5 standard specifically addresses token passing and ensures that only one device transmits at a time, mandating rules for maintaining the network's integrity and efficiency.
Network Performance
Network performance refers to how well a network carries out its intended functions. This involves several factors, such as bit rate, propagation delay, and network topology.

In the exercise, we assessed network performance by calculating delay in terms of bits, which is a direct measure of how much data is in transit. Performance can be influenced by the choice of protocols, the physical properties of the network, and the overall design.

Maintaining low delays and high data rates is critical for high-performance networks. Understanding how to adjust parameters such as propagation delay and bit rate helps in optimizing network efficiency and reliability.
Digital Communication
Digital communication involves the transmission of data in a digital format. It relies on various metrics like bit rate and delays to measure efficiency and performance.

The exercise highlights key principles in digital communication by asking to calculate bit delays based on bit rates and propagation delay.

Digital communication systems aim to be fast and reliable. Engineers must consider factors like propagation speed, distance, and data formats. This ensures that the network can handle the required data load and maintain functionality across different conditions.

By mastering concepts like bit rate and propagation delay, students can design and troubleshoot effective digital communication systems.

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Most popular questions from this chapter

Suppose you are designing a sliding window protocol for a 1-Mbps point-to- point link to the moon, which has a one-way latency of \(1.25\) seconds. Assuming that each frame carries \(1 \mathrm{~KB}\) of data, what is the minimum number of bits you need for the sequence number?

Consider an ARQ protocol that uses only negative acknowledgments (NAKs), but no positive acknowledgments (ACKs). Describe what timeouts would need to be scheduled. Explain why an ACK-based protocol is usually preferred to a NAK- based protocol.

What kinds of problems can arise when two hosts on the same Ethernet share the same hardware address? Describe what happens and why that behavior is a problem.

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 \(500 \mathrm{~m}\) ? $$ \begin{array}{l|l} \hline \text { Item } & \text { Delay } \\ \hline \text { Coaxial cable } & \text { propagation speed } .77 c \\ \text { Link/drop cable } & \text { propagation speed .65c } \\ \text { Repeaters } & \text { approximately 0.6 } \mu \text { s each } \\ \text { Transceivers } & \text { approximately 0.2 } \mu \text { s each } \\ \hline \end{array} $$

Give some details of how you might augment the sliding window protocol with flow control by having ACKs carry additional information that reduces the SWS as the receiver runs out of buffer space. Illustrate your protocol with a timeline for a transmission; assume the initial sWS and RWS are 4, the link speed is instantaneous, and the receiver can free buffers at the rate of one per second (i.e., the receiver is the bottleneck). Show what happens at \(T=0, T=1, \ldots, T=4 \mathrm{sec}-\) onds.

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