/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 Consider a closed-loop network (... [FREE SOLUTION] | 91Ó°ÊÓ

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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

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

Step by step solution

01

- 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.
02

- Calculate Transmission Time

Determine the time it takes to transmit the packet using the bandwidth. Transmission time = \( \frac{ \text{Packet size (bits)} }{ \text{Bandwidth (bits/second)} } \) \[ \text{Transmission time} = \frac{2000}{100 \times 10^6} = 20 \mu s \]
03

- 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.
04

- 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.
05

- 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 \[ \text{Total transmission time} = 20 \text{ microseconds} + 4 \text{ microseconds} = 24 \text{ microseconds} \]
06

- 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.

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Key Concepts

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

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.

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Most popular questions from this chapter

Suppose that a certain communications protocol involves a per-packet overhead of 100 bytes for headers and framing. We send 1 million bytes of data using this protocol; however, one data byte is corrupted and the entire packet containing it is thus lost. Give the total number of overhead + loss bytes for packet data sizes of \(1000,5000,10,000\), and 20,000 bytes. Which size is optimal?

One property of addresses is that they are unique; if two nodes had the same address it would be impossible to distinguish between them. What other properties might be useful for network addresses to have? Can you think of any situations in which network (or postal or telephone) addresses might not be unique?

Consider a simple protocol for transferring files over a link. After some initial negotiation, A sends data packets of size \(1 \mathrm{~KB}\) to B; B then replies with an acknowledgment. A always waits for each ACK before sending the next data packet; this is known as stop-and-wait. Packets that are overdue are presumed lost and are retransmitted. (a) In the absence of any packet losses or duplications, explain why it is not necessary to include any "sequence number" data in the packet headers. (b) Suppose that the link can lose occasional packets, but that packets that do arrive always arrive in the order sent. Is a 2-bit sequence number (that is, \(N\) mod 4) enough for \(\mathrm{A}\) and \(\mathrm{B}\) to detect and resend any lost packets? Is a 1-bit sequence number enough? (c) Now suppose that the link can deliver out of order, and that sometimes a packet can be delivered as much as 1 minute after subsequent packets. How does this change the sequence number requirements?

The Unix utility traceroute, or its Windows equivalent tracert, can be used to find the sequence of routers through which a message is routed. Use this to find the path from your site to some others. How well does the number of hops correlate with the RTT times from ping? How well does the number of hops correlate with geographical distance?

Calculate the total time required to transfer a \(1000-\mathrm{KB}\) file in the following cases, assuming an RTT of \(100 \mathrm{~ms}\), a packet size of \(1 \mathrm{~KB}\) and an initial \(2 \times\) RTT of "handshaking" before data is sent. (a) The bandwidth is \(1.5 \mathrm{Mbps}\), and data packets can be sent continuously. (b) The bandwidth is \(1.5 \mathrm{Mbps}\), but after we finish sending each data packet we must wait one RTT before sending the next. (c) The bandwidth is "infinite," meaning that we take transmit time to be zero, and up to 20 packets can be sent per RTT. (d) The bandwidth is infinite, and during the first RTT we can send one packet \(\left(2^{1-1}\right)\), during the second RTT we can send two packets \(\left(2^{2-1}\right)\), during the third we can send four \(\left(2^{3-1}\right)\), and so on. (A justification for such an exponential increase will be given in Chapter \(6 .)\)

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