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How "wide" is a bit on a 1-Gbps link? How long is a bit in copper wire, where the speed of propagation is \(2.3 \times 10^{8} \mathrm{~m} / \mathrm{s}\) ?

Short Answer

Expert verified
0.23 meters.

Step by step solution

01

- Understand the Problem

Determine how long a single bit takes to travel on a 1-Gbps link with a propagation speed of \(2.3 \times 10^8 \text{ meters/second}\).
02

- Calculate Bit Time

The bit rate is 1 Gbps (\(10^9 \text{ bits/second}\)). Therefore, the time to transmit one bit is the inverse of the bit rate: \[ \text{Bit Time} = \frac{1}{\text{Bit Rate}} = \frac{1}{10^9 \text{ bits/second}} \].
03

- Time Calculation

Solving the above equation: \[ \text{Bit Time} = 10^{-9} \text{ seconds (or nanoseconds)} \].
04

- Calculate Bit Length

To find the length of a bit in the wire, multiply the propagation speed by the bit time: \[ \text{Bit Length} = \text{Propagation Speed} \times \text{Bit Time} \].
05

- Length Calculation

Using the given propagation speed \(2.3 \times 10^8 \text{ meters/second}\) and bit time \(10^{-9} \text{ seconds}\): \[ \text{Bit Length} = 2.3 \times 10^8 \text{ meters/second} \times 10^{-9} \text{ seconds} \]. This simplifies to: \[ \text{Bit Length} = 0.23 \text{ meters} \].

Key Concepts

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

bit rate
Bit rate is the number of bits transmitted per second in a communication channel. In this exercise, we deal with a 1-Gbps (Gigabit per second) link. This means that the channel transmits 1 billion bits every second. Understanding bit rate is vital because it directly influences how much data can be transferred in a given time frame. The formula for bit rate is simple: it's the total number of bits divided by the time in seconds. So, for our 1-Gbps link, the bit rate is:
  • 1 Gbps = \(1 \times 10^9\) bits per second.

Knowing the bit rate helps in determining other key parameters, such as bit time, which is critical for calculating how
propagation speed
Propagation speed refers to the speed at which a signal travels through a medium, such as copper wire or fiber optic cable. In the context of this exercise, the propagation speed is given as \(2.3 \times 10^8\) meters per second. This is the speed at which an electrical signal moves through the copper wire. To put it into perspective, light travels at a speed of approximately \(3 \times 10^8\) meters per second, so our signal is traveling at a speed that is quite fast but somewhat slower than light. Understanding propagation speed is crucial for calculating how long it takes for a signal to travel a certain distance. It helps in determining the bit length, which is the physical distance a single bit occupies in the medium.
bit time
Bit time is the duration it takes for one bit to be transmitted and it can be calculated using the bit rate. For a 1-Gbps link, the bit time is the inverse of the bit rate. The formula used is:
  • \text{Bit Time} = \frac{1}{\text{Bit Rate}}

In our exercise, the bit rate is 1 Gbps (\(10^9\) bits per second), so the bit time is:
  • \text{Bit Time} = \frac{1}{10^9 \text{ bits/second}} = 10^{-9}$ or 1 nanosecond.

To determine how

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Most popular questions from this chapter

Assume you wish to transfer an \(n\)-byte file along a path composed of the source, destination, seven point-to-point links, and five switches. Suppose each link has a propagation delay of \(2 \mathrm{~ms}\), bandwidth of \(4 \mathrm{Mbps}\), and that the switches support both circuit and packet switching. Thus you can either break the file up into 1-KB packets, or set up a circuit through the switches and send the file as one contiguous bit stream. Suppose that packets have 24 bytes of packet header information and 1000 bytes of payload, that store-and-forward packet processing at each switch incurs a 1 -ms delay after the packet has been completely received, that packets may be sent continuously without waiting for acknowledgments, and that circuit setup requires a 1-KB message to make one round-trip on the path incurring a 1-ms delay at each switch after the message has been completely received. Assume switches introduce no delay to data traversing a circuit. You may also assume that file size is a multiple of 1000 bytes. (a) For what file size \(n\) bytes is the total number of bytes sent across the network less for circuits than for packets? (b) For what file size \(n\) bytes is the total latency incurred before the entire file arrives at the destination less for circuits than for packets? (c) How sensitive are these results to the number of switches along the path? To the bandwidth of the links? To the ratio of packet size to packet header size? (d) How accurate do you think this model of the relative merits of circuits and packets is? Does it ignore important considerations that discredit one or the other approach? If so, what are they?

Suppose hosts A and B are connected by a link. Host A continuously transmits the current time from a high-precision clock, at a regular rate, fast enough to consume all the available bandwidth. Host \(\mathrm{B}\) reads these time values and writes them each paired with its own time from a local clock synchronized with A's. Give qualitative examples of B's output assuming the link has (a) high bandwidth, high latency, low jitter (b) low bandwidth, high latency, high jitter (c) high bandwidth, low latency, low jitter, occasional lost data For example, a link with zero jitter, a bandwidth high enough to write on every other clock tick, and a latency of 1 tick might yield something like \((0000,0001)\), \((0002,0003),(0004,0005)\).

Give an example of a situation in which multicast addresses might be beneficial.

Suppose a host has a 1-MB file that is to be sent to another host. The file takes 1 second of CPU time to compress \(50 \%\), or 2 seconds to compress \(60 \%\). (a) Calculate the bandwidth at which each compression option takes the same total compression + transmission time. (b) Explain why latency does not affect your answer.

Modify the simplex-talk socket program so that it uses UDP as the transport protocol, rather than TCP. You will have to change SOCK_STREAM to SOCK_DGRAM in both client and server. Then, in the server, remove the calls to listen() and ac\(\operatorname{cept}()\), and replace the two nested loops at the end with a single loop that calls recv() with socket s. Finally, see what happens when two such UDP clients simultaneously connect to the same UDP server, and compare this to the TCP behavior.

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