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How long does it take to transmit \(x\) KB over a \(y\)-Mbps link? Give your answer as a ratio of \(x\) and \(y\).

Short Answer

Expert verified
\(\frac{8x}{y \times 1000}\) seconds

Step by step solution

01

Convert KB to Kbits

First, convert the data size from kilobytes (KB) to kilobits (Kbits). Since 1 byte = 8 bits, multiply the given data size by 8. So, if the data size is given as \(x\) KB, it will be \(8x\) Kbits.
02

Calculate the Transmission Time

Now, use the link speed in Mbps to calculate the transmission time. The link speed given is \(y\) Mbps, which means \(y\) megabits per second. To find the transmission time in seconds, divide the total data size in Kbits by the link speed in Mbps.The formula for the transmission time \(T\) is: \[ T = \frac{8x \text{ Kbits}}{y \text{ Mbps}} \] Since 1 Mbps = 1,000 Kbits, you need to divide the numerator by y*1000 to convert the denominator appropriately.
03

Simplify the Expression

Simplify the formula to find the final time taken: \[ T = \frac{8x}{y \times 1000} \text{ seconds} \] So, the transmission time is given as the ratio \( \frac{8x}{y \times 1000}\) seconds.

Key Concepts

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

Understanding Data Size Conversion
Before diving into the calculation of transmission time, it's important to understand how to convert data sizes. Data sizes in computer systems are typically measured in bytes (B) or bits (b). Here's a quick guide to converting kilobytes (KB) to kilobits (Kbits):

1. A byte consists of 8 bits. When you have data in kilobytes (KB), multiply by 8 to convert it to kilobits (Kbits).
2. For example, if you have a file size of 2 KB:
- 2 KB x 8 = 16 Kbits

This step is crucial because transmission speeds are often given in megabits per second (Mbps) and not in megabytes per second (MBps). Converting data size to kilobits ensures that we use the correct units in our formula for transmission time.
Understanding Transmission Speed
Transmission speed refers to the rate at which data is transmitted from one point to another in a given amount of time. This speed is often given in megabits per second (Mbps), which stands for millions of bits per second. Here are some key points to consider:

- 1 Mbps equals 1,000 Kbits per second.
- A higher transmission speed means faster data transfer.

For example, a transmission speed of 10 Mbps means that 10 million bits can be transmitted every second. Knowing the transmission speed helps us determine how quickly data can be sent over a network.
Formula for Transmission Time
Once you've converted the data size and understood the transmission speed, you can calculate the transmission time using a simple formula. The key to solving this problem is the formula:

\[ T = \frac{8x \text{ Kbits}}{y \text{ Mbps}} \tag{1} \]

Where:
- \( T \) is the transmission time in seconds
- \( x \) is the data size in kilobytes (KB)
- \( y \) is the transmission speed in megabits per second (Mbps)

To simplify it further, since 1 Mbps = 1,000 Kbits, we can rewrite the formula as:

\[ T = \frac{8x}{y \times 1000} \text{ seconds} \tag{2} \]

This means the time it takes to transmit the data is given by the ratio of \( \frac{8x}{y \times 1000} \) seconds. This formula helps you quickly determine how long it will take to send a certain amount of data over a network with a given speed.

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Most popular questions from this chapter

Suppose a 128-Kbps point-to-point link is set up between Earth and a rover on Mars. The distance from Earth to Mars (when they are closest together) is approximately \(55 \mathrm{Gm}\), and data travels over the link at the speed of light-3 \(\times 10^{8} \mathrm{~m} / \mathrm{s}\). (a) Calculate the minimum RTT for the link. (b) Calculate the delay \(\times\) bandwidth product for the link. (c) A camera on the rover takes pictures of its surroundings and sends these to Earth. How quickly after a picture is taken can it reach Mission Control on Earth? Assume that each image is \(5 \mathrm{Mb}\) in size.

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 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.

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?

Calculate the bandwidth \(x\) delay product for the following links. Use one-way delay, measured from first bit sent to first bit received. (a) 10-Mbps Ethernet with a delay of \(10 \mu \mathrm{s}\). (b) 10-Mbps Ethernet with a single store-and-forward switch like that of Exercise \(18(\mathrm{a})\), packet size 5000 bits, and \(10 \mu\) ser link propagation delay. (c) \(1.5\)-Mbps T1 link, with a transcontinental one-way delay of \(50 \mathrm{~ms}\). (d) \(1.5-\mathrm{Mbps} \mathrm{T} 1\) link through a satellite in geosynchronous orbit, \(35,900 \mathrm{~km}\) high. The only delay is speed-of- light propagation delay.

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