/*! 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 14 How long does it take to transmi... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

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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 100-Mbps point-to-point link is being set up between Earth and a new lunar colony. The distance from the moon to Earth is approximately \(385,000 \mathrm{~km}\), 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) Using the RTT as the delay, calculate the delay \(\times\) bandwidth product for the link. (c) What is the significance of the delay \(\times\) bandwidth product computed in (b)? (d) A camera on the lunar base takes pictures of Earth and saves them in digital format to disk. Suppose Mission Control on Earth wishes to download the most current image, which is \(25 \mathrm{MB}\). What is the minimum amount of time that will elapse between when the request for the data goes out and the transfer is finished?

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.

What differences in traffic patterns account for the fact that STDM is a costeffective form of multiplexing for a voice telephone network and FDM is a costeffective form of multiplexing for television and radio networks, yet we reject both as not being cost-effective for a general-purpose computer network?

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?

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 \((0000,0001)\), \((0002,0003),(0004,0005)\).

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