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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

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?

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

Consider a point-to-point link \(2 \mathrm{~km}\) in length. At what bandwidth would propagation delay (at a speed of \(2 \times 10^{8} \mathrm{~m} / \mathrm{s}\) ) equal transmit delay for 100 -byte packets? What about 512 -byte packets?

Calculate the latency (from first bit sent to last bit received) for the following: (a) 10-Mbps Ethernet with a single store-and-forward switch in the path, and a packet size of 5000 bits. Assume that each link introduces a propagation delay of \(10 \mu \mathrm{s}\) and that the switch begins retransmitting immediately after it has finished receiving the packet. (b) Same as (a) but with three switches. (c) Same as (a) but assume the switch implements "cut-through" switching: It is able to begin retransmitting the packet after the first 200 bits have been received.

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?

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