What is the minimum time required to download a video file with a size equal to 700MB, if the connection bandwidth is 512 Kbps? To make your web page display quickly (in less 1 sec... What is the minimum time required to download a video file with a size equal to 700MB, if the connection bandwidth is 512 Kbps? To make your web page display quickly (in less 1 second), what is the maximum size of the front page (including images), if the mean BW for the users is 256 kbps? You want to have a DSL connection but you don’t know which BW you need to select. If you always download videos with 700 MB, which connection bandwidth is needed if you want to download this video file in less than an hour?
Understand the Problem
The question is asking for calculations related to download times and bandwidth requirements for video files and web pages. It focuses on understanding the relationships between file size, bandwidth (BW), and time in a network context.
Answer
1. 10.94 seconds 2. 0.25 MB 3. 1555.56 bps
Answer for screen readers
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The minimum time required to download the video file is approximately 10.94 seconds.
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The maximum size of the front page that can be displayed in less than 1 second is 0.25 MB (or 256 KB).
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The required bandwidth to download the video file in less than an hour is approximately 1555.56 bps.
Steps to Solve
- Calculate download time for video file
To find the minimum time required to download a video file, use the formula: $$ \text{Time (seconds)} = \frac{\text{File size (bits)}}{\text{Bandwidth (bps)}} $$
First, convert the file size from megabytes to bits:
- $700 \text{ MB} = 700 \times 8 \times 10^6 \text{ bits} = 5.6 \times 10^6 \text{ bits}$
Now, apply the formula with a bandwidth of $512 \text{ Kbps} = 512 \times 10^3 \text{ bps}$.
$$ \text{Time} = \frac{5.6 \times 10^6 \text{ bits}}{512 \times 10^3 \text{ bps}} $$
- Calculate maximum size for webpage
To ensure the web page displays in less than 1 second, we use the same formula:
Set the time to less than or equal to 1 second, rearranging the formula gives:
$$ \text{Maximum size (bits)} = \text{Bandwidth (bps)} \times \text{Time (seconds)} $$
Substituting values: $$ \text{Maximum size} = 256 \times 10^3 \text{ bps} \times 1 \text{ second} $$
Convert the bits back to megabytes by dividing by $8 \times 10^6$.
- Calculate required bandwidth for video download
To download a 700 MB video file in less than 1 hour (3600 seconds), we use: $$ \text{Bandwidth (bps)} = \frac{\text{File size (bits)}}{\text{Time (seconds)}} $$
Convert the file size to bits: $$ \text{File size} = 700 \text{ MB} = 5.6 \times 10^6 \text{ bits} $$
Now calculate the required bandwidth: $$ \text{Bandwidth} = \frac{5.6 \times 10^6 \text{ bits}}{3600 \text{ seconds}} $$
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The minimum time required to download the video file is approximately 10.94 seconds.
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The maximum size of the front page that can be displayed in less than 1 second is 0.25 MB (or 256 KB).
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The required bandwidth to download the video file in less than an hour is approximately 1555.56 bps.
More Information
- Download time increases with file size and decreases with higher bandwidth.
- The efficiency of webpage loading heavily depends on the balance between size and bandwidth.
- For large files, ensuring adequate bandwidth can significantly reduce wait times.
Tips
- Confusing megabytes (MB) and bits (b): Remember that $1 \text{ MB} = 8 \text{ Megabits}$.
- Ignoring the conversions when applying the formulas for size or bandwidth calculations.
- Miscalculating time based on incorrect bandwidth values.
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