1. What is the minimum time required to download a video file with a size equal to 700MB, if the connection bandwidth is 512 Kbps? 2. To make your web page display quickly (in less... 1. What is the minimum time required to download a video file with a size equal to 700MB, if the connection bandwidth is 512 Kbps? 2. To make your web page display quickly (in less than 1 second), what is the maximum size of the front page (including images), if the mean bandwidth for the users is 256 Kbps? 3. You want to have a DSL connection but you don’t know which bandwidth 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? 4. A file is downloaded to a home computer using a 56 kbps modem connected to an Internet Service Provider. If the download completes in 2 minutes, estimate the maximum size of data downloaded. Read more

Steps to Solve

1. Convert file size to bits

   The file size is given in megabytes (MB). We convert it to bits to match the bandwidth units.

   Since $1 \text{MB} = 8 \times 10^6 \text{bits}$, therefore:

   [ 700 \text{MB} = 700 \times 8 \times 10^6 \text{bits} = 5.6 \times 10^6 \text{bits} ]

2. Calculate download time in seconds

   To find the minimum time required to download, we use the formula:

   [ \text{Time (seconds)} = \frac{\text{File Size (bits)}}{\text{Bandwidth (bps)}} ]

   Plugging in the values, where bandwidth ($BW$) is $512 \text{Kbps} = 512 \times 10^3 \text{bps}$:

   [ \text{Time} = \frac{5.6 \times 10^6 \text{bits}}{512 \times 10^3 \text{bps}} ]

3. Perform the division to find the time

   Performing the calculation:

   [ \text{Time} = \frac{5.6 \times 10^6}{512 \times 10^3} \approx 10.94 \text{ seconds} ]

4. Maximum front page size for web display

   Given the bandwidth is $256 \text{Kbps}$, we calculate the maximum size for a quick display (in 1 second):

   [ \text{Maximum Size} = \text{Bandwidth (bps)} \times \text{Time (seconds)} ]

   Where $256 \text{Kbps} = 256 \times 10^3 \text{bps}$:

   [ \text{Maximum Size} = 256 \times 10^3 \text{bps} \times 1 \text{ second} = 256 \times 10^3 \text{bits} ]

5. Calculate maximum size in megabytes

   Convert the bit result to megabytes:

   [ \text{Size (MB)} = \frac{256 \times 10^3 \text{bits}}{8 \times 10^6 \text{bits/MB}} = 0.032 \text{ MB} = 32 \text{ KB} ]

6. Determine required bandwidth for 700MB in less than an hour

   For a 700MB file in less than 1 hour (3600 seconds):

   Calculate total bits:

   [ 700 \text{MB} = 700 \times 8 \times 10^6 \text{bits} ]

   Set up the equation:

   [ \text{Required BW (bps)} = \frac{\text{File Size (bits)}}{\text{Time (seconds)}} = \frac{700 \times 8 \times 10^6 \text{bits}}{3600 \text{seconds}} ]

7. Perform the calculation for bandwidth

   Calculate:

   [ \text{Required BW} \approx \frac{5.6 \times 10^6 \text{bits}}{3600} \approx 1556.7 \text{bps} \approx 1.56 \text{Kbps} ]

8. Calculate maximum size downloaded with a 56K modem in 2 minutes

   The bandwidth is $56 \text{Kbps} = 56 \times 10^3 \text{bps}$.

   Calculate the size for 2 minutes (120 seconds):

   [ \text{Size} = 56 \times 10^3 \text{bps} \times 120 \text{ seconds} = 6720000 \text{ bits} ]

9. Convert this size to megabits

   Convert bits to megabits:

   [ \text{Size (Mbit)} = \frac{6720000 \text{ bits}}{10^6} = 6.72 \text{Mbit} ]