Is there a easy way to calculate time for data transfer?

Question

Are there any sites or easy methods of finding out it would take for file transfer over different speed links?

For example, I need to find out how long it would take to download/transfer file of 75G with a transfer rate of 2 Mbps.

Answer 1

Answer : 3 Days 15 Hours 22 Minutes 52.8 Seconds.

Hope this site will help you.

Answer 2

75GB can be either 75000000000 bytes (in SI units), or 80530636800 bytes (in IEC units)

A byte is 8 bits, so we are talking either

600000000000 bits

or 644245094400 bits

A Mbps is either

1000000 bits per second (SI)

or

1048576 bits per second (IEC)

So divide the number of bits by the number of bits per second to get the number of seconds:

600000000000  / 2000000 = 300000 = 83.33 hours or 3.47 days
644245094400  / 2097152 = 307200 = 85.33 hours or 3.56 days

Answer 3

It depends on what you actually mean by Mbps, but I would reccomend Wolfram Alpha for things like this - it can do a lot more than you think. Anyway, here's a screenie of your query:

Answer 4

300,000 seconds, assuming you meant Gigabytes and Megabits/second.

8,000 Mb   75 GB   600,000 x 1 s   
-------- x ----- = ------------- = 300,000 s
    1 GB    2 Mb         2
           -----
            1 s

Answer 5

Here is a method I came up with that is easy to remember for rough estimation.

Short answer

Multiply the file size in gigabytes by 9000 and then divide it by the speed of your connection in megabits per second. The resulting number is in seconds, which you can then convert into days, hours, and minutes using your favorite conversion program.

Example

75 GB × 9000 / 2 Mb/s = 337500 s ≈ 3 days 22 hours

Explanation

Consider overhead imposed by the Ethernet, IP, and TCP protocols. The IPv4 header is 20 octets long, not counting any option headers. The TCP header is also 20 octets long, not counting any option headers. An Ethernet connection can carry 1500 octets of data per 1538 octet frame (1542 if VLAN tagging is used) using Ethernet II framing (for PPPoE encapsulated connections, only 1492 octets). Then, do the math:

[ 1500 o (Ethernet data) - 20 o (IPv4) - 20 o (TCP) ] / 1538 o (Ethernet frame) ≈ 95% efficiency

[1024 Mio/Gio] * [8 Mb/Mio] / 94.92848% ≈ 8630

9000 is the nearest multiple of 1000; it is both easier to remember and less optimistic.

Answer 6

It's 3 days, 22 hours. Things not to forget:

1) Data sizes are measured in binary units, line speeds in decimal units.

2) There are 8 bits in a byte.

3) The line has to carry address and control information, not just data.

4) You can't pack the line 100% with information.