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

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.

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What did you want the end result to be measured in (units)? – iglvzx Nov 24 '11 at 5:12
There seems to be calls to close this question as too localized. Perhaps the specifics of the question are, but the answers don't need to be, which makes the question more valuable. – Paul Nov 24 '11 at 8:27
hours is basically what i need to break it down too. thank you. – jdamae Nov 27 '11 at 18:21

Answer : 3 Days 15 Hours 22 Minutes 52.8 Seconds.

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great site. thank you! – jdamae Nov 24 '11 at 5:05
@jdmae There seems to be discrepancy with that website and the other answers here. I recommend Wolfram|Alpha for quick conversions. For example, your question can be entered as 75 gigabytes at 2 megabits per second – iglvzx Nov 24 '11 at 5:28
thanks for you input everybody. I am looking at everyone's answerss – jdamae Nov 27 '11 at 18:21

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
``````
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You forgot that line speeds are measured in decimal units while file sizes are measured in binary units. You also forgot that the line has to carry address and control information as well as data. – David Schwartz Nov 24 '11 at 7:56

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
``````
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I am trying to understand why Desingh got `314573 seconds`. – iglvzx Nov 24 '11 at 5:13
The site is using 2048000 bits per second which is not 2Mbps in IEC or SI units (which isn't to say it isn't used as a speed on an E1 line). – Paul Nov 24 '11 at 5:22
You are forgetting that the line has to carry address and control information as well as data. – David Schwartz Nov 24 '11 at 7:58
This isn't forgotten, it just isn't calculable with the information provided. – Paul Nov 24 '11 at 8:26

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.

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Can you break down your answer? How did you calculate it? I'm intrigued by this not-so-straightforward situation. – iglvzx Nov 24 '11 at 8:30
75*1024*1024*1024*8/2000000/.95/60/60 = 94.18 hours. The factors are: 75GB, 1024*1024*1024 bytes per GB, 8 bits per byte, 2Mbps, 95% TCP/line efficiency, 60 seconds in a minute, 60 minutes in an hour. – David Schwartz Nov 25 '11 at 8:12

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:

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

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.

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Thanks. Interesting. I like to think it terms of rough estimation for this anyways. – jdamae Nov 27 '11 at 18:19

There is small program for this (Tembo Transfer Time Estimator). After usage it I got: 75 GB on 2 Mbps and network efficiency 90% will be transfered in 3 days, 22 hours, 48 minutes, 53.33 seconds with average speed 230.4 KB/s.

Check it yourself at http://tembolab.pl/products/transfer-time-estimator.html.

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