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Converting from bandwidth to traffic gives different results depending on the location of the operators?

This should be a stupid question, but nevertheless it seems curious to me:

Let's say I have a constant download of 128 Kbps.

How much disk space will be consumed in megabytes in one hour?

128 x 60 x 60/8/1024 = 56.25 MB

But

128 x 60 x 60/1000/8 = 57.6 MB

So what's the correct way to calculate this?

Thanks!

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Ok, I found out the official explanation from Symantec on this issue: http://seer.entsupport.symantec.com/docs/274171.htm

The idea seems to be to convert from bits to bytes per calculation as quickly as possible, and then the usual division of 1024 appears.



I just hope this is standard procedure and not Symantec: 1).

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In one calculation, you divide by 1000, and in another, you divide by 1024. It shouldn't come as a surprise that you get different numbers.



Officially, the International Electrotechnical Commission's standardization body tried to nudge "kibibyte" as an alternative to "kilobyte" when you talk about the 1024 version. But if you use it, people will laugh at you.

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Remember that there is an overhead in any transmission. There may be dropped packets, etc. There is usually some upstream traffic as well, as your computer is receiving packets. Finally, since the packets may be received out of order, the packets themselves contain "extra" data for all recipients to restore the data in the correct order.

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